Find $e^{z} = 1+ \sqrt{3} i$ for the exponential function in complex analysis

Question

I am working through a book on complex analysis. Now they give an example when introducing the exponential function.

We write $$e^{z} = e^{x} e^{iy},$$ where $x$ is the real part and $y$ the imaginary part of $z$.

Now I need to find the numbers $x+iy$ such that $$ e^{z} = 1 + \sqrt{3}i,$$ and they start by rewriting this into $$e^{x}e^{iy} = 2 e^{i \pi /3}.$$

Now I get the left part of the rewriting but how do they arrrive at $1+\sqrt{3}i= 2e^{i \pi /3}$ ?

I get the rest of the example, hence there is no need for finding the actual $x+iy$ for which the equation holds.

Thanks.

Answer 1

$e^{i\frac {\pi} 3}=\cos (\frac {\pi} 3)+i\sin \frac {\pi} 3= \frac 1 2+i\frac {\sqrt 3} 2$

Answer 2

I believe you've seen Euler's equation somewhere in that book.

Long story short, it says that all outputs of $f(x)=e^{ix}$ lie on the unit circle in the complex plane.

Calculating the polar co-ordinates for $1+\sqrt{3}$ gives $(2, \dfrac{\pi}{3}^c)$

And scaling it down by 2 places it on the unit circle. Hence by Euler's equation it is an output from that very angle i.e. $e^{i\dfrac{\pi}{3}}$, twice that is the original number which is $2e^{i\frac{\pi}{3}}$

But to be complete I think $z=\ln{2}+\dfrac{\pi}{3}i$