4  Complex Numbers

Definition 1: Complex Numbers

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

\[ \mathbb{C}:= \mathbb{R}+ i \mathbb{R}:= \{x + i y \, \colon \,x, y \in \mathbb{R}\} \,. \] For a complex number \[ z = x + i y \in \mathbb{C} \] 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.

Definition 2: Addition and multiplication in \(\mathbb{C}\)

Let \(z_1,z_2 \in \mathbb{C}\), so that \[ z_1 = x_1 + i y_1 \,\,, \quad z_2 = x_2 + i y_2 \,\, , \] for some \(x_1,x_2,y_1,y_2 \in \mathbb{R}\):

1. The sum of \(z_1\) and \(z_2\) is \[ z_1 + z_2 :=\left(x_{1}+x_{2}\right) + i \left(y_{1}+y_{2}\right) \,. \]

2. The multiplication of \(z_1\) and \(z_2\) is \[\begin{align*} z_1 \cdot z_2 := \left(x_{1} \cdot x_{2} - y_{1} \cdot y_{2}\right) + i \left(x_{1} \cdot y_{2} + x_{2} \cdot y_{1} \right) \,, \end{align*}\]

Example 3

Question. Compute \(zw\), where \[ z = -2+3 i \,, \quad w = 1- i \,. \]

Solution. 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: We just need to recall that \(i^2\) has to be replaced with \(-1\): \[\begin{align*} z \cdot w & = (-2+3 i) \cdot (1 - i) \\ & = - 2 + 2i + 3i - 3 i^2 \\ & = (-2 + 3 ) + ( 2 + 3 ) i \\ & = 1 + 5 i \, . \end{align*}\]

Proposition 4: 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 \[ - z := -x - i y \,. \]

Proposition 5: 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 \(z^{-1}=w\).

Example 6

Question. Let \(z = 3 + 2 i\). Compute \(z^{-1}\).

Solution. By the formula in Propostion 5 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: \[\begin{align*} (3+2 i)^{-1} & = \frac{1}{3+2 i} \\ & = \frac{1}{3+2 i} \, \frac{3-2 i}{3-2 i} \\ & = \frac{3-2 i}{3^2+2^2} \\ & = \frac{3}{13}-\frac{2}{13} i \,, \end{align*}\] and obtain the same result.

Theorem 7

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

Example 8

Question. Let \(w=1+i\) and \(z=3-i\). Compute \(\frac{w}{z}\).

Solution. We compute \(w/z\) using the two options we have:

1. Using the formula for the inverse from Proposition 5 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 \[\begin{align*} \frac{w}{z} & = \frac{1+i}{3-i} \\ & = \frac{1+i}{3-i} \frac{3+i}{3+i} \\ & = \frac{3-1+(3+1) i}{3^2+1^2} \\ & =\frac{2}{10}+\frac{4}{10} i \\ & = \frac{1}{5}+\frac{2}{5} i \end{align*}\]

Definition 9: Complex conjugate

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

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

Theorem 10

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

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

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

4.1 The complex plane

Definition 11: Modulus

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

Definition 12: 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 \[ |z_1 - z_2| \,. \]

Theorem 13

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

Example 14

Question. Compute the distance between \[ z = 2-4 i \,, \quad w = -5+i \,. \]

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

Theorem 15

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

Theorem 16: Triangle inequality in \(\mathbb{C}\)

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

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

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

Definition 17: 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 \[ \theta := \arg (z) \,. \]

Example 18

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

Theorem 19: 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 \[ \rho := |z| = \sqrt{x^2 + y^2} \,, \quad \theta := \arg(z) \,. \]

Definition 20: Trigonometric form

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

Example 21

Question. Suppose that \(z \in \mathbb{C}\) has polar coordinates \[ \rho = \sqrt{8}\,, \quad \theta = \frac{3}{4} \pi \,. \] Therefore, the trigonometric form of \(z\) is \[ z = \sqrt{8} \left[ \cos \left( \frac{3}{4} \pi \right) + i \sin \left( \frac{3}{4} \pi \right) \right] \,. \] Write \(z\) in cartesian form.

Solution. We have \[\begin{gather*} x = \rho \cos (\theta) = \sqrt{8} \cos \left( \frac{3}{4} \pi \right) = - \sqrt{8} \cdot \frac{\sqrt{2}}{2} = -2 \\ y = \rho \sin (\theta) = \sqrt{8} \sin \left( \frac{3}{4} \pi \right) = \sqrt{8} \cdot \frac{\sqrt{2}}{2} = 2 \,. \end{gather*}\] Therefore, the cartesian form of \(z\) is \[ z = x + i y = - 2 + 2 i \,. \]

Corollary 22: 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 \(\tan\).

