Complex but Simple equation.

Question

Im( e^z ) = 1

I've reached rsin(t) = 1 but I am stuck here, where r is radius and t is argument.

There may be use of logs here I might be missing.

Answer 1

Assuming $z=a+bi$ with $a,b\in\mathbb{R}$:

$$\Im\left(e^{a+bi}\right)=1\Longleftrightarrow$$ $$-\Re\left(ie^{a+bi}\right)=1\Longleftrightarrow$$ $$i\left(-e^{a+bi}+\Re\left(e^{a+bi}\right)\right)=1\Longleftrightarrow$$ $$-ie^{a+bi}+i\Re\left(e^{a+bi}\right)=1\Longleftrightarrow$$ $$-ie^{a+bi}+ie^a\cos(b)=1\Longleftrightarrow$$ $$e^a\sin(b)=1$$

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Solving the last part we get:

$$a=\ln\left(\csc(b)\right)+2i\pi n$$ $$b=\pi-\arcsin\left(\frac{1}{e^a}\right)+2\pi n\space\space\vee\space\space b=\arcsin\left(\frac{1}{e^a}\right)+2\pi n$$

With $n\in\mathbb{Z}$

Answer 2

$$Im(e^{x+yi})=e^x\sin y=1\\ x=-\ln (\sin y)\\ z=-\ln (\sin t)+ti$$