5  Complex Numbers

We have seen that \(\sqrt{x}\) exists in \(\mathbb{R}\) for all for \(x \geq 0\). We defined \[ \sqrt{x}:= \alpha \,, \quad \alpha := \sup \{ t \in \mathbb{R}\, \colon \,t^2 < x \}\,, \] and proved that \[ \alpha^2 = x \,. \] This procedure was possible for \(x \geq 0\).

Question 1

Is there a number \(\alpha \in \mathbb{R}\) such that \[ \alpha^{2}=-1 \,\, ? \tag{5.1}\]

The answer to the above question is no. This is because \(\mathbb{R}\) is an ordered field, and from axiom (MO) it follows that: \[ x^2 \geq 0 \,, \quad \forall \, x \in \mathbb{R}\,. \] However we would still like to solve equation (5.1) somehow. To do this, we introduce the imaginary numbers or complex numbers. We define \(i\) to be that number such that \[ i^{2}=-1 \,. \] Formally, we can also think of \(i=\sqrt{-1}\). We can use this speacial number to define the square root of a negative number \(x<0\): \[ \sqrt{x}: = i \sqrt{-x} \,. \] Note that \(\sqrt{-x}\) is properly defined in \(\mathbb{R}\), because \(-x>0\) if \(x<0\).

5.1 The field \(\mathbb{C}\)

We would like to be able to do calculations with the newly introduced complex numbers, and investigate their properties. We can introduce them rigorously as a field, as we did for \(\mathbb{R}\).

Definition 2: Complex Numbers

The set of complex numbers \(\mathbb{C}\) is defined as

\[ \mathbb{C}:= \mathbb{R}\oplus i \mathbb{R}:= \{x \oplus i y \, \colon \,x, y \in \mathbb{R}\} \,. \]

In the above the symbol \(\oplus\) is used to denote the pair \[ x \oplus i y = (x,y) \] with \(x,y \in \mathbb{R}\). This means that \(x\) and \(y\) play different roles.

Definition 3

For a complex number \[ z = x \oplus i y \in \mathbb{C} \] we say that

• \(x\) is the real part of \(z\), and denote it by \[ x = \operatorname{Re}(z) \]
• \(y\) is the imaginary part of \(z\), and denote it by \[ y = \operatorname{Im}(z) \]

We say that

• If \(\operatorname{Re}z = 0\) then \(z\) is a purely imaginary number.
• If \(\operatorname{Im}z = 0\) then \(z\) is a real number.

In order to make the set \(\mathbb{C}\) into a field, we first have to define the two binary operations of addition \(+\) and multiplication \(\cdot\), \[ + , \cdot \,\, \colon \mathbb{C}\times \mathbb{C}\to \mathbb{C}\,. \] Then we need to prove that these operations satisfy all the field axioms.

Definition 4: Addition in \(\mathbb{C}\)

Let \(z_1,z_2 \in \mathbb{C}\), so that \[ z_1 = x_1 \oplus i y_1 \,\,, \quad z_2 = x_2 \oplus i y_2 \,\, , \] for some \(x_1,x_2,y_1,y_2 \in \mathbb{R}\). We define the sum of \(z_1\) and \(z_2\) as \[ z_1 + z_2 = \left( x_{1} \oplus i y_{1} \right)+ \left(x_{2} \oplus i y_{2} \right):=\left(x_{1}+x_{2}\right) \oplus i \left(y_{1}+y_{2}\right) \] where the \(+\) symbol on the right hand side is the addition operator in \(\mathbb{R}\).

Clearly, \(z_1 + z_2\) as defined above is an element of \(\mathbb{C}\). Therefore \(+\) defines a binary operation over \(\mathbb{C}\).

Notation 5

From the above definition, we have that, for all \(x, y \in \mathbb{R}\),

\[ (x \oplus i 0 )+(0 \oplus i y)= x \oplus i y . \] To simplify notation, we will write \[ x \oplus i 0 = x \,, \quad \, 0 \oplus i y = i y \] and \[ x \oplus i y = x + i y \,. \] We will also often swap \(i\) and \(y\), writing equivalently \[ x + i y = x + y i \,. \]

We now want to define multiplication between complex numbers.

Remark 6: Formal calculation for multiplication in \(\mathbb{C}\)

How to define multiplication in \(\mathbb{C}\)? Whatever the definition may be, at least it has to give that that \[ i^{2}=i \cdot i=-1 \,. \] Keeping the above in mind, let us do some formal calculations: For \(z_1 = x_1 + i y_1, z_2 = x_2 + i y_2\) we have

\[\begin{align*} z_1 \cdot z_2 & = \left(x_{1}+i y_{1} \right) \cdot \left(x_{2}+ i y_{2} \right) \\ & = x_{1} \cdot x_{2} + x_{1} \cdot i y_{2} + x_{2} \cdot i y_{1} + y_{1} \cdot i^2 y_{2} \\ & = \left(x_{1} \cdot x_{2} - y_{1} \cdot y_{2} \right) + i \left(x_{1} \cdot y_{2} + x_{2} \cdot y_{1} \right) \end{align*}\]

Remark 6 motivates the following definition of multiplication.

Definition 7: Multiplication in \(\mathbb{C}\)

Let \(z_1,z_2 \in \mathbb{C}\), so that \[ z_1 = x_1 \oplus i y_1 \,\,, \quad z_2 = x_2 \oplus i y_2 \,\, , \] for some \(x_1,x_2,y_1,y_2 \in \mathbb{R}\). We define the multiplication of \(z_1\) and \(z_2\) as \[ z_1 \cdot z_2 = \left(x_{1}+i y_{1} \right) \cdot\left(x_{2}+ i y_{2} \right) := \left(x_{1} \cdot x_{2} - y_{1} \cdot y_{2}\right) + i \left(x_{1} \cdot y_{2} + x_{2} \cdot y_{1} \right) \,, \] where the operations \(+\) and \(\cdot\) on the right hand side are the operations in \(\mathbb{R}\).

Clearly, \(z_1 \cdot z_2\) as defined above is an element of \(\mathbb{C}\). Therefore \(\cdot\) defines a binary operation over \(\mathbb{C}\).

Remark 8

To check that we have given a good definition of product, we should have that \[ i^2 = - 1\,, \] as expected. Indeed:

\[\begin{align*} i^{2} & = (0+ 1 i) \cdot (0+ 1 i) \\ & = (0 \cdot 0 - 1 \cdot 1) + (0 \cdot 1 + 0 \cdot 1) i = - 1 \,. \end{align*}\]

Important

In view of Remark 8, we see that he formal calculations in Remark 6 are compatible with the definition of multiplication of complex numbers. Therefore, it is not necessary to memorize the multiplication formula, but it suffices to carry out calculations as usual, and replace \(i^{2}\) by \(-1\).

Example 9

Suppose we want to multiply the complex numbers \[ z = -2+3 i \,, \quad w = 1- i \,. \] Using the definition we compute \[\begin{align*} z \cdot w & = (-2+3 i) \cdot (1 - i) \\ & = (-2-(-3))+(2+3) i \\ & = 1 + 5 i \, . \end{align*}\] Alternatively, we can proceed formally as in Remark 6. We just need to recall that \(i^2\) has to be replaced with \(-1\): \[\begin{align*} z \cdot w & = (-2+3 i) \cdot (1 - i) \\ & = - 2 + 2i + 3i - 3 i^2 \\ & = (-2 + 3 ) + ( 2 + 3 ) i \\ & = 1 + 5 i \, . \end{align*}\]

We now want to check that \[ (\mathbb{C}, + , \cdot) \] is a field. All the field axioms are trivial to check, except for the existence of additive and multiplicative inverses.

Proposition 10: Additive inverse in \(\mathbb{C}\)

The neutral element of addition in \(\mathbb{C}\) is the number \[ 0 := 0 + 0 i \,. \] For any \(z = x + i y \in \mathbb{C}\), the unique additive inverse is given by \[ - z := -x - i y \,. \]

The proof is immediate and is left as an exercise. The multiplication requires more care.