Example 23

Question. Compute the arguments of the complex numbers \[ z = 3+4 i\,, \quad \bar{z} = 3-4 i \,, \quad - \bar{z} = -3+4 i \,, \quad - z= -3-4 i \,. \]

Solution. Using the formula for \(\arg\) in Corollary 22 we have \[\begin{align*} \arg (3+4 i) & =\arctan \left(\frac{4}{3}\right) \\ \arg (3-4 i) & =\arctan \left(-\frac{4}{3}\right) = -\arctan \left(\frac{4}{3}\right) \\ \arg (-3+4 i) & = \arctan \left(-\frac{4}{3}\right) + \pi = - \arctan \left(\frac{4}{3}\right) + \pi \\ \arg (-3-4 i) & = \arctan \left(\frac{4}{3}\right) - \pi \end{align*}\]

Theorem 24: Euler’s identity

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

Theorem 25

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

Theorem 26

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

Definition 27: Exponential form

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

Example 28

Question. Write the number \[ z=-2+2 i \] in exponential form.

Solution. From Example 21 we know that \(z = -2+2i\) 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 29: Periodicity of exponential

For all \(k \in \mathbb{Z}\) we have \[ e^{i \theta}=e^{i(\theta+2 \pi k)} \,, \tag{4.1}\] meaning that the complex exponential is \(2\pi\)-periodic.

Proposition 30

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 \(n \in \mathbb{N}\).

Example 31

Question. Compute \((-2+2 i)^{4}\).

Solution. We have two possibilities:

1. Use the binomial theorem: \[\begin{align*} (-2+2 i)^{4} & =(-2)^{4}+\left(\begin{array}{l} 4 \\ 1 \end{array}\right)(-2)^{3} \cdot 2 i+\left(\begin{array}{l} 4 \\ 2 \end{array}\right)(-2)^{2} \cdot(2 i)^{2} \\ & \quad +\left(\begin{array}{l} 4 \\ 3 \end{array}\right)(-2) \cdot(2 i)^{3}+(2 i)^{4} \\ & =16-4 \cdot 8 \cdot 2 i-6 \cdot 4 \cdot 4+4 \cdot 2 \cdot 8 i+16 \\ & =16-64 i-96+64 i+16=-64 \,. \end{align*}\]

2. A much simpler calculation is possible by using the exponential form: We know that \[ -2+2 i = \sqrt{8} e^{i \frac{3}{4} \pi} \] by Example 28. 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 \(e^{i\theta}\) and Euler’s identity.

Definition 32: Complex exponential

The complex exponential of \(z \in \mathbb{C}\) is defined as \[ e^z = |z| e^{i\theta} \,, \quad \theta = \arg(z) \,. \]

Theorem 33

Let \(z,w \in \mathbb{C}\). Then \[ e^{z+w} = e^z e^w \,, \quad (e^z)^{w} = e^{zw} \,. \tag{4.2}\]

Example 34

Question. Compute \(i^{i}\).

Solution. 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 \[ i^{i}=\left(e^{i \frac{\pi}{2}}\right)^{i}=e^{i^2\frac{\pi}{2}}=e^{-\frac{\pi}{2}} \,. \]

4.2 Fundamental Theorem of Algebra

Theorem 35: Fundamental theorem of algebra

Let \(p_{n}(z)\) be a polynomial of degree \(n\) with complex coefficients, i.e., \[ p_{n}(z)=a_{n} z^{n}+a_{n-1} z^{n-1}+\ldots+a_{1} z+a_{0}, \] for some coefficients \(a_{n}, \ldots, a_{0} \in \mathbb{C}\) with \(a_{n} \neq 0\). There exist \[ z_{1}, \ldots, z_{n} \in \mathbb{C} \] solutions to the polynomial equation \[ p_n(z) = a_{n} z^{n}+a_{n-1} z^{n-1}+\ldots+a_{1} z+a_{0}=0 \,. \tag{4.3}\] In particular, \(p_n\) factorizes as \[ p_{n}(z)=a_{n}\left(z-z_{1}\right)\left(z-z_{2}\right) \cdots\left(z-z_{n}\right) \,. \tag{4.4}\]

Example 36

Question. Find all the complex solutions to \[ z^{2}=-1 \tag{4.5}\]

Solution. The equation \(z^2 = -1\) is equivalent to \[ p(z) = 0 \,, \quad p(z):=z^{2}+1 \,. \] Since \(p\) has degree \(n=2\), the Fundamental Theorem of Algebra tells us that there are two solutions to (4.5). We have already seen that these two solutions are \(z=i\) and \(z=-i\). Then \(p\) factorizes as \[ p(z) = z^{2} + 1 = (z-i)(z+i) \,. \]

Example 37

Question. Find all the complex solutions to \[ z^{4}-1=0 \,. \tag{4.6}\]

