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Find all solutions of $\sin (z) = 2$.

Here are the things I did:

1) By definition: $\sin z =\dfrac{e^{iz} − e^{−iz}}{2i}= 2$.
Multiply $2i$ to the equation and make it quadratic:
$e^{2iz} -4ie^{iz}-1=0$

2) Solve the quadratic:
$e^{iz} = 2i±\sqrt{3}i$
$= (2±\sqrt{3})i$
$= (2±\sqrt{3})e^{\frac{1}{2}\pi i}$

$\therefore$ $e^{iz} = (2±\sqrt{3})e^{\frac{1}{2}\pi i}$

3) I read the solution book, it says:
$iz = \ln(2±\sqrt{3}) + 2k\pi +\dfrac{1}{2}\pi i, k \in \mathbb{Z}$.

Why would $iz = \ln(2±\sqrt{3})+...$ but not $iz = \ln((2±\sqrt{3})i)$? where did the $i$ go
Where did $2k\pi +\dfrac{1}{2}\pi i$ come from?

Thank you.

Apurv
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samheihey
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1 Answers1

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So $e^z = e^{a+ib} = e^a e^{ib}$. But $e^{ib_1} = e^{ib_2}$ exactly when $b_2-b_1 = 2k\pi$ where $k \in \mathbb Z$ (here i choose $a>0$ - as I can always do).

So lets assume we wanted to create an inverse to the exponential called $\ln $, but what value should we assign to $\ln(w)$? It should give the solutions the the equation $e^{z} =w = |w|e^{i\arg(w)}$. Thus using the above we have that $e^a = |w|$ and that $b= \arg(w)+2\pi k$ where $k \in \mathbb Z$. Hence we define $\ln$ by: $$\ln(z) = \ln(|z|) +i \arg(z)+2\pi k i, \quad k \in \mathbb Z $$

In your case you have that: $e^{iz} = (2 \pm \sqrt{3})e^{\frac{1}{2}\pi i}$ thus using the above definition you would get: $$iz = \ln(2\pm \sqrt{3})+i \frac{1}{2}\pi +2\pi k i$$

Martin
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