Remark 11: Formal calculation for multiplicative inverse

Let us first carry our some formal calculations. Let \[ z = x+iy \in \mathbb{C}\,, \quad z \neq 0 \,. \] First, note that \[ z \cdot 1 = (x + iy) \cdot (1 + 0 i) = x + iy = z \,, \] and therefore \(1\) is the neutral element of multiplication. Thus, the inverse of \(z\) should be a complex number \(z^{-1} \in \mathbb{C}\) such that \[ z \cdot z^{-1} = 1 \,. \] We would like to define \[ z^{-1} = \frac{1}{x + iy} \,. \] Such number does not belong to \(\mathbb{C}\), as it is not of the form \(a + ib\) for some \(a,b \in \mathbb{R}\). However it is what the inverse should look like. Proceeding formally: \[\begin{align*} \frac{1}{x+ i y} & = \frac{1}{x+ iy} \cdot 1 \\ & = \frac{1}{x+ i y} \cdot \frac{x- i y}{x - i y} \\ & =\frac{x - i y}{x^{2}-(iy)^{2}} \\ & =\frac{x- iy}{x^{2}+y^{2}} \\ & = \frac{x}{x^{2}+y^{2}}+i \, \frac{-y}{x^{2}+y^{2}} \,. \end{align*}\] The right hand side is an element of \(\mathbb{C}\), and looks like a good candidate for \(z^{-1}\).

Motivated by the above remark, we define inverses in \(\mathbb{C}\) in the following way.

Proposition 12: Multiplicative inverse in \(\mathbb{C}\)

The neutral element of multiplication in \(\mathbb{C}\) is the number \[ 1 := 1 + 0 i \,. \] For any \(z = x + i y \in \mathbb{C}\), the unique multiplicative inverse is given by \[ z^{-1} := \frac{x}{x^{2}+y^{2}}+ i \, \frac{-y}{x^{2}+y^{2}} \,. \]

Proof

It is immediate to check that \(1\) is the neutral element of multiplication in \(\mathbb{C}\). For the remaining part of the statement, set \[ w := \frac{x}{x^{2}+y^{2}}+ i \, \frac{-y}{x^{2}+y^{2}} \,. \] We need to check that \(z \cdot w = 1\) \[\begin{align*} z \cdot w & = (x+i y) \cdot \left(\frac{x}{x^{2}+y^{2}}+ i \, \frac{-y}{x^{2}+y^{2}} \right) \\ & = \left(\frac{x^{2}}{x^{2}+y^{2}} - \frac{y \cdot(-y)}{x^{2}+y^{2}}\right) + i \, \left(\frac{x \cdot(-y)}{x^{2}+y^{2}}+\frac{x y}{x^{2}+y^{2}}\right) \\ & =1 \,, \end{align*}\] so indeed \(z^{-1}=w\).

Important

It is not necessary to memorize the formula for \(z^{-1}\). Indeed one can just remember the trick of multiplying by \[ 1 = \frac{x - iy }{x - iy} \,, \] and proceed formally, as done in Remark 11.

Example 13

Let \(z = 3 + 2 i\). We want to compute \(z^{-1}\). By the formula in Propostion 12 we immediately get \[ z^{-1} = \frac{3}{3^{2}+2^{2}} + \, \frac{-2}{3^{2}+2^{2}} \, i = \frac{3}{13}-\frac{2}{13} \, i \,. \] Alternatively, we can proceed formally as in Remark 11 \[\begin{align*} (3+2 i)^{-1} & = \frac{1}{3+2 i} \\ & = \frac{1}{3+2 i} \, \frac{3-2 i}{3-2 i} \\ & = \frac{3-2 i}{3^2+2^2} \\ & = \frac{3}{13}-\frac{2}{13} i \,, \end{align*}\] and obtain the same result.

We can now prove that \(\mathbb{C}\) is a field.

Theorem 14

\((\mathbb{C},+, \cdot)\) is a field.

Proof

We need to check that all field axioms hold. For the addition we have

• (A1) To show that \(+\) is commutative, note that \[\begin{align*} (x+iy)+(a+ ib) & = (x+a)+ i (y+b) \\ & = (a + x)+ i (b+y) \\ & = (a + i b ) + (x + i y) \,, \end{align*}\] where we used Definition 4 in the first and last equality, and the commutative property of the real numbers (which holds since by definition \(\mathbb{R}\) is a field) in the second equality. Associativity can be checked in the same way.

• (A2) The neutral element of addition is \(0\), as stated in Proposition 10.

• (A3) Existence of additive inverses is given by Proposition 10.

For multiplication we have:

• (M1) Commutativity and associativity of product in \(\mathbb{C}\) can be checked using Definition 7 and commutativity and associativity of sum and multiplication in \(\mathbb{R}\).

• (M2) The neutral element of multiplication is \(1\), as stated in Propostion 12.

• (M3) Existence of multiplicative inverses is guaranteed by Proposition 12.

Finally one should check the associative property (AM). This is left as an exercise.

5.1.1 Division in \(\mathbb{C}\)

Suppose we want to divide two complex numbers \(w,z \in \mathbb{C}, z \neq 0\), with \[ z = x + i y \,, \quad w = a + ib \,. \] We have two options:

1. Use the formula for the inverse from Proposition 12 and compute \[ z^{-1} := \frac{x}{x^{2}+y^{2}}+ i \, \frac{-y}{x^{2}+y^{2}} \,. \] Then we use the multiplication formula of Definition 7 to compute \[\begin{align*} \frac{w}{z} & = w \cdot z^{-1} \\ & = (a + i b) \, \cdot \left( \frac{x}{x^{2}+y^{2}}+ i \, \frac{-y}{x^{2}+y^{2}} \right) \\ & = \frac{ (ax + by) + i (bx - ay) }{ x^2 + y^2 } \end{align*}\]

2. Proceed formally as in Remark 11, using the multiplication by \(1\) trick. We would have \[\begin{align*} \frac{w}{z} & = \frac{ a + i b }{ x + i y } \\ & = \frac{ a + i b }{ x + i y } \, \frac{ x - i y }{ x - i y } \\ & = \frac{ (ax + by) + i (bx - ay) }{ x^2 + y^2 } \end{align*}\]

Example 15

Let \(w=1+i\) and \(z=3-i\). We compute \(\frac{w}{z}\) using the two options we have:

1. Using the formula for the inverse from Proposition 12 we compute \[\begin{align*} z^{-1} & = \frac{x}{x^{2}+y^{2}} + i \, \frac{-y}{x^{2}+y^{2}} \\ & = \frac{3}{3^2 + 1^2} - i \, \frac{-1}{3^2 + 1^2} \\ & = \frac{3}{10} + \frac{1}{10} \, i \end{align*}\] and therefore \[\begin{align*} \frac{w}{z} & = w \cdot z^{-1} \\ & = (1 + i) \, \left( \frac{3}{10} + \frac{1}{10} \, i \right) \\ & = \left(\frac{3}{10}-\frac{1}{10}\right)+\left(\frac{1}{10}+\frac{3}{10}\right) i \\ & = \frac{2}{10}+\frac{4}{10} i \\ & = \frac{1}{5}+\frac{2}{5} i \end{align*}\]

2. We proceed formally, using the multiplication by \(1\) trick. We have \[\begin{align*} \frac{w}{z} & = \frac{1+i}{3-i} \\ & = \frac{1+i}{3-i} \frac{3+i}{3+i} \\ & = \frac{3-1+(3+1) i}{3^2+1^2} \\ & =\frac{2}{10}+\frac{4}{10} i \\ & = \frac{1}{5}+\frac{2}{5} i \end{align*}\]

5.1.2 \(\mathbb{C}\) is not ordered

We have seen that \((\mathbb{C},+,\cdot)\) is a field. One might wonder whether \(\mathbb{C}\) is also an ordered field. It turns out that this is not the case.

Theorem 16

The field \((\mathbb{C},+,\cdot)\) is not ordered.

Proof

Suppose that \(\mathbb{C}\) is an ordered field, that is, there exists an order relation \(\leq\) on \(\mathbb{C}\) compatible with the operations \(+\) and \(\cdot\). By axiom (MO) it follows that for all elements \(z \in \mathbb{C}, z \neq 0\), we have that \(z^{2}>0\). But since \(i^{2}=-1<0\), we get a contradiction.