Solution The associated polynomial equation is \[ p(z) = 0 \,, \quad p(z) := z^4 - 1 \,. \] Since \(p\) has degree \(n=4\), the Fundamental Theorem of Algebra tells us that there are \(4\) solutions to (4.6). Let us find such solutions. We use the well known identity \[ a^2-b^2 = (a+b)(a-b) \,, \quad \forall \, a,b \in \mathbb{R}\,, \] to factorize \(p\). We get: \[ p(z) = (z^4-1) = (z^2+1)(z^2-1) \,. \] We know that \[ z^2 + 1 = 0 \] has solutions \(z = \pm i\). Instead
\[ z^2 - 1 = 0 \] has solutions \(x = \pm 1\). Hence, the four solutions of (4.6) are given by \[ z=1,-1, i,-i \,, \] and \(p\) factorizes as \[ p(z) = z^4 - 1 = (z-1)(z+1)(z-i)(z+i) \,. \]

Definition 38

Suppose that the polynomial \(p_n\) factorizes as \[ p_n(z) = a_n (z-z_1)^{k_1} (z-z_2)^{k_2} \cdots (z-z_m)^{k_m} \] 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 \[ n = k_1 + \ldots + k_m = \sum_{i=1}^m k_i \,. \] Note that \(z_i\) is solves the equation \[ p_n(z) = 0 \] exactly \(k_i\) times. We call \(k_i\) the multiplicity of the solution \(z_i\).

Example 39

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

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

4.3 Solving polynomial equations

Proposition 40: Quadratic formula

Let \(a, b, c \in \mathbb{R}, a \neq 0\) and consider the equation \[ ax^2 + bx + c = 0 \,. \tag{4.7}\] Define \[ \Delta := b^2 - 4 ac \in \mathbb{R}\,. \] The following hold:

1. If \(\Delta > 0\) then (4.7) has two distinct real solutions \(z_1,z_2 \in \mathbb{R}\) given by \[ z_1 = \frac{-b - \sqrt{\Delta}}{2 a} \,, \quad z_2 = \frac{-b + \sqrt{\Delta}}{2 a} \,. \]
2. If \(\Delta = 0\) then (4.7) has one real solution \(z \in \mathbb{R}\) with multiplicity \(2\). Such solution is given by \[ z = z_1 = z_2 = \frac{-b}{2 a} \,. \]
3. If \(\Delta < 0\) then (4.7) has two distinct complex solutions \(z_1,z_2 \in \mathbb{C}\) given by \[ z_1 = \frac{-b - i\sqrt{-\Delta}}{2 a} \,, \quad z_2 = \frac{-b + i\sqrt{-\Delta}}{2 a} \,, \] where \(\sqrt{-\Delta} \in \mathbb{R}\), since \(-\Delta>0\).

In all cases, the polynomial at (4.7) factorizes as \[ a z^{2}+b z+c = a (z-z_1)(z-z_2) \,. \]

Example 41

Question. Solve the following equations:

1. \(3 z^{2}-6 z+2 = 0\)
2. \(4 z^{2}-8 z+4=0\)
3. \(z^{2}+2 z+3=0\)

Solution.

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

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

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

Proposition 42: Quadratic formula with complex coefficients

Let \(a, b, c \in \mathbb{C}, a \neq 0\). The two solutions to \[ a z^{2}+b z + c=0 \] are given by \[ z_1 = \frac{-b + S_1}{2 a} \,, \quad z_2 = \frac{-b + S_2}{2 a} \,, \] where \(S_1\) and \(S_2\) are the two solutions to \[ z^2 = \Delta \,, \quad \Delta := b^2 - 4ac \,. \]

Example 43

Question Find all the solutions to \[ \frac{1}{2} z^{2}-(3+i) z+(4-i)=0 \,. \tag{4.8}\]

Solution. 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 \[ z^2 = \Delta = 8i \,. \tag{4.9}\] We look for solutions of the form \(z=a+ i b\). 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 (4.9) are \[ S_{1} = 2+2 i \,, \quad S_{2}=-2-2 i \,. \] Hence the solutions to (4.8) are \[\begin{align*} z_1 & = \frac{3+i+S_{1}}{2 \cdot \frac{1}{2}} = 3 + i + S_{1} \\ & = 3 + i + 2 + 2i = 5 + 3i \,, \end{align*}\] and \[\begin{align*} z_2 & = \frac{3+i+S_{2}}{2 \cdot \frac{1}{2}} = 3+i+S_{2} \\ & = 3+i -2 - 2i = 1 - i \,. \end{align*}\] In particular, the given polynomial factorizes as \[\begin{align*} \frac{1}{2} z^{2}-(3+i) z+(4-i) & = \frac12 (z - z_1)(z-z_2) \\ & = \frac12 (z - 5 - 3i)(z - 1 + i) \,. \end{align*}\]

Example 44

Question. Consider the equation \[ z^{3}-7 z^{2}+6 z=0 \,. \]

1. Check whether \(z=0,1,-1\) are solutions.
2. Using your answer from Point 1, and polynomial division, find all the solutions.

