## Coursera: analysis complex kind Week 3.3. Complex exponential function.

### Background: $e^{i\theta} = cos(\theta) + i sin (\theta)$

Hence, $e^{i(\theta + 2k \pi)} = cos(\theta + 2k \pi) + i sin (\theta + 2k \pi) = cos(\theta) + i sin (\theta) = e^{i\theta}$

Overall: $e^{i(\theta + 2k \pi)} = e^{i\theta}$

### Definition

f(z) = $e^x(cos y + i sin y)$ = $e^{(x + iy)}$

When y = 0, this is same as real exponential function

Properties $| e^z | = |e^x | |e^{iy} | = e^x$ (since $e^{i\theta}$ sits on the unit circle, thus has size = 1)

Also, $arg(e^z) = arg(e^{x+iy}) = arg(e^{x}.e^{iy}) = arg(e^{x}) + arg(e^{iy})$

from above, $e^x$ is the modulus or size of radius of $e^z$, then $(e^{x}.e^{iy})$ is in polar form $r.e^{i\theta}$ and hence y is argument of $e^{(x + iy)}$

Note: $arg(z_1z_2) = arg(z_1) + arg(z_2)$ (from earlier lecture – ref required) is not used in determining $arg(e^{x+iy})$ and do not confuse this with $arg({x+iy})$ $e^{z + 2{\pi}i} = e^{z}$ $e^{z + w} = e^{z}.e^{w}$ by setting z = x+iy and w= u + iv $\frac{1}{e^z} = e^{-z }$, since ${e^z}.e^{-z } = e^0 = 1$ $e^z = 1 \iff e^{x+iy} = e^x.e^{iy} = 1$ $\implies$ modulus (length) = 1 so x =0, and argument = $2k \pi$ $e^z = 1 \iff z = 2ik \pi$

Hence , when $e^z = e^w$ then $e^{z - w} = 1$ therefore z – w = $2ik \pi$

So z  = w + $2ik \pi$ results in $e^z = e^w$

<< to be done : mapping w = $e^z$ >>