Hence, it is not possible to compare two complex numbers.

5.1.3 Completeness of \(\mathbb{C}\)

One might also wonder whether \(\mathbb{C}\) is complete. Our definition of completeness uses the notion of supremum, which only makes sense if the field is ordered. This is not the case for \(\mathbb{C}\) as we have seen in Theorem 16.

Still, it is possible to give a different definition of completeness using the notion of Cauchy sequence. In ordered fields, this new definition of completeness is equivalent to the definition which uses the supremum.

The new definition of completeness with Cauchy sequences also makes sense in non-ordered fields. We will see that \(\mathbb{C}\) is a complete field, according to this new definition.

5.2 Complex conjugates

When computing inverses, we used the trick to multiply by \(1\): \[ z^{-1} = \frac{1}{z} \cdot 1 = \frac{1}{x + i y} \cdot \frac{x-iy}{x-iy} \,. \] The complex number \(x-iy\) is obtained by changing the sign to the imaginary part of \(z = x+iy\). We give a name to this operation.

Definition 17: Complex conjugate

Let \(z=x+iy\). We call the complex conjugate of \(z\), denoted by \(\bar{z}\), the complex number

\[ \bar{z}=x- i y \, . \]

Example 18

We have the following conjugates: \[\begin{align*} &\overline{3+4 i} = 3 - 4 i \, , & \quad & \overline{3-4 i} = 3+4 i\,, \\ & \overline{-3+4 i} = -3 - 4 i \,, & \quad & \overline{-3-4 i}=-3+4 i \,, \\ & \overline{3} = 3 \, , & \quad & \overline{4 i}=-4 i \, . \end{align*}\]

Complex conjugates have the following properties:

Theorem 19

For all \(z_1, z_2 \in \mathbb{C}\) it holds:

• \(\overline{z_1 + z_2 }=\overline{z_1}+\overline{z_2}\)

• \(\overline{z_1 \cdot z_2}=\overline{z_1} \cdot \overline{z_2}\)

Proof

Let \(z_{1}, z_{2} \in \mathbb{C}\). Then \[ z_{1}=x_{1}+i y_{1} \,, \quad z_{2} = x_{2} + i y_{2} \,, \] for some \(x_{1}, y_{1}, x_{2}, y_{2} \in \mathbb{R}\).

• Using the definition of addition in \(\mathbb{C}\) and of conjugate, \[\begin{align*} \overline{z_{1}+z_{2}} & =\overline{\left(x_{1}+i y_{1} \right)+\left(x_{2}+ i y_{2} \right)} \\ & =\overline{\left(x_{1}+x_{2}\right)+ i \left(y_{1}+y_{2}\right)} \\ & =\left(x_{1}+x_{2}\right)- i \left(y_{1}+y_{2}\right) \\ & =\left(x_{1}- i y_{1} \right) + \left(x_{2}-i y_{2} \right) \\ & =\overline{x_{1}+ i y_{1} } + \overline{x_{2}+i y_{2} } \\ & =\overline{z_{1}}+\overline{z_{2}} \,. \end{align*}\]

• Using the definition of multiplication in \(\mathbb{C}\) and of conjugate,

\[\begin{align*} \overline{z_{1} \cdot z_{2}} & = \overline{\left(x_{1}+i y_{1} \right) \cdot\left(x_{2}+ i y_{2} \right)} \\ & = \overline{\left(x_{1} x_{2}-y_{1} y_{2}\right)+ i \left(x_{1} y_{2}+x_{2} y_{1}\right)} \\ & = \left(x_{1} x_{2}-y_{1} y_{2}\right)- i \left(x_{1} y_{2}+x_{2} y_{1}\right) \\ & =\left(x_{1}-i y_{1} \right) \cdot\left(x_{2}- i y_{2} \right) \\ & =\overline{z_{1}} \cdot \overline{z_{2}} \end{align*}\]

Example 20

Let \(z_{1}=3-4 i\) and \(z_{2}=-2+5 i\). Then

• Let us check that \[ \overline{z_{1}+z_{2}}=\overline{z_{1}}+\overline{z_{2}} \] Indeed, we have \[ z_1 + z_2 = 1 + i \quad \implies \quad \overline{z_1 + z_2} = 1 - i \, . \] On the other hand \[ \overline{z_{1}} = 3 + 4i \,, \quad \overline{z_{2}} = -2 -5i \quad \implies \quad \overline{z_{1}} + \overline{z_{2}}= 1-i \,. \]

• Let us check that \[ \overline{z_{1} \cdot z_{2}}=\overline{z_{1}} \cdot \overline{z_{2}} \] Indeed, \[\begin{align*} z_{1} \cdot z_{2} & = (3+4i) \cdot (-2 + 5i) \\ & = (- 6 + 20)+ (8 + 15 )i \\ & = 14 + 23 i \end{align*}\] so that \[ \overline{z_{1} \cdot z_{2}} = 14 - 23 i \] On the other hand: \[\begin{align*} \overline{z_{1}} \cdot \overline{z_{2}} & =(3+4 i) \cdot(-2-5 i) \\ & = (-6+20)+(-15-8) i \\ & = 14 - 23 i \end{align*}\]

5.3 The complex plane

We can depict a real number \(x\) as a point on the one-dimensional real line \(\mathbb{R}\). The distance between two real numbers \(x, y \in \mathbb{R}\) on the real line is given by \(|x-y|\), see Figure 5.1.

Figure 5.1: Two points \(x\) and \(y\) on the real line \(\mathbb{R}\). Their distance is \(|x-y|\).

We would like to do something similar for the complex numbers, but the point \[ z = x + iy \,, \quad x , \, y \in \mathbb{R}\,. \] We therefore depict \(z= z + i y\) in the two-dimensional plane at the point with (Cartesian) coordinates \((x, y)\). This two-dimensional plane in which we can depict all complex numbers is called the complex plane. The origin of such plane, with coordinates \((0,0)\), corresponds to the complex number \[ 0+0 i = 0 \,, \] see Figure 5.2.

Figure 5.2: A point \(z = x + iy \in \mathbb{C}\) can be represented on the complex plane by the point of coordinates \((x,y)\). The distance between \(z\) and \(0\) is given by \(|z| = \sqrt{z^2 + y^2}\).

5.3.1 Distance on \(\mathbb{C}\)

The Cartesian representation allows us to introduce a distance between two complex numbers. Let us start with the distance between a complex number \(z=x+i y\) and \(0\). By Pythagoras Theorem this distance is given by \[ \sqrt{x^{2}+y^{2}} \,, \] see Figure 5.2. We give a name to this quantity.

Definition 21: Modulus

The modulus of a complex number \(z=x+i y\) is defined by \[ |z|: = \sqrt{x^{2}+y^{2}} \,. \]

Note that the distance between \(z\) and \(0\) is always a non-negative number.

Remark 22: Modulus of Real numbers

A real number \(x \in \mathbb{R}\) can be written as \[ x = x+ 0 i \in \mathbb{C}\,. \] Hence the modulus of \(x\) is given by \[ |x|=\sqrt{x^{2}+0^{2}}=\sqrt{x^{2}} \,. \] The above coincides with the absolute value of \(x\). This explains why the notation for modulus in \(\mathbb{C}\) is the same as the one for absolute value in \(\mathbb{R}\).

We can now define the distance between two complex numbers.

Definition 23: Distance in \(\mathbb{C}\)

Given \(z_1,z_2 \in \mathbb{C}\), we define the distance between \(z_1\) and \(z_2\) as the quantity \[ |z_1 - z_2| \,. \]

The geometric intuition of why the quantity \(|z_1 - z_2|\) is defined as the distance between \(z_1\) and \(z_2\) is given in Figure 5.3.

Figure 5.3: The difference \(z_1 - z_2\) of the two points \(z_1,z_2 \in \mathbb{C}\) is given by the magenta vector. We define \(|z_1-z_2|\) as the distance between \(z_1\) and \(z_2\).

