Find z for ez=1+i√3: Solution Explained

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Homework Statement


Find all values of z such that ez=1+i√3

Homework Equations


The Attempt at a Solution


I have no idea how to do this. I was going to start with ez=exeiy and try to figure something out from that, but I'm not seeing anything. I checked the solution shown below, but I'm really confused as to how they went from the equation given to the first step. An explanation would be really helpful.
 

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alexcc17 said:

Homework Statement


Find all values of z such that ez=1+i√3


Homework Equations





The Attempt at a Solution


I have no idea how to do this. I was going to start with ez=exeiy and try to figure something out from that, but I'm not seeing anything. I checked the solution shown below, but I'm really confused as to how they went from the equation given to the first step. An explanation would be really helpful.

If you write ##1 + i \sqrt{3} = r e^{i \theta}## (where you can easily figure out ##r## and ##\theta##) then the equation is ##e^{z} \equiv e^x e^{iy} = r e^{i \theta}##.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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