Lecture 3: Elementary Complex Functions

Elementary complex functions

Starting from this lecture, we study functions between sets of complex numbers.

Definition 1 (Functions). Let $X$ and $Y$ be two sets. A function $f$ from $X$ to $Y$ , often written as $f: X\to Y$ , is a binary relation between $X$ and $Y$ that associate each $x$ in a subset of $X$ to a unique $y\in Y$ . The set of $x\in X$ on which $f$ is defined is called the domain of the function, denoted by $Dom(f)$ . The set of all $y$ values, $\{y \in Y \mid y=f(x),\ x\in X\}$ , is called the range of the function $f$ , which we denote by $Rang(f)$ .

In calculus, we have seen many elementary functions that are real-valued, meaning that the range of those functions are subsets of $\mathbb R$ . Examples of these elementary functions include, but are not limited to, the polynomial functions, the exponential function, the trignometric functions, and the logarithmic function. We now briefly introduce the complex equivalence of those functions.

Polynomial functions

Definition 2 (The complex power function). The usual power function $f(z)=z^n$ ( $n\in\mathbb Z, n\ge 0$ ) is defined on the complex plane $\mathbb C$ as follows: $\forall z\in \mathbb C,\ \ z^n:= \left\{ \begin{array}{cc} 1, & n=0\\ \underbrace{z\cdot z \cdots z}_{n}, & n\ge 1 \end{array} \right.$

For instance, let $f(z)=z^3$ , then when $z=2+i$ , we have $f(2+i)=(2+i)^3=(2+i)(2+i)(2+i)=2+11i$ . For any $n\ge 0$ , this function is well defined at any point $z\in\mathbb C$ , therefore we can take $Dom(f)=\mathbb C$ .

Example 3 (Equivalent representation of a complex function). Note that for any given complex function $f(z)$ ( $z=x+iy$ ), we can separate its real and imaginary components as follows: $f(z)=f(x+iy)=u(x,y)+iv(x,y)$ where $x$ , $y$ , $u(x,y)$ and $v(x,y)$ are all real-valued.

Take the example $f(z)=z^3$ , $z\in\mathbb C$ , we can write $f(z)=f(x+iy)=(x+iy)^3=(x^3-3xy^2)+i(3x^2y-y^3)$ . Here $u(x, y)=x^3-3xy^2$ and $v(x, y)=3x^2y-y^3$ .

Definition 4 (Polynomial Function). The polynomial function of $n$ -th order, often denoted by $P_n(z)$ , is defined as $P_n(z) := a_0+a_1 z + \cdots + a_n z^n = \sum_{k=0}^n a_k z^k,$ where $a_k\in \mathbb C$ , $k=0,1,...,n$ , are complex numbers.

A rational function $f(z)$ is a function of the form $f(z)=\dfrac{P_n(z)}{Q_m(z)}$ where $P_n$ and $Q_m$ are polynomials of order $n$ and $m$ respectively, and $Q_m(z)\neq 0$ .

Here is a simple polynomial function: $f(z) = 2z^3+4z+1, \ \ z\in\mathbb C.$

Here is a simple rational function: $f(z) = \dfrac{1}{1+z}, \ \ z\neq -1$ We can write this rational function into the $u(x,y)+iv(x,y)$ form as $f(z)=\dfrac{1}{1+z}= \dfrac{1+\bar z}{|1+z|^2} = \dfrac{1+x-iy}{(x+1)^2+y^2}=\dfrac{1+x}{(x+1)^2+y^2}-i\dfrac{y}{(x+1)^2+y^2}.$

The exponential function

Definition 5 (Exponential Function). For any $z=x+iy$ , $x, y\in\mathbb R$ , we define the complex exponential function $e^z$ as $e^z\equiv e^{x+iy} := e^x e^{iy}=e^x(\cos y + i\sin y).$

Note that this definition uses Euler’s formula.

It is straightforward to prove the following result.

Theorem 6. (i) $\forall z_1, z_2\in\mathbb C$ , we have that $e^{z_1}e^{z_2}=e^{z_1+z_2}.$ (ii) $\forall z\in\mathbb C$ and $k\in \mathbb N$ , we have that $(e^z)^k=e^{kz}.$

Proof. (i) Let $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$ , then $\begin{gathered} e^{z_1}e^{z_2}=e^{x_1}e^{iy}e^{x_2}e^{iy_2}=e^{x_1+x_2}(\cos y_1+i\sin y_1)(\cos y_2+i\sin y_2) \\ =e^{x_1+x_2}(\cos (y_1+y_2)+i\sin (y_1+y_2))=e^{x_1+x_2}e^{i(y_1+y_2)}=e^{z_1+z_2}. \end{gathered}$ (ii) We apply (i) $k-1$ times with $z_1=z_2=z$ . ◻

The exponential function is very useful in the theory and applications of complex variables. Here are some examples.

Example 7. To compute $(1+i)^9$ , we first observe that $z=1+i=\sqrt{2}e^{i\pi/4}$ . Therefore, $z^9=(\sqrt{2}e^{i\pi/4})^9=16\sqrt{2}e^{i9\pi/4}= 16\sqrt{2}(\cos(9\pi/4)+i\sin(9\pi/4))=16(1+i)$ .

Example 8. Let’s try to find complex solutions to the equation: $z^2+2z+5=0.$ We first observe that we can rewrite the equation as $(z+1)^2=-4.$ Let $w=z+1=|w|e^{i\theta}$ . Use the fact that $-4=-4+0i=4(-1+0i)=4e^{(1+2k)\pi i}$ $k\in\mathbb Z$ , we can further simplify the equation to $|w|^2e^{i2\theta}=4e^{(1+2k)\pi i}.$ Therefore $|w|=2$ and $2\theta=(1+2k)\pi\implies \theta=(1/2+k)\pi$ . Therefore, $w=2e^{(1/2+k)\pi i}, \quad k\in\mathbb Z.$ There are two distinct values of $w$ : $w_1=2e^{1/2 \pi i}=2i$ and $w_2=2e^{3/2 \pi i}=-2i$ . Thus $z$ has two distinct values $z_1=w-1=-1+2i$ and $z_2=w_2-1=-1-2i$ .

