Logs of a complex number

Question

Write a solution in Cartesian for of

What should come next?

Answer 1

r is simply just the modulus of the number on the right side. Arg(z) on the other hand is the angle from the positive real axis to the point on the complex plane, in this case -1. So, looking from the positive real axis, we now get Arg(z) = $\pi$. Also, your formulation is off. $e^w = z$ has an inverse $\ln ( \vert z \vert ) + i(Arg(z) + 2 \pi n)$, and you forgot the $i$. Hopefully that clears some stuff up.

If you want the answer to the problem, then $$ e^w = z \Rightarrow w = \ln(\vert z \vert) + i(Arg(z)+2\pi n)$$ Now since $e^w = -1$, then we know that $\ln( \vert z \vert)$ = $\ln (1)$ which is simply zero, so we look at the imaginary part. Arg(z) = $\pi$ because the angle measured from the positive real axis to -1 is $\pi$, so now we have $$w = 0 + i(\pi + 2 \pi n),$$ which now gives us $$w = 0 + i((2n+1)\pi).$$ QED

Answer 2

$$z=e^w=e^{a+ib}=e^ae^{ib}$$

Since $|z|=1$ we have $e^a=1$, i.e. $a=0$. And since $\arg\{z\}=(2k+1)\pi$ (because $z=-1$) you get $b=(2k+1)\pi$, and consequently

$$w=ib=i(2k+1)\pi,\quad k\in\mathbb{Z}$$