Complex exponential functions

Complex numbers: Complex functions

Complex exponential functions

We have already introduced the imaginary power $e^{\,\mathrm{i}\,\varphi}$ for real $\varphi$ . It is not difficult to guess from this what the definition of $e^z$ should be for a complex number $z$ . If $z=x+\mathrm{i}\,y$ , then naturally you want that the calculation rule $e^{x+\mathrm{i}\,y}=e^x\cdot e^{\,\mathrm{i}\,y}$ remains in effect (Recall that we want to maintain as many properties of real functions as possible). But then the only possible definition for the complex exponential function $e^z$ , also referred to as $\exp(z)$ , is as follows:

If $z=x+\mathrm{i}\,y$ is in standard form, then we define $\begin{aligned}e^z &=e^x\cdot e^{\;\mathrm{i}\,y }\\ &= e^x\cdot\bigl(\cos(y)+\mathrm{i}\,\sin(y)\bigr)\end{aligned}$

In other words, for every complex number $z$ we define the complex number $e^z$ by $\begin{cases} |\e^z| &= e^{\mathrm{Re}(z)}\\ \mathrm{arg}(e^z) &=\mathrm{Im}(z)\quad (\mathrm{mod\;} 2\pi)\end{cases}$ The function that maps $z$ onto the complex number $e^z$ is called called the complex exponential function and we also denote it as $\exp$ , so $\exp(z)=e^z$ .

If $z$ is a real number, then: $|e^z|=e^z$ and $\mathrm{Arg}(z)=0$ . In this case, the complex exponential function value equal to the well-known, real power of $e$ .

Complex power with a positive real base With the complex exponential function in our hands we can also define powers with other positive real bases.

Let $g$ be a positive real number and $z$ a complex number, then we can write $g^z = e^{\ln(g)\cdot z}$

Via the real exponential function, logarithm, and trigonometric functions and a scientific calculator you can calculate function values.

Write the number $e^{1.0-3.0\,\mathrm{i}}$ in standard form and use a calculator to compute the answer to three decimal places.

$\begin{aligned}e^{1.0-3.0\,\mathrm{i}}&=e^{1.0}\cdot e^{-3.0\,\mathrm{i}} & \color{blue}{\text{calculation rules for powers}}\\ &=e^{1.0}\cdot \bigl(\cos(-3.0)+\mathrm{i}\sin(-3.0)\bigr) &\color{blue}{\text{Euler's formula}} \\ &\approx -2.691-0.384\,\mathrm{i} & \color{blue}{\text{calculator}}\end{aligned}$

New example

The complex exponential function has many features that we can expect. We give two examples.

For any positive real number $g$ , integer $n$ , and each pair of complex numbers $z$ and $w$ the following holds: $\begin{aligned}r^{z}\cdot r^{w}&=r^{z+w}\\ \left(r^z\right)^n&=r^{n\cdot z}\end{aligned}$

The complex exponential function is equal to its own derivative, in other words: $\frac{\dd }{\dd z}(e^z)=e^z$

But there are also differences between the real exponential function and the complex exponential function. From the first multiplication rule we obtain the following statement:

The complex exponential function is periodic with period $2\pi\,\mathrm{i}$ .

After all, for every integer $k$ the following statement is valid: $e^{2\,k\,\pi\,\mathrm{i}}=\left(e^{\pi\,\mathrm{i}}\right)^{2k}=(-1)^{2k}=1$ So: $e^{z+2\,k\,\pi\,\mathrm{i}}=e^z\cdot e^{2\,k\,\pi\,\mathrm{i}}=e^z$

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In other words, the last rule for the complex exponential function means that the equation $e^z=1$ in $\mathbb{C}$ has infinitely many solutions. The general solution can be written as $z=0\quad(\mathrm{mod}\;2\,\pi\,\ii)$ Look at more examples of complex exponential equations that are exactly solvable.

Solve the following equation exactly in $\mathbb{C}$ and write the answer in standard form: $e^z=-1+\sqrt{3}\,\ii$

In order to solve the equation $\e^z=-1+\sqrt{3}\,\ii$ , we set the absolute values and the arguments of the complex numbers on the left-hand and right-hand side equal to each other. We first write $z$ in standard form as $z=x+y\,\ii$ , with real unknowns $x$ and $y$ .

First we set the absolute values equal to each other on the left-hand and right-hand side. On the left-hand side we have: $|e^z|=e^{x}$ On the right-hand side we find: $|-1+\sqrt{3}\,\ii|=\sqrt{(-1)^2+(\sqrt{3})^2}=\sqrt{4}=2$ Equating the absolute values leads to the equation $\e^x=2$ The exact solution of this equation is and can be written as $x=\ln(2)$ Hereafter we set the arguments of the complex numbers on left-hand and right-hand side equal to each other. On the left-hand side we get: $\mathrm{arg}(e^z)=\mathrm{Im}(z)=y\quad (\mathrm{mod\;} 2\pi)$ On the right-hand side special values of real and imaginary parts lead to $-1+\sqrt{3}\,\ii=2\cdot\bigl(\cos(\tfrac{2}{3}\pi)+\ii\,\sin(\tfrac{2}{3}\pi)\bigr)$ Thus $\mathrm{Arg}(-1+\sqrt{3}\,\ii)=\tfrac{2}{3}\pi$ So $y=\tfrac{2}{3}\pi+k\cdot 2\pi\qquad\text{for some integer }k$
In summary: there are infinitely many solutions of the equation $\e^z=-1+\sqrt{3}\,\ii$ but each solution can be written as $z=\ln(2)+(\tfrac{2}{3}\pi+k\cdot 2\pi)\,\ii \qquad\text{for some integer }k$ The general solution can be written as $z=\ln(2)+\tfrac{2}{3}\pi\,\ii\quad (\mathrm{mod\;} 2\,\pi\,\ii)$
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A shorter, but equivalent route starts with the writing of the right-hand side in polar form, say $r\, e^{\ii\,\phi}$ .
But if $e^z=r\, e^{\ii\,\phi}$ then it follows that $e^z=e^{\ln(r)+\ii\,\phi}$ So $z=\ln(r)+\ii\,\phi\quad(\text{mod }2\,\pi\,\ii)$ In this exercise: $-1+\sqrt{3}\,\ii=2\,e^{\tfrac{2}{3}\pi\,\ii}$ So $z=\ln(2)+\tfrac{2}{3}\pi\,\ii\quad (\mathrm{mod\;} 2\,\pi\,\ii)$

New example