Theorem 24

Given \(z_1,z_2 \in \mathbb{C}\), we have \[ |z_1 - z_2| = \sqrt{ (x_1 - x_2)^2 + (y_1 - y_2)^2 } \,. \]

Proof

We have \[ z_1 - z_2 = (x_1 - x_2) + i (y_1 - y_2) \,. \] Therefore, by definition of modulus, \[ |z_1 - z_2| = \sqrt{ (x_1 - x_2)^2 + (y_1 - y_2)^2 } \,. \]

Example 25

The distance between \[ z = 2-4 i \,, \quad w = -5+i \] is given by \[\begin{align*} |z- w| & = |(2-4 i)-(-5+i)| \\ & = |7-5 i| \\ & =\sqrt{7^{2}+(-5)^{2}} \\ & =\sqrt{74} \end{align*}\]

5.3.2 Properties of modulus

The modulus has the following properties.

Theorem 26

Let \(z, z_{1}, z_{2} \in \mathbb{C}\). Then

1. \(\left|z_1 \cdot z_2\right|=\left|z_{1}\right|\left|z_{2}\right|\)

2. \(\left|z^{n}\right|=|z|^{n}\) for all \(n \in \mathbb{N}\)

3. \(z \cdot \bar{z}=|z|^{2}\)

Proof

Part 1. We have \[\begin{align*} z_1 \cdot z_2 & = (x_1 + i y_1) \cdot (x_2 + i y_2) \\ & = (x_1 x_2 - y_1 y_2) + i (x_2 y_1 + x_1 y_2) \end{align*}\] and therefore \[\begin{align*} |z_1 \cdot z_2| & = \sqrt{ (x_1 x_2 - y_1 y_2)^2 + (x_2 y_1 + x_1 y_2)^2 }\\ & = \sqrt{ x_1^2 x_2^2 + y_1^2 y_2^2 + x_2^2 y_1^2 + x_1^2 y_2^2 } \,. \end{align*}\] On the other hand, \[ |z_1| = \sqrt{x_1^2 + y_1^2} \,, \quad |z_2| = \sqrt{x_2^2 + y_2^2} \] so that \[\begin{align*} |z_1| |z_2| & = \sqrt{x_1^2 + y_1^2} \sqrt{x_2^2 + y_2^2} \\ & = \sqrt{ x_1^2 x_2^2 + y_1^2 y_2^2 + x_2^2 y_1^2 + x_1^2 y_2^2 } \end{align*}\] proving that \(\left|z_1 \cdot z_2\right|=\left|z_{1}\right|\left|z_{2}\right|\).

Part 2. Exercise. It easily follows from Point 1 and induction.

Part 3. Let \(z=x+ i y\) for some \(x, y \in \mathbb{R}\). Then,

\[\begin{align*} z \cdot \bar{z} & = (x+i y)(x- i y) \\ & = x^{2}-(i y )^{2} \\ & = x^{2}+y^{2} \\ & =|z|^{2} \end{align*}\]

The modulus in \(\mathbb{C}\) satisfies the triangle inequality.

Theorem 27: Triangle inequality in \(\mathbb{C}\)

For all \(x, y, z \in \mathbb{C}\),

1. \(|x+y| \leq|x|+|y|\)

2. \(|x-z| \leq|x-y|+|y-z|\)

Proof

Part 1. Suppose that \(x=a+ib\) and \(y=c+ i d\) for \(a, b, c, d \in \mathbb{R}\). Then, \[ |x+y|=|(a+c)+i (b+d) |=\sqrt{(a+c)^{2}+(b+d)^{2}} \,. \] Therefore the inequality \[ |x+y| \leq|x|+|y| \tag{5.2}\] is equivalent to \[ \sqrt{(a+c)^{2}+(b+d)^{2}} \leq \sqrt{a^2 + b^2} + \sqrt{c^2 + d^2}\,. \tag{5.3}\] Now note that, for \(A,B \in \mathbb{R}\), we have that \[ A^2 \leq B^2 \quad \implies \quad |A| \leq |B| \,. \tag{5.4}\] In the two reverse implications \(\impliedby\) below we will use (5.4): \[\begin{align*} & \, \sqrt{(a+c)^{2}+(b+d)^{2}} \leq \sqrt{a^{2}+b^{2}}+\sqrt{c^{2}+d^{2}} \\ \impliedby & \, (a+c)^{2}+(b+d)^{2} \leq\left(\sqrt{a^{2}+b^{2}}+\sqrt{c^{2}+d^{2}}\right)^{2} \\ \iff & \, a^{2}+2 a c+c^{2}+b^{2}+2 b d+d^{2} \leq a^{2}+b^{2}+2 \sqrt{a^{2}+b^{2}} \sqrt{c^{2}+d^{2}}+c^{2}+d^{2} \\ \iff & \, a c+b d \leq \sqrt{a^{2}+b^{2}} \sqrt{c^{2}+d^{2}} \\ \impliedby & \, (a c+b d)^{2} \leq\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right) \\ \iff & \, a^{2} c^{2}+2 a b c d+b^{2} d^{2} \leq a^{2} c^{2}+a^{2} d^{2}+b^{2} c^{2}+b^{2} d^{2} \\ \iff & \, a^{2} d^{2}+b^{2} c^{2}-2 a b c d \geq 0 \\ \iff & \, (a d-b c)^{2} \geq 0 . \end{align*}\]

This last statement is clearly true, since \(a d-b c \in \mathbb{R}\). Therefore (5.3) holds, and so (5.2) follows.

Part 2. Using (5.2) we estimate \[ |x-z|=|x-y+y-z| \leq|x-y|+|y-z| . \]

Remark 28: Geometric interpretation of triangle inequality

We finally have a justification of why the inequality \[ |x-z| \leq |x-y| + |y - z| \] is called triangle inequality: By drawing three points \(x, y, z \in \mathbb{C}\) in the complex plane, the distance between \(x\) and \(z\) is shorter than the distance to go from \(x\) to \(z\) via the point \(y\), see Figure 5.4.

Figure 5.4: Let \(x,y,z \in \mathbb{C}\). The distance between \(x\) and \(z\) is shorter than the distance to go from \(x\) to \(z\) via the point \(y\).

5.4 Polar coordinates

We have seen that we can identify a complex number \(z=x+iy\) by a point in the complex plane with Cartesian coordinates \((x, y)\). We can also specify the point \((x,y)\) by using the so-called polar coordinates \((\rho, \theta)\), where

• \(\rho\) is the distance between \(z\) and the origin \[ \rho = |z| = \sqrt{x^2 + y^2} \]

• \(\theta\) is the angle between the line connecting the origin and \(z\) and the positive real axis, see Figure 5.5.

Figure 5.5: Polar coordinates \((\rho,\theta)\) for the complex number \(z \in \mathbb{C}\).

We give such angle a name.

Definition 29: Argument

Let \(z \in \mathbb{C}\). The angle \(\theta\) between the line connecting the origin and \(z\) and the positive real axis is called the argument of \(z\), and is denoted by \[ \theta := \arg (z) \,. \]

Warning

We always use angles in radians, not degrees. Make sure your calculator is set to radians if you want to use it to compute angles.

Remark 30: Principal Value

The argument of a complex number is not uniquely defined. We can always add an integer number of times \(2 \pi\) to the argument to specify the same point. We usually use the convention to choose the argument in the interval \((-\pi, \pi]\). This is called the principal value of the argument function. Therefore the complex numbers in the upper half plane have a positive argument, and in the lower half plane have a negative argument.

Example 31

We have the following arguments: \[\begin{align*} & \arg (1) = 0 & \quad & \arg (i) = \frac{\pi}{2} \\ & \arg (-1) = \pi & \quad & \arg (-i) = -\frac{\pi}{2} \\ & \arg (1+i)=\frac{1}{4} \pi & \quad & \arg (-1-i)=-\frac{3}{4} \pi \end{align*}\]

We can represent non-zero complex numbers in polar coordinates.

Theorem 32: Polar coordinates

Let \(z \in \mathbb{C}\) with \(z = x + i y\) and \(z \neq 0\). Then \[ x = \rho \cos(\theta) \,, \quad y = \rho \sin(\theta) \,, \] where \[ \rho = \sqrt{x^2 + y^2} \,, \quad \theta = \arg(z) \,. \]

The proof of Theorem 32 is trivial, and is based on basic trigonometry and definition of \(\arg(z)\). Complex numbers in polar form can be useful. We give a name to such polar form.

