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There are a number of solutions to this problem online that use identities I have not been taught. Here is where I am in relation to my own coursework:

$ \sin(z) = 2 $

$ \exp(iz) - \exp(-iz) = 4i $

$ \exp(2iz) - 1 = 4i \cdot \exp (iz) $

Then, setting $w = \exp(iz),$ I get:

$ w^2 - 4iw -1 = 0$

I can then use the quadratic equation to find:

$ w = i(2 \pm \sqrt 3 )$

So therefore,

$\exp(iz) = w = i(2 \pm \sqrt 3 ) $ implies

$ e^{-y}\cos(x) = 0 $, thus $ x = \frac{\pi}{2} $ $ ie^{-y}\sin(x) = i(2 \pm \sqrt 3 ) $ so $ y = -\ln( 2 \pm \sqrt 3 ) $

So I have come up with $ z = \frac{\pi}{2} - i \ln( 2 \pm \sqrt 3 )$

But the back of the book has $ z = \frac{\pi}{2} \pm i \ln( 2 + \sqrt 3 ) +2n\pi$

Now, the $+2n\pi$ I understand because sin is periodic, but how did the plus/minus come out of the natural log? There is no identity for $\ln(a+b)$ that I am aware of. I believe I screwed up something in the calculations, but for the life of me cannot figure out what. If someone could point me in the right direction, I would appreciate it.

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    $\begingroup$ Please use \exp, \ln, \sin etc. You'll see what a nice difference that makes. $\endgroup$ – Namaste Feb 26 '15 at 15:33
  • $\begingroup$ Why does $e^{-y}\cos x=0$ implies $x={\pi\over2}+2\pi k$ and not $x={\pi\over 2}+\pi k$? @Tom $\endgroup$ – J. Doe Mar 12 at 15:30
  • $\begingroup$ @J.Doe If you're referring to Tom K's comment under Kelvin Soh's answer, the answer to your question is yes, but this cannot be post as a decent answer since it's too short. $\endgroup$ – Saad Mar 13 at 1:03
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Your answer is actually identical to the book's solution (after accounting for $2n\pi$).

The key identity: $$ \frac{1}{2+\sqrt{3}} = 2 - \sqrt{3}.$$

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    $\begingroup$ Tom: it's not really an identity, as in knowing this exact equation. The equation can be obtained by multiplying the numerator and denominator of the left hand side by $$\frac{2-\sqrt 3}{2-\sqrt 3}$$ $$\frac 1{2+\sqrt 3}\cdot\frac {2-\sqrt 3}{2-\sqrt 3} = \frac {2-\sqrt 3}{4-3} = 2-\sqrt 3$$ It's simply rationalizing the denominator. $\endgroup$ – Namaste Feb 26 '15 at 15:41
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    $\begingroup$ So, $ (2 - \sqrt3) = (2+\sqrt3)^{-1} $ and this allows me to say for that second case $ \ln(2-\sqrt3) = -\ln(2+\sqrt3) $. Then I can combine these two cases for the imaginary part as $ \pm i\ln(2+\sqrt3) $ , correct? $\endgroup$ – Tom K Feb 26 '15 at 15:48
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    $\begingroup$ @am@Why : It is indeed an identity. Your argument showing how to prove it by rationalizing the denominator demonstrates that it is an identity. What does the word "identity" mean to you if this is not an identity. $\endgroup$ – Michael Hardy Feb 26 '15 at 15:52
  • $\begingroup$ I'd have demonstrated it thus: $(2+\sqrt3)(2-\sqrt3)$ $=(a+b)(a-b)$ $=a^2-b^2$ $= 4 -3=1$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Feb 26 '15 at 15:54
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    $\begingroup$ @Michael Why being pedantic? I meant no harm. My comment was aimed at showing Tom how he can use rationalization of the denominator in any situation in which it is applicable, that it wasn't an "identity" specific to this particular occurrence. Please put your efforts into being constructive and not into being argumentative. $\endgroup$ – Namaste Feb 26 '15 at 15:56
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Setting $w=e^{iz},$ we need to solve the equation $w^2-4iw-1=0.$ The solutions to this quadratic equation are $w=i(2+\sqrt 3)$ and $w=i(2-\sqrt 3).$

Let's deal with the first solution. We need to find $z=x+iy$ such that $e^{iz}= e^{ix}e^{-y}= i(2+\sqrt 3).$ This implies $\cos x =0.$ As you point out, that has solution set $\pi/2 + n\pi, n\in \mathbb Z.$ But there is another implication: In order to get $2+\sqrt 3$ as the imaginary part, we have to delete all $\pi/2 + n\pi$ for $n$ odd, as they lead to negative imaginary values. This is why we end up with $\pi/2 + 2n\pi.$

At this point, let's say goodbye to the original post for inspiration, as things are a little cloudy there. The easiest way to do this is write $e^{ix}e^{-y}=i(2+\sqrt 3)= e^{i\pi/2}(2+\sqrt 3).$ This tells us that $x= \pi/2 +2n\pi,$ and $-y= \ln(2+\sqrt 3).$

Solving for $z$ in the case $w=i(2-\sqrt 3)$ is the same. So in all, the solutions to the original problem are $z = (\pi/2 +2n\pi) -i\ln(2\pm\sqrt 3),n\in \mathbb Z.$

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Let us solve $$e^{-y+ix}=i(2\pm\sqrt3),\quad (x,y)\in \mathbb R^2.$$ One can get \begin{cases} e^{-y}\cos x = 0\\ e^{-y}\sin x = 2\pm\sqrt3. \end{cases} It seems that $\sin x=\pm1.$ What's happened? $$\cos x = 0\rightarrow x=k\pi+\frac\pi2 = 2n\pi\color{red}\pm\frac\pi2.$$ $$(e^{-y}\sin x = 2\pm\sqrt3 >0) \wedge (e^{-y} > 0)\rightarrow \sin x >0 \rightarrow x = 2\pi\color{red}+\frac\pi2,$$ $$e^y = \color{red}+(2\pm\sqrt3),$$ $$y= \ln(2\pm\sqrt3) =\pm \ln(2+\sqrt3)=\pm \ln(2-\sqrt3),$$ $$\boxed{x+iy = \frac\pi2+2\pi n \pm \ln(2+\sqrt3)}$$ or $$\boxed{x+iy = \frac\pi2+2\pi n \pm \ln(2-\sqrt3)}.$$

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