Complex number manipulation

Question

Considering that we have $w = e^z$ how can we be convinced that $z=\log\left(r\right) + i\theta$ ? is there a way to get to this without using $w=re^\left(i\theta \right)$?

Answer 1

If you're happy to assume that $w = r(\cos \theta + i \sin \theta)$ (since that's polar form), then we might proceed as follows. Let $z = x + iy$ with $x, y \in \mathbb R$. Then $e^z = e^{x + iy} = e^x(\cos y + i \sin y)$ by definition of the complex exponential. Then by equating real and imaginary parts of $e^z$ and $w$, we get \begin{align*} r\cos \theta &= e^x \cos y \\ r\sin \theta &= e^x \sin y \end{align*} Squaring and adding these together, $r^2 = e^{2x}$, so $e^x = \pm r$. If $x$ is to be real, then $e^x$ should be positive, so take $e^x = r \implies x = \log r$. Then we can cancel the $e^x$s, getting \begin{align*} \cos \theta &= \cos y \\ \sin \theta &= \sin y \end{align*} You can check if you want that the only possible solutions to this are $y = \theta + 2\pi n$ for some $n \in \mathbb Z$. Therefore, $w = \log r + i(\theta + 2\pi n)$ (which is the only thing we can deduce from $w = e^z$, since for example $e^{i\pi} = e^{3i\pi}$.)

Answer 2

Note

$$z=\ln w = \ln (re^{i\theta} )=\ln r + \ln e^{i\theta}=\ln r + {i\theta }$$

Answer 3

Write $w$ in polar coordinates: $w = r e^{i \theta}$, where $r = |w|$ and $\theta = \arg w$.
Now consider this other complex number, $z = \log r + i\theta$. What is $e^z$ ? Compute $$ e^{z}= e^{\log r + i\theta} = e^{\log r}\;e^{i\theta} = r e^{i\theta} = w . $$

Note
This fails when $w=0$ so $r=0$ and $\log r$ is undefined. Zero does not belong to the range of the complex exponential function.