Definition 33: Trigonometric form

Let \(z \in \mathbb{C}\). The trigonometric form of \(z\) is \[ z = |z| \left[ \cos (\theta)+i \sin (\theta) \right] \,, \] where \(\theta = \arg(z)\).

Let us make an example.

Example 34

Suppose that we have polar coordinates \[ \rho = \sqrt{8}\,, \quad \theta = \frac{3}{4} \pi \] We compute \[\begin{gather*} x = \rho \cos (\theta) = \sqrt{8} \cos \left( \frac{3}{4} \pi \right) = - \frac{\sqrt{8}\sqrt{2}}{2} = -2 \\ y = \rho \sin (\theta) = \sqrt{8} \sin \left( \frac{3}{4} \pi \right) = \frac{\sqrt{8}\sqrt{2}}{2} = 2 \,. \end{gather*}\] The complex number \(z\) corresponding to the polar coordinates \((\rho,\theta)\) is \[ z = x + i y = - 2 + 2 i \,. \] The trigonometric form of \(z\) is \[ z = \sqrt{8} \left[ \cos \left( \frac{3}{4} \pi \right) + i \sin \left( \frac{3}{4} \pi \right) \right] \,. \]

As a consequence of Theorem 32 we obtain a formula for computing the argument.

Corollary 35: Computing \(\arg(z)\)

Let \(z \in \mathbb{C}\) with \(z = x + i y\) and \(z \neq 0\). Then \[ \arg(z) = \begin{cases} \arctan \left( \dfrac{y}{x} \right) & \,\, \mbox{ if } x>0 \\ \arctan \left( \dfrac{y}{x} \right) + \pi & \,\, \mbox{ if } x<0\,\, \mbox{and} \,\, y \geq 0 \\ \arctan \left( \dfrac{y}{x} \right) - \pi & \,\, \mbox{ if } x<0\,\, \mbox{and} \,\, y < 0 \\ \dfrac{\pi}{2} & \,\, \mbox{ if } x=0\,\, \mbox{and} \,\, y > 0 \\ -\dfrac{\pi}{2} & \,\, \mbox{ if } x=0\,\, \mbox{and} \,\, y < 0 \\ \end{cases} \] where \(\arctan\) is the inverse of \(\tan\).

Proof

Using the polar coordinates formulas from Theorem 32 we have \[ \frac{y}{x} = \frac{\rho \sin(\theta)}{\rho \cos(\theta)} = \tan (\theta) \,. \] The thesis can be obtained by carefully inverting the tangent.

Example 36

We want to compute the arguments of the complex numbers \[ 3+4 i\,, \quad 3-4 i \,, \quad -3+4 i \,, \quad -3-4 i \,. \] Using the formula for \(\arg\) in Corollary 35 we have \[\begin{align*} \arg (3+4 i) & =\arctan \left(\frac{4}{3}\right) \\ \arg (3-4 i) & =\arctan \left(-\frac{4}{3}\right) = -\arctan \left(\frac{4}{3}\right) \\ \arg (-3+4 i) & = \arctan \left(-\frac{4}{3}\right) + \pi = - \arctan \left(\frac{4}{3}\right) + \pi \\ \arg (-3-4 i) & = \arctan \left(\frac{4}{3}\right) - \pi \end{align*}\]

5.5 Exponential form

We have seen that we can represent complex numbers in

• Cartesian form
• Trigonometric form

We now introduce a third way of representing complex numbers: the exponential form. For this, we need Euler’s identity:

Theorem 37: Euler’s identity

For all \(\theta \in \mathbb{R}\) it holds \[ e^{i \theta}=\cos (\theta)+ i \sin (\theta) \,. \]

Proof

The proof of this theorem uses Taylor power series. Note that we have not introduced what series are, yet, so we just assume that everything below makes sense and actually exists. We have the following Taylor series at \(x_{0}=0\) that you might know from calculus: \[\begin{align*} e^{x} & =1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\frac{x^{4}}{4 !}+\frac{x^{5}}{5 !}+\frac{x^{6}}{6 !}+\frac{x^{7}}{7 !}+\ldots \\ \sin (x) & =\frac{x}{1 !}-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}-\frac{x^{7}}{7 !}+\ldots \\ \cos (x) & =1-\frac{x^{2}}{2 !}+\frac{x^{4}}{4 !}-\frac{x^{6}}{6 !}+\ldots \end{align*}\]

The above identities also holds for \(x \in \mathbb{C}\). Hence we can substitute \(x=i \theta\) in the series for \(e^{x}\) to obtain \[\begin{align*} e^{i \theta} & =1+i \theta+\frac{(i \theta)^{2}}{2 !}+\frac{(i \theta)^{3}}{3 !}+\frac{(i \theta)^{4}}{4 !}+\frac{(i \theta)^{5}}{5 !}+\frac{(i \theta)^{6}}{6 !}+\frac{(i \theta)^{7}}{7 !} \ldots \\ & =1+i \theta-\frac{\theta^{2}}{2 !}-i \frac{\theta^{3}}{3 !}+\frac{\theta^{4}}{4 !}+i \frac{\theta^{5}}{5 !}-\frac{\theta^{6}}{6 !}-i \frac{\theta^{7}}{7 !}+\ldots \\ & =\cos (\theta)+i \sin (\theta), \end{align*}\]

where we used that \(i^{2}=-1\) in the second equality, and the third equality follows by observing that all terms with an even power of \(\theta\) are exactly the terms in the expansion of \(\cos (\theta)\) and all terms with an odd power of \(\theta\) are exactly the terms in the expansion of \(\sin (\theta)\) multiplied by \(i\).

Theorem 38

For all \(\theta \in \mathbb{R}\) it holds \[ \left|e^{i \theta}\right|=1 \,. \]

Proof

From Euler’s identity in Theorem 37 we get \[ \left|e^{i \theta}\right|=|\cos (\theta)+i \sin (\theta)|=\sqrt{\cos ^{2}(\theta)+\sin ^{2}(\theta)}=1 \,. \]

Theorem 39

Let \(z \in \mathbb{C}\) with \(z = x+iy\) and \(z \neq 0\). Then \[ z = \rho e^{i\theta}\,, \] where \[ \rho = \sqrt{x^2 + y^2} = |z| \,, \quad \theta = \arg(z) \,. \]

Proof

By Theorem 32 we have \[ x = \rho \cos(\theta)\,, \quad y = \rho \sin(\theta) \,. \] Hence \[\begin{align*} z & = x + i y \\ & = \rho \cos(\theta) + i \rho \sin(\theta) \\ & = \rho e^{i\theta} \,, \end{align*}\] where in the last line we used Euler’s identity in Theorem 37.

Definition 40: Exponential form

A complex number \(z \in \mathbb{C}\) is in exponential form if \[ z= \rho e^{i\theta} = |z| \, e^{i \arg(z)} \,. \]

Example 41

From Example 34 we know that \[ z=-2+2 i \] can be written in trigonometric form as \[ z = \sqrt{8} \left[ \cos \left( \frac{3}{4} \pi \right) + i \sin \left( \frac{3}{4} \pi \right) \right] \,. \] By Euler’s identity we hence obtain the exponential form \[ z=\sqrt{8} e^{i \frac{3}{4} \pi} \,. \]

Remark 42: Periodicity of exponential

For all \(k \in \mathbb{Z}\) we have \[ e^{i \theta}=e^{i(\theta+2 \pi k)} \,. \tag{5.5}\] As we did for the principal value of the argument, also for the exponential form we select \(\theta \in(-\pi, \pi]\). In particular equation (5.5) is saying that the complex exponential is \(2\pi\)-periodic.

Equation (5.5) follows immediately by Euler’s identity and periodicity of \(\cos\) and \(\sin\), since \[\begin{align*} e^{i(\theta+2 \pi k)} & = \cos(\theta+2 \pi k) + i \sin (\theta+2 \pi k) \\ & = \cos(\theta) + i \sin (\theta) = e^{i \theta} \,. \end{align*}\]

The exponential for is very useful for computing products and powers of complex numbers.

