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I am trying to solve a question which asks to find all the solutions of $e^{iz}=1+i$. Here is what I have done:

$z = x + yi$
$e^{i(x+yi)} = 1+i$
$e^{xi-y}=1+i$
$e^{xi-y}=\sqrt2e^{(\frac\pi4 +2n\pi)i}$
$e^{-y}=\sqrt2$
$y=-\frac12ln2$

$x=\frac\pi4 +2n\pi$

$Z = i(x+yi)$
$Z = i((\frac\pi4 +2n\pi)+ (-\frac12ln2)i)$
$Z = \frac12ln2 + (\frac\pi4 +2n\pi)i$

The correct answer according to the marking scheme is:
$Z = (\frac\pi4 +2n\pi) - i\frac\pi4$

This differs slightly from my answer. Could someone please explain where I have gone wrong in my working?

Did
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abruzzi26
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    Probably two mistakes: 1. Why is your $z=x+iy$ becoming $Z=i(x+iy)$? 2. Did you copy faithfully the imaginary part of the solution $z$ proposed in your solutions sheet? – Did Oct 03 '15 at 11:25
  • oops! I copied out the wrong answer. Sorry, my bad. It was silly of me to add the extra i. Thanks for responding :) – abruzzi26 Oct 03 '15 at 14:22

1 Answers1

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Your answer is correct except for the extra $i$ you wrote. The answer is $$ z = x + i y = (\pi/4 + 2n \pi) + i ((-1/2) \ln 2).$$ You can check this answer is correct by plugging it in: Since $$iz = (1/2) \ln 2 + i (\pi/4 + 2n \pi),$$ we do have $$ e^{iz} = e^{(1/2)\ln 2} e^{i(\pi/4 + 2n \pi)} = \sqrt{2} e^{i(\pi/4 + 2n \pi)} = 1 + i$$ The answer $z = (\pi/4 + 2 n \pi) - i \pi/4$ in the marking scheme is wrong, because you can check to see you get $$iz = \pi/4 + i (\pi/4 + 2n \pi),$$ so $$ e^{iz} = e^{\pi/4} e^{i(\pi/4 + 2n \pi)} = e^{\pi/4} (1 + i)/\sqrt{2},$$ which does NOT equal $1 + i$.

Curiosity
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  • oops! I kind of copied out the wrong answer. The correct answer is indeed the correct one. It was silly of me to put the extra i. Thanks for the response :) – abruzzi26 Oct 03 '15 at 14:21
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    I just came across other similar questions, but with some modifications using the absolute notation. Example, $e^z=|1+i|$, $|e^z=|1+i|$, $e^{|z|}=|1+i|$ and $e^{i|z|}=|1+i|$. I am completely lost here - I do not understand how to approach these kinds of problems. Could you please guide me on how to solve these? I would really really be grateful. Thanks – abruzzi26 Oct 03 '15 at 14:32