Lecture 20 - Complex Exponential, Logarithm, and Powers

Complex Exponential

Definition of a complex exponential is given here.

Proposition

The function $\exp$ satisfies the following:

1. $\exp (z)=e^z$ is entire and $\frac{d{e}^z}{dz}=z$,
2. $e^{z_1}{e}^{z_2}=e^{z_1+z_2}$
3. $e^{z+2\pi in}=e^z$ for $n\in \mathbb{Z}$

Proof

Proposition 1

Writing $z=x+iy$ we get
$\frac{\partial e^z}{\partial x}=e^x(\cos (y)+i\sin (y)),$
$\frac{\partial e^z}{\partial y}=e^x(-\sin (y)+i\cos (y)).$
These are continuous and
$\frac{\partial e^z}{\partial x}=-i\frac{\partial e^z}{\partial y}.$
Hence $e^z$ is holomorphic for any $z$ by the Cauchy-Riemann Equations.

Proposition 2

We have
$e^{z_1}{e}^{z_2}=e^{x_1}{e}^{x_2}{e}^{i{y}_1}{e}^{i{y}_2}$
$=e^{x_1+x_2}{e}^{i(y_1+y_2)}$
$=e^{z_1+z_2}$
Where we used the previous result for $e^{i\varphi }$

Proposition 3

This follows from the periodicity of $\cos (y)$ and $\sin (y)$.

Another possible definition of $e^z$

$e^z={\sum }_{n=0}^{\infty }\frac{z^n}{n!}$
This leads to Euler’s Formula:
$e^{ix}=\cos (x)+i\sin (x)$

The Complex Logarithm

In the real case, $\log$ is the inverse function of the $\exp$. It is mapped as a one-to-one function $\exp :\mathbb{R}\to \mathbb{R}$. This means that:
$\ln (\exp (x))=x,\forall x\in \mathbb{R},$
$\exp (\ln (y))=y,\forall y>0.$
However, this is not the case in the complex case, $e^z$, since $e^{z+2\pi in}=e^z$, for $n\in \mathbb{Z}$.

We characterize all solutions to the equation $e^w=z$.

Lemma

Fix $z\in \mathbb{C}$. Then all solutions $w$ to $e^w=z$ are of the form
$w=\ln |z|+i\arg (z)$
for any argument of $z$.

Proof

Write $w=x+iy$. Then we have
$z=|z|\times e^{i\arg (z)}.$
Comparing with the polar form of $e^w$, we get that:
$e^x=|z|,e^{i\arg (z)}=e^{iy}.$
The first gives $x=\ln |z|$, while the second gives $\arg (z)=\arg (y)+\text{multiples of}2\pi$.

If we want to define a complex log we have two possibilities:

1. We can consider multi-valued functions (or Riemann Surfaces),
2. We can make it into a single-valued function by making a imposing some restrictions on the argument.

Principal Argument Definition

Given a non-zero complex number $z\in \mathbb{C}$, its Principal Argument $\mathrm{Arg}z$ is the unique argument for $z$ such that
$-\pi <\mathrm{Arg}z\le \pi .$
Then we define the log as follows

Principal Logarithm Definition

The principal logarithm is the function $\mathrm{Log}:\mathbb{C}\setminus \{0\}\to \mathbb{C}$ is defined by
$\mathrm{Log}z=\ln |z|+i\mathrm{Arg}z.$

For $z=x>0$

We get $\mathrm{Log}x=\ln x$. (The right hand side is the usual log for real numbers)

Remark

Here we use the following notations:

• $\ln x$ is the usual log for real numbers
• $\mathrm{Log}z$ is the Principal Logarithm, as introduced
• $\log z$ refers to other possible choices for the logarithm

It tuns out that $\mathrm{Log}z$ is also Holomorphic.

Complex Powers

We use the complex exponential and the complex logarithm to define the complex powers of complex numbers.

Example

Let us show that the complex power
$z^{\frac{1}{2}}=e^{\frac{1}{2}\mathrm{Log}z}$
gives the usual square-root for positive real numbers (so $z=x>0$).

We have
$x^{\frac{1}{2}}=e^{\frac{1}{2}\mathrm{Log}x}=e^{\frac{1}{2}(\ln |x|+i\mathrm{Arg}x)}.$
We have $|x|=x$, since $x>0$, and $\mathrm{Arg}x=0$. Then
$x^{\frac{1}{2}}=e^{\frac{1}{2}\ln x}=e^{\ln x^{\frac{1}{2}}}=\sqrt{x}$
gives back the expected result.

What happens if we make another choice for the logarithm? Equivalently, another choice for the argument, such as $\arg x=2\pi$.
Then, writing $\log z$ for the corresponding logarithm, we have
$x^{\frac{1}{2}}=e^{\frac{1}{2}\log x}=e^{\frac{1}{2}(\ln x+i\arg x)}$
$=e^{\frac{1}{2}\ln x+i\pi }$
$=e^{i\pi }\times e^{\frac{1}{2}\ln x}$
$=-\sqrt{x}$
This is related to $x^2$ as it is not one-to-one.
All different choices fr the $\log$/$\arg$ lead to such different choices for the inner functions.

Remark

Note that for $n\in \mathbb{Z}$ there is no ambiguity in defining $z^n$. Indeed, we have
$z^n=e^{n\mathrm{Log}z}=e^{n(\ln |z|+i(\theta +2\pi k))},k\in \mathbb{Z}$
$=e^{n(\ln |z|+i\theta )}$
because $n\times k$ is again an integer (and $e^{i2\pi n}=1$).

Example - Roots of Unity

Let’s look at an example of solving polynomial equations

We want to find all solutions to the equation $z^3=1$

Writing $z=n{e}^{i\theta }$ we must have
$n^3{e}^{3i\theta }=1.$
Then $n^3=1$, which gives $n=1$. (Since $n\ge 0$)
For angles we must have
$3i\theta =2\pi in,n\in \mathbb{Z}.$
This is rewritten as $\theta =\frac{2\pi }{3}n$.
This leads to
$z_0=e^{i\frac{2\pi }{3}\times 0}=1,z_1=e^{i\frac{2\pi }{3}\times 1}=-\frac{1}{2}+i\frac{\sqrt{3}}{2},z_3=e^{i\frac{2\pi }{3}\times 2}=-\frac{1}{2}-i\frac{\sqrt{3}}{2}.$