Proposition 43

Let \(z, z_1,z_2 \in \mathbb{C}\) and suppose that \[ z= \rho e^{i \theta} \,, \quad z_1 = \rho_1 e^{i \theta_1}\,, \quad z_2 = \rho_2 e^{i \theta_2}\,. \] We have \[ z_1 \cdot z_2 = \rho_1 \rho_2 e^{i (\theta_1 + \theta_2)} \,, \quad z^n = \rho^n e^{i n \theta} \,, \] for all \(n \in \mathbb{N}\).

The proof follows immediately by the properties of the exponential. Let us see some applications of Propostion 43.

Example 44

Suppose we want to compute \((-2+2 i)^{4}\). We could do this by means of the binomial theorem: \[\begin{align*} (-2+2 i)^{4} & =(-2)^{4}+\left(\begin{array}{l} 4 \\ 1 \end{array}\right)(-2)^{3} \cdot 2 i+\left(\begin{array}{l} 4 \\ 2 \end{array}\right)(-2)^{2} \cdot(2 i)^{2}+\left(\begin{array}{l} 4 \\ 3 \end{array}\right)(-2) \cdot(2 i)^{3}+(2 i)^{4} \\ & =16-4 \cdot 8 \cdot 2 i-6 \cdot 4 \cdot 4+4 \cdot 2 \cdot 8 i+16 \\ & =16-64 i-96+64 i+16=-64 \,. \end{align*}\]

Using the exponential form simplifies this calculation. Indeed, we know that \[ -2+2 i = \sqrt{8} e^{i \frac{3}{4} \pi} \] by Example 41. Hence \[ (-2+2 i)^{4}=\left(\sqrt{8} e^{i \frac{3}{4} \pi}\right)^{4}=8^{2} e^{i 3 \pi}=-64 \,, \] where we used that \[ e^{i 3 \pi} = e^{i \pi} = \cos(\pi) + i \sin(\pi) = - 1 \] by \(2\pi\) periodicity of \(e^z\) and Euler’s identity.

Example 45

Suppose we want to compute \[ i^{i} \,. \] It is not even clear how to do this calculation in Cartesian form. However, we know that \[ |i| = 1 \,, \quad \arg(i) = \frac{\pi}{2} \,. \] Hence we can write \(i\) in exponential form \[ i= |i| e^{i\arg(i)} = e^{i \frac{\pi}{2}} \,. \] Therefore \[ i^{i}=\left(e^{i \frac{\pi}{2}}\right)^{i}=e^{i^2\frac{\pi}{2}}=e^{-\frac{\pi}{2}} \,. \]

5.6 Fundamental Theorem of Algebra

We started the introduction to complex numbers with the following question:

Question 46

Is there a number \(x \in \mathbb{R}\) such that \[ x^{2}=-1 \,\, ? \tag{5.6}\]

The answer is no. For this reason we introduced the complex number \(i\), which satisfies \[ i^{2}=-1 \,. \] Therefore (5.6) has solution in \(\mathbb{C}\), with \(x = i\). We also have that \[ (-i)^{2}=(-1)^{2} i^{2}=-1 \,. \] Hence (5.6) has two solutions in \(\mathbb{C}\), given by \[ x_1 = i \,, \quad x_2 = - i \,. \]

It turns out that the set \(\mathbb{C}\) is so large that we are not only able to solve (5.6), but in fact any polynomial equation.

Theorem 47: Fundamental theorem of algebra

Let \(p_{n}(x)\) be a polynomial of degree \(n\) with complex coefficients, i.e., \[ p_{n}(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{1} x+a_{0}, \] for some coefficients \(a_{n}, \ldots, a_{0} \in \mathbb{C}\) with \(a_{n} \neq 0\). Then there exist \(z_{1}, \ldots, z_{n} \in \mathbb{C}\) such that \[ p_{n}(x)=a_{n}\left(x-z_{1}\right)\left(x-z_{2}\right) \cdots\left(x-z_{n}\right) \,. \tag{5.7}\] Hence, the polynomial equation \[ a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{1} x+a_{0}=0 \tag{5.8}\] has solutions \(z_{1}, \ldots, z_{n}\).

Theorem 47 says that every polynomial of degree \(n\) has \(n\) zeros, sometimes also called roots, i.e., \(n\) solutions to (5.8). We call the expression (5.7) a factorization of the polynomial \(p_n\).

Several proofs of Theorem 47 exist in the literature, but they all use mathematical tools which are out of reach for now. Therefore we will not show a proof. For example one can prove Theorem 47 by

• Liouville’s theorem (complex analysis)
• Homotopy arguments (general topology)
• Fundamental Theorem of Galois Theory (algebra)

Example 48

The equation \[ x^{2}=-1 \tag{5.9}\] is equivalent to \[ p(x) = 0 \,, \quad p(x):=x^{2}+1 \,. \] Since \(p\) has degree \(n=2\), the Fundamental Theorem of Algebra tells us that there are two solutions to (5.9). We have already seen that these two solutions are \(x=i\) and \(x=-i\). Then \[ p(x) = x^{2} + 1 = (x-i)(x+i) \,. \]

Example 49

Suppose we want to solve \[ x^{4}-1=0 \,. \tag{5.10}\] The associated polynomial equation is \[ p(x) = 0 \,, \quad p(x) := x^4 - 1 \,. \] Since \(p\) has degree \(n=4\), the Fundamental Theorem of Algebra tells us that there are \(4\) solutions to (5.10).

Let us find such solutions. We use the well known formula \[ a^2-b^2 = (a+b)(a-b) \,, \quad \forall \, a,b \in \mathbb{R}\,, \] to factorize \(p\). We get: \[ p(x) = (x^4-1) = (x^2+1)(x^2-1) \,. \] We know that \[ x^2 + 1 = 0 \] has solutions \(x = \pm i\). Instead
\[ x^2 - 1 = 0 \] has solutions \(x = \pm 1\). Hence, the four solutions of (5.10) are given by \(x=1,-1, i,-i\) and \[ p(x) = x^4 - 1 = (x-1)(x+1)(x-i)(x+i) \,. \]

Definition 50

Suppose that the polynomial \(p_n\) factorizes as \[ p_n(x) = a_n (x-z_1)^{k_1} (x-z_2)^{k_2} \cdots (x-z_m)^{k_m} \] with \(a_n \neq 0\), \(z_1,\ldots, z_m \in \mathbb{C}\) and \(k_1, \ldots , k_m \in \mathbb{N}\), \(k_i \geq 1\). In this case \(p_n\) has degree \[ n = k_1 + \ldots + k_m = \sum_{i=1}^m k_i \,. \] We have that \(z_i\) is solves the equation \[ p_n(x) = 0 \] exactly \(k_i\) times. We call \(k_i\) the multiplicity of the solution \(z_i\).

Example 51

The equation \[ (x-1)(x-2)^{2}(x+i)^{3}=0 \] has 6 solutions:

• \(x=1\) with multiplicity \(1\)
• \(x=2\) with multiplicity \(2\)
• \(x=-i\) with multiplicity \(3\)

5.7 Solving polynomial equations

The non-factorized version of the polynomial \[ p(x) = (x-1)(x-2)^{2}(x+i)^{3} \] from Example 51 is \[\begin{align*} p(x) = & x^{6}-(5-3 i) x^{5}+(5-15 i) x^{4} \\ & + (11+23 i) x^{3}-(24+7 i) x^{2}+(12-8 i) x+4 i \end{align*}\] We therefore have the following natural question.

Question 52

The Fundamental Theorem of Algebra states that \[ p_n(x) = 0 \tag{5.11}\] has \(n\) complex solutions. How do we find such solutions?

The answer is that there is no general way to solve (5.11) when \(n \geq 5\). This is the content of the Abel-Ruffini Theorem.

Theorem 53: Abel-Ruffini

There is no elementary solution formula to the polynomial equation \[ p_n(x) = 0 \] with \(p_n\) polynomial of degree \(n\), with \(n \geq 5\).

Similarly to the Fundamental Theorem of Algebra, the proof of the Abel-Ruffini Theorem is out of reach for our current mathematical knowledge. A proof can be carried out, for example, using Galois Theory.