Example 9. Let’s try to find all the complex solutions to the equation: $e^z=-1.$ We first observe that $-1 = \cos(\pi+2k\pi)+i\sin(\pi+2k\pi) = e^{i(1+2k)\pi}, \ \ k\in\mathbb Z.$ Therefore the equation can be rewritten as $e^z= e^{i(1+2k)\pi},$ which implies $z=i(1+2k)\pi, \ \ k\in \mathbb Z.$

Trigonometric functions

We first observe that from Euler’s formula, we immediately have $\forall y\in\mathbb R, \qquad \cos y=\dfrac{e^{iy}+e^{-iy}}{2}, \quad \mbox{and},\quad \sin y=\dfrac{e^{iy}-e^{-iy}}{2i}.$

This motivates the following definition of complex trigonometric functions.

Definition 10. For any complex variable $z$ , we define the complex trigonometric sine $(\sin z)$ , cosine $(\cos z)$ , tangent $(\tan z)$ and cotangent $(\cot z)$ functions as follows: $\sin z := \frac{e^{iz}-e^{-iz}}{2i},$ $\cos z := \frac{e^{iz}+e^{-iz}}{2},$ $\tan z : \frac{\sin z}{\cos z}, \ \ \mbox{when}\ \ \cos z\neq 0$ $\cot z := \frac{\cos z}{\sin z}, \ \ \mbox{when}\ \ \sin z\neq 0$

The trignometric functions we defined above shares many nice properties with their real-valued counterparts. For instance, one can prove the trignometric identies: $\sin^2 z + \cos^2 z=1$ , $\forall z\in\mathbb C$ , and $\sin(z+z')=\sin z\cos z' + \cos z\sin z'$ (these are left as exercises). However, there are also some properties that makes the complex version of those functions special. For instance, it is well-known that $-1\le \sin x\le 1$ $\forall x\in\mathbb R$ . This is NOT true for $\sin z$ when $z\in\mathbb C$ . To see that, let $z=x+iy$ , $x, y\in\mathbb R$ . Then $\sin z=\dfrac{e^{iz}-e^{-iz}}{2i}=\dfrac{e^{-y}e^{ix}-e^ye^{-ix}}{2i}=\dfrac{(e^{-y}-e^y)\cos x+i(e^{-y}+e^y)\sin x}{2i}.$ It is easy to see that, for a general $x$ , both the real and the imaginary parts of $\sin z$ blow up when $y\to -\infty$ or $y\to +\infty$ .

The logarithmic function

The complex logarithmic function is defined as the inverse of the exponential function, just as in the real variable case.

Definition 11. For the complex variable $z$ , the complex logarithmic function of $z$ , denoted by $\ln z$ , is defined through the relation $e^{\ln z}=z.$

Note that from the definition, we have $e^{\ln z}=z=|z|e^{i(\theta+2k\pi)} = e^{\ln|z|+i(\theta_0+2k\pi)}, \quad \theta_0\in(-\pi, \pi],\quad k\in\mathbb Z.$ This implies that $\ln z =\ln|z|+i(\theta_0+2k\pi), \quad \theta_0\in(-\pi, \pi],\quad k\in\mathbb Z.$ This relation can be directly used as the definition of the complex logarithmic function. It shows that $\ln z$ is a “multi-valued” function. We have to specify which $k$ we take every time we use this function. The function with different $k$ are called different branches of the function.

Guided review and additional examples

Learning goals

Relate the complex exponential to real exponential and trigonometric functions.
Solve exponential and trigonometric equations while accounting for periodicity.
Distinguish the multivalued logarithm from a selected branch of the logarithm.
Recognize how branch choices affect powers such as $z^a$ .

Concept connection: periodicity creates branches

Euler’s formula gives $e^{x+iy}=e^x(\cos y+i\sin y)$ . The exponential is therefore periodic with period $2\pi i$ . Reversing the exponential cannot produce a single value on all of $\mathbb C\setminus\{0\}$ ; a logarithm branch is a consistent local choice of argument.

Worked example 1: solve $e^z=-1$

Write $z=x+iy$ . Taking moduli gives $e^x=1$ , so $x=0$ . The point $-1$ has arguments $(2k+1)\pi$ . Hence

$z=(2k+1)\pi i,\qquad k\in\mathbb Z.$

Substitution confirms that $e^{(2k+1)\pi i}=-1$ .

Worked example 2: understand $i^i$

Complex powers are defined using a logarithm. Since

$\log i=i\left(\frac{\pi}{2}+2\pi k\right),$

the multivalued expression is

$i^i=\exp(i\log i)=\exp\left(-\frac{\pi}{2}-2\pi k\right), \qquad k\in\mathbb Z.$

All these values are positive real numbers. Using the principal logarithm selects the principal value $e^{-\pi/2}$ .

Check your understanding

Why can the principal logarithm fail to satisfy $\operatorname{Log}(z_1z_2)=\operatorname{Log}z_1+\operatorname{Log}z_2$ ?

Show the answer

Principal arguments are restricted to a fixed interval. For $z_1=z_2=-1$ , the right side is $2\pi i$ , while the left side is $\operatorname{Log}(1)=0$ . The two values differ by $2\pi i$ , reflecting the exponential’s periodicity.