Solution.

1. By direct inspection we see that \(z=0\) and \(z=1\) are solutions.

2. Since \(z = 0\) is a solution, we can factorize \[ z^{3}-7 z^{2}+6 z=z\left(z^{2}-7 z+6\right) \,. \] We could now use the quadratic formula on the term \(z^{2}-7 z+6\) to find the remaining two roots. However, we have already observed that \(z=1\) is a solution. Therefore \(z-1\) divides \(z^{2}-7 z+6\). Using polynomial long division, we find that \[ \frac{z^{2}-7 z+6}{z-1}=z-6 \,. \] Therefore the last solution is \(z=6\), and \[ z^{3}-7 z^{2}+6 z=z(z-1)(z-6) \,. \]

Example 45

Question. Find all the complex solutions to \[ z^{3}-7 z+6=0 \,. \]

Solution. It is easy to see \(z=1\) is a solution. This means that \(z-1\) divides \(z^{3}-7 z+6\). By using polynomial long division, we compute that \[ \frac{z^{3}-7 z+6}{z-1}=z^{2}+z-6 \,. \] We are now left to solve \[ z^{2}+z-6 = 0\,. \] Using the quadratic formula, we see that the above is solved by \(z=2\) and \(z=-3\). Therefore the given polynomial factorizes as \[ z^{3}-7 z+6 = (z-1)(z-2)(z+3) \,. \]

4.4 Roots

Theorem 46

Let \(n \in \mathbb{N}\) and consider the equation \[ z^{n}=1 \,. \tag{4.10}\] All the \(n\) solutions to (4.10) are given by \[ z_k = \exp \left(i \frac{2 \pi k}{n}\right) \,, \quad k = 0, \ldots, n-1 \,, \] where \(\exp(x)\) denotes \(e^x\).

Definition 47

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

Example 48

Question. Find all the solutions to \[ z^{4}=1 \,. \]

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

Example 49

Question. Find all the solutions to \[ z^{3}=1 \,. \]

Solution. The \(3\) solutions are given by \[ z_k = \exp \left( i \frac{2 \pi k}{3} \right) \,, \] for \(k=0,1,2\). We compute: \[ z_0=e^{i 0}=1, \quad z_1=e^{i \frac{2 \pi}{3}}, \quad z_2=e^{i \frac{4 \pi}{3}} . \] We can write \(z_1\) and \(z_2\) in cartesian 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 \[ z_2 = e^{i \frac{4 \pi}{3}}=\cos \left(\frac{4 \pi}{3}\right)+i \sin \left(\frac{4 \pi}{3}\right)=-\frac{1}{2}-\frac{\sqrt{3}}{2} i \,. \]

Theorem 50

Let \(n \in \mathbb{N}\), \(c \in \mathbb{C}\) and consider the equation \[ z^{n}=c \,. \tag{4.11}\] All the \(n\) solutions to (4.11) 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 \(\theta = \arg(c)\).

Example 51

Question. Find all the \(z \in \mathbb{C}\) such that \[ z^{5}=-32 \,. \]

Solution. Let \(c = -32\). We have \[ |c| = |-32|=32=2^{5}\,, \quad \theta = \arg (-32)=\pi \,. \] The \(5\) solutions are given by \[ z_k = \left(2^{5}\right)^{\frac{1}{5}} \exp \left(i \pi \, \frac{1+2 k}{5} \right) \,, \quad k \in \mathbb{Z}\,, \] for \(k=0,1,2,3,4\). We get \[\begin{align*} z_0 & = 2 e^{i \frac{\pi}{5}} \, & \quad & z_1 = 2 e^{i \frac{3 \pi}{5}} \\ z_2 & = 2 e^{i \pi}=-2 \, & \quad & z_3=2 e^{i \frac{7 \pi}{5}} \\ z_4 &= 2 e^{i \frac{9 \pi}{5}} & \quad & \end{align*}\]

Example 52

Question. Find all the \(z \in \mathbb{C}\) such that \[ z^{4}=9\left(\cos \left(\frac{\pi}{3}\right)+i \sin \left(\frac{\pi}{3}\right)\right) \,. \]

Solution. 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 \[ |c| = 9 \,, \quad \theta = \arg(c) = \frac{\pi}{3} \,. \] The \(4\) 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 \(k=0,1,2,3\). We compute \[\begin{align*} z_0 & = \sqrt{3} e^{i \pi \frac{1}{12}} & \quad & z_1 = \sqrt{3} e^{i \pi \frac{7}{12}} \\ z_2 & = \sqrt{3} e^{i \pi \frac{13}{12}} & \quad & z_3 = \sqrt{3} e^{i \pi \frac{19}{12}} \end{align*}\]