There are however explicit formulas for solving (5.11) when \(p_n\) has degree \(n=2,3,4\). For \(n=2\) we can use the well-known quadratic formula.

Proposition 54: Quadratic formula

Let \(a, b, c \in \mathbb{R}, a \neq 0\) and consider the equation \[ a x^{2}+b x+c=0 \,. \tag{5.12}\] Define \[ \Delta := b^2 - 4 ac \,. \] The following hold:

• If \(\Delta > 0\) then (5.12) has two distinct real solutions given by \[ x_1 = \frac{-b - \sqrt{\Delta}}{2 a} \,, \quad x_2 = \frac{-b + \sqrt{\Delta}}{2 a} \,. \]
• If \(\Delta = 0\) then (5.12) has one solution with multiplicity \(2\). Such solution is given by \[ x_1 = \frac{-b}{2 a} \,. \]
• If \(\Delta < 0\) then (5.12) has two distinct complex solutions given by \[ x_1 = \frac{-b - i\sqrt{-\Delta}}{2 a} \,, \quad x_2 = \frac{-b + i\sqrt{-\Delta}}{2 a} \,, \] where \(\sqrt{-\Delta}\) is a real number, since \(-\Delta>0\).

Moreover, if \(\Delta \neq 0\) we have \[ a x^{2}+b x+c = a (x-x_1)(x-x_2) \,, \] while if \(\Delta = 0\) then \[ a x^{2}+b x+c = a (x-x_1)^2 \,. \]

Example 55

Suppose we want to solve \[ 3 x^{2}-6 x+2 = 0 \,. \] We have that \[ \Delta = (-6)^{2}-4 \cdot 3 \cdot 2 = 12 > 0 \] Therefore the equation has two distinct real solutions, given by \[ x=\frac{-(-6) \pm \sqrt{12}}{2 \cdot 3}=\frac{6 \pm \sqrt{12}}{6}=1 \pm \frac{\sqrt{3}}{3} \] In particular we have the factorization \[ 3 x^{2}-6 x+2 = 3 \left( x - 1 - \frac{\sqrt{3}}{3} \right) \left( x - 1 + \frac{\sqrt{3}}{3} \right) \,. \]

Example 56

Suppose we want to solve \[ 4 x^{2}-8 x+4=0 \,. \] We have that \[ \Delta = (-8)^{2}-4 \cdot 4 \cdot 4 = 0 \,. \] Therefore there exists one solution with multiplicity \(2\). This is given by \[ x=\frac{-(-8)}{2 \cdot 4} = 1 \,. \] In particular we have the factorization \[ 4 x^{2}-8 x+4 = 4 (x-1)^2 \,. \]

Example 57

Consider \[ x^{2}+2 x+3=0 \,. \] We have \[ \Delta = 2^{2}-4 \cdot 1 \cdot 3 = - 8 < 0 \,. \] Therefore there are two complex solutions given by \[ x=\frac{-2 \pm i \sqrt{8}}{2 \cdot 1} = -1 \pm i \sqrt{2} \,. \] In particular we have the factorization \[ x^{2}+2 x+3 = (x + 1 - i \sqrt{2}) (x +1 + i\sqrt{2}) \,. \]

So far we have considered the polynomial equation \[ ax^2 + bx + c = 0 \,, \] for \(a,b,c \in \mathbb{R}\) and \(a \neq 0\).

Question 58

What if \(a,b,c \in \mathbb{C}\)?

If \(a, b, c \in \mathbb{C}\) then we might have \[ \Delta := b^{2}-4 a c \in \mathbb{C}\,. \] Therefore it is not clear how to compute \[ \sqrt{\Delta} \,. \] However, we can still use the quadratic equation.

Proposition 59: Generalization of quadratci formula

Let \(a, b, c \in \mathbb{C}, a \neq 0\). The two solutions to \[ a x^{2}+b x+c=0 \] are given by \[ x_1 = \frac{-b + S_1}{2 a} \,, \quad x_2 = \frac{-b + S_2}{2 a} \,, \] where \(S_1\) and \(S_2\) are the two solutions to \[ z^2 = \Delta \,, \quad \Delta := b^2 - 4ac \,. \]

Remark 60

Suppose that \(\Delta \in \mathbb{R}\). The equation \[ z^2 = \Delta \] has the following solutions:

• If \(\Delta>0\) there are two real solutions \[ S_1 = -\sqrt{\Delta} \,, \quad S_2 = \sqrt{\Delta} \]
• If \(\Delta = 0\) then \(0\) is the only solution with multiplicity \(2\). Hence \[ S_1 = S_2 = 0 \]
• If \(\Delta<0\) there are two complex solutions \[ S_1 = - i \sqrt{-\Delta} \,, \quad S_2 = i \sqrt{-\Delta} \]

Therefore the solutions \[ x_1 = \frac{-b + S_1}{2 a} \,, \quad x_2 = \frac{-b + S_2}{2 a} \,, \] given in Proposition 54 coincide with the ones given in Proposition 59.

Example 61

Let us see an application of Proposition 59. Consider the equation \[ \frac{1}{2} x^{2}-(3+i) x+(4-i)=0 \,. \tag{5.13}\] We have \[\begin{align*} \Delta & = (-(3+i))^{2}-4 \cdot \frac{1}{2} \cdot(4-i) \\ & = 8+6 i-8+2 i \\ & =8 i \,. \end{align*}\] Therefore \(\Delta \in \mathbb{C}\). We have to find solutions \(S_1\) and \(S_2\) to the equation \[ z^2 = \Delta = 8i \,. \tag{5.14}\] We look for solutions of the form \(z=x+ i y\). Then we must have that \[ z^{2}=(a+ ib)^{2}=a^{2}-b^{2}+2 a b i = 8 i \,. \] Thus \[ a^{2}-b^{2}=0 \,, \quad 2 a b = 8 \,. \] From the first equation we conclude that \(|a|=|b|\). From the second equation we have that \(ab=4\), and therefore \(a\) and \(b\) must have the same sign. Hence \(a=b\), and \[ 2 a b = 8 \quad \implies \quad a = b = \pm 2 \,. \] From this we conclude that the solutions to (5.14) are \[ S_{1} = 2+2 i \,, \quad S_{2}=-2-2 i \,. \] Hence the solutions to (5.13) are \[\begin{align*} x_1 & = \frac{3+i+S_{1}}{2 \cdot \frac{1}{2}} = 3 + i + S_{1} \\ & = 3 + i + 2 + 2i = 5 + 3i \,, \end{align*}\] and \[\begin{align*} x_2 & = \frac{3+i+S_{2}}{2 \cdot \frac{1}{2}} = 3+i+S_{2} \\ & = 3+i -2 - 2i = 1 - i \,. \end{align*}\]

In the above example it was a bit laborious to compute \(S_{1}\) and \(S_{2}\). In the next section we will see an easier way to solve problems of the form \(z^2 = \Delta\).

Remark 62: Polynomial equations of order \(n=3,4\)

For polynomial equations \[ p_n(x) = 0 \] with \(p_n\) of degree \(n=3,4\) similar methods exist. However the solution formulas for such equations are really complicated, and we do not cover them here.

Still, it is sometimes possible to solve equations of degree higher than 2, in case it is obvious from inspection that a certain number is a solution, e.g., when \(x=-1,0,1\) is a solution.

Example 63

Consider the equation \[ x^{3}-7 x^{2}+6 x=0 \,. \] It is clear that \(x=0\) is a solution and that we can write \[ x^{3}-7 x^{2}+6 x=x\left(x^{2}-7 x+6\right) \,. \] We could now use the quadratic formula to find the remaining two roots, but we can also directly observe that also \(x=1\) is a solution, so that \(x-1\) divides \(x^{2}-7 x+6\). Using polynomial long division, we find that \[ \frac{x^{2}-7 x+6}{x-1}=x-6 \,, \] see Figure 5.6. Therefore the last solution is \(x=6\), and \[ x^{3}-7 x^{2}+6 x=x(x-1)(x-6) \,. \]

Figure 5.6: Polynomial long division between \(x^{2}-7 x+6\) and \(x-1\).

Example of polynomial long division between \(6x^3+5x^2-7\) and \(3x^2-2x-1\).

Example 64

Suppose we want to solve \[ x^{3}-7 x+6=0 \,. \] It is easy to see \(x=1\) is a solution. This means that \(x-1\) divides \(x^{3}-7 x+6\), so we can compute by using polynomial long division,

\[ \frac{x^{3}-7 x+6}{x-1}=x^{2}+x-6 \,, \] see Figure 5.7. For the remaining two solutions, we can use the quadratic formula to obtain that also \(x=2\) and \(x=-3\) are solutions. Thus \[ x^{3}-7 x+6 = (x-1)(x-2)(x+3) \,. \]

Figure 5.7: Polynomial long division between \(x^{3}-7 x+6\) and \(x-1\).

5.8 Roots of unity

Problem

Let \(n \in \mathbb{N}\). We want to find all complex solutions to \[ z^{n}=1 \,. \tag{5.15}\]

Note that \(z=1\) is always a solution to (5.15) if \(n\) is even. In such case also \(z=-1\) is a solution. If we were only looking for solutions in \(\mathbb{R}\), these two would be the only solutions.

However, the Fundamental Theorem of Algebra, see Theorem 47, tells us that there are \(n\) complex solutions to (5.15).

Question 65

Is there a way to find all \(n\) solutions?

Example 66

We have seen in Example 49 that the solutions to \[ x^{4}=1 \] are \(x=\) \(-1,1, i,-i\). However we deduced this with a procedure which does not seem to generalize well to other exponents.

The trick to find all \(n\) solutions to (5.15) is to use the exponential form.

Theorem 67

Let \(n \in \mathbb{N}\) and consider the equation \[ z^{n}=1 \,. \tag{5.16}\] All the \(n\) solutions to (5.16) are given by \[ z_k = \exp \left(i \frac{2 \pi k}{n}\right) \,, \quad k = 0, \ldots, n-1 \,, \] where \(\exp(x)\) denotes \(e^x\).

Proof

Rewrite \(1\) in exponential form: \[ 1=|1|e^{i \arg(1)} =e^{i 2 \pi k} \, , \quad k \in \mathbb{Z}\,. \] Therefore (5.16) is equivalent to \[ z^n = e^{i 2 \pi k} \,. \] By the properties of the exponential, we see that the above is solved by \[ z_k = \exp \left(i \frac{2 \pi k}{n}\right), \quad k \in \mathbb{Z}\,. \] By choosing \(k = 0, \ldots, n-1\) we obtain \(n\) different solutions.

Definition 68

The solutions to \[ z^{n}=1 \] are called the roots of unity.

Example 69

The solutions to \[ z^{4}=1 \] are given by \[ z_k = \exp \left(i \frac{2 \pi k}{4} \right) = \exp \left(i \frac{\pi k}{2} \right) \,. \] By taking \(k=0,1,2,3\), we obtain the four solutions \[\begin{align*} z_0 & = e^{i 0} = 1 \,, & \quad & z_1 = e^{i \frac{\pi}{2}}=i \,, \\ z_2 & = e^{i \pi}=-1 \,, & \quad & z_3 = e^{i \frac{3 \pi}{2}}=-i \, . \end{align*}\] Note that for \(k=4\) we would again get the solution \(z=e^{i 2 \pi}=1\).

Example 70

The solutions to \[ z^{3}=1 \] are given by \[ z_k = \exp \left( i \frac{2 \pi k}{3} \right) \,. \] By taking \(k=0,1,2\), we obtain the three solutions \[ z_0=e^{i 0}=1, \quad z_1=e^{i \frac{2 \pi}{3}}, \quad z_2=e^{i \frac{4 \pi}{3}} . \] We can also convert \(z_1\) and \(z_2\) to trigonometric form: \[ z_1 = e^{i \frac{2 \pi}{3}}=\cos \left(\frac{2 \pi}{3}\right)+i \sin \left(\frac{2 \pi}{3}\right)=-\frac{1}{2}+\frac{\sqrt{3}}{2} i \] and \[ z_2 = e^{i \frac{4 \pi}{3}}=\cos \left(\frac{4 \pi}{3}\right)+i \sin \left(\frac{4 \pi}{3}\right)=-\frac{1}{2}-\frac{\sqrt{3}}{2} i \]

5.9 Roots in \(\mathbb{C}\)

Problem

Let \(n \in \mathbb{N}\) and \(c \in \mathbb{C}\). We want to find the \(n\)-th roots of \(c\). This means we want to find all complex solutions to \[ z^{n}=c \,. \]

The Fundamental Theorem of Algebra ensures that the above has \(n\) complex solutions. To find these solutions, we pass to the exponential form.

Theorem 71

Let \(n \in \mathbb{N}\), \(c \in \mathbb{C}\) and consider the equation \[ z^{n}=c \,. \tag{5.17}\] All the \(n\) solutions to (5.17) are given by \[ z_k = \sqrt[n]{|c|} \, \exp \left(i \, \frac{\theta + 2 \pi k}{n}\right) \,, \quad k = 0, \ldots, n-1 \,, \] where \(\sqrt[n]{|c|}\) is the \(n\)-th root of the real number \(|c|\), and \(\theta = \arg(c)\).

Proof

Write \(c\) in exponential form: \[ c=|c|e^{i \theta} =|c| e^{i(\theta + 2 \pi k)} \, , \quad k \in \mathbb{Z}\,, \] where \(\theta = \arg(c)\). Therefore (5.17) is equivalent to \[ z^n = |c| e^{i(\theta + 2 \pi k)} \,. \] By the properties of the exponential, we see that the above is solved by \[ z_k= \sqrt[n]{|c|} \exp \left(i \, \frac{\theta + 2 \pi k}{n}\right), \quad k \in \mathbb{Z}\,. \] By choosing \(k = 0, \ldots, n-1\) we obtain \(n\) different solutions.

Example 72

We want to find all the \(z \in \mathbb{C}\) such that \[ z^{5}=-32 \,. \] Let \(c = -32\). We have \[ |c| = |-32|=32=2^{5}\,, \quad \theta = \arg (-32)=\pi \,. \] Hence, the solutions are given by \[ z_k = \left(2^{5}\right)^{\frac{1}{5}} \exp \left(i \pi \, \frac{1+2 k}{5} \right) \,, \quad k \in \mathbb{Z}\,. \]

By taking \(k=0,1,2,3,4\) we get the solutions \[\begin{align*} z_0 & = 2 e^{i \frac{\pi}{5}} \, & \quad & z_1 = 2 e^{i \frac{3 \pi}{5}} \\ z_2 & = 2 e^{i \pi}=-2 \, & \quad & z_3=2 e^{i \frac{7 \pi}{5}} \\ z_4 &= 2 e^{i \frac{9 \pi}{5}} & \quad & \end{align*}\]

Example 73

We want to find all the \(z \in \mathbb{C}\) such that \[ z^{4}=9\left(\cos \left(\frac{\pi}{3}\right)+i \sin \left(\frac{\pi}{3}\right)\right) \,. \] Set \[ c:=9\left(\cos \left(\frac{\pi}{3}\right)+i \sin \left(\frac{\pi}{3}\right)\right) \,. \] The complex number \(c\) is already in the trigonometric form, so that we can immediately obtain \[ |c| = 9 \,, \quad \theta = \arg(c) = \frac{\pi}{3} \,. \] Hence the solutions are given by \[\begin{align*} z_k & = \sqrt[4]{9} \, \exp \left( i \, \frac{ \pi/3 + 2 \pi k}{4} \right) \\ & = \sqrt{3} \exp \left( i \pi \, \frac{1+6 k}{12} \right) \end{align*}\] for \(k \in \mathbb{Z}\). Choosing \(k=0,1,2,3\) gives the \(4\) solutions \[\begin{align*} z_0 & = \sqrt{3} e^{i \pi \frac{1}{12}} & \quad & z_1 = \sqrt{3} e^{i \pi \frac{7}{12}} \\ z_2 & = \sqrt{3} e^{i \pi \frac{13}{12}} & \quad & z_3 = \sqrt{3} e^{i \pi \frac{19}{12}} \end{align*}\]