# Given $\sin z=5$. Find $e^{iz}$ (Complex Trigonometric Function)

Given $sin$ $z=5$. Find $e^{iz}$.

Here is what I have done:

\begin{align} \sin z &= \frac{e^{iz}-e^{-iz}}{2i}=\frac{e^{i(x+iy)}-e^{-i(x+iy)}}{2i}\\ &=\frac{e^{-y}(\cos x+i\sin x)-e^{y}(\cos x-i\sin x)}{2i}\\ &=\frac{e^{-y}\sin x+e^{y}\sin x}{2}+\frac{i(e^{y}\cos x-e^{-y}\cos x)}{2} \end{align} So if $\sin z = 5 = 5+0i$ then by equating the real and imaginary parts, $$\frac{e^{y}\sin x+e^{-y}\sin x}{2}=5\\ \frac{e^{y}\cos x-e^{-y}\cos x}{2}=0$$ Let us have the real part. $$\frac{e^{y}\sin x+e^{-y}\sin x}{2} = \frac{\sin x(e^{y}+e^{-y})}{2}=5$$ From this we know that $y\neq 0$ because if $y=0$ then $\sin x=5$ which is not possible.

Let us have the imaginary part, $$\frac{e^{y}\cos x-e^{-y}\cos x}{2}=\frac{\cos x(e^{y}-e^{-y})}{2}=0$$ Since we know that $y\neq 0$ then $(e^{y}-e^{-y})\neq 0$ thus $\cos x=0 \implies x=\frac{\pi}{2}+n\pi$ where $n$ is an integer.

Going back to the real part: $$\frac{\sin x(e^{y}+e^{-y})}{2}=5 \\ \sin x(e^{y}+e^{-y})=10$$ We know now that $x=\frac{\pi}{2}+n\pi$ thus, $$(-1)^{n}(e^{y}+e^{-y})=10$$ where $n$ is an integer.

We know that since $10$ is positive and $(e^{y}+e^{-y})$ will not be negative then we must have $(-1)^{n}$ as positive also. It implies $n$ is even.

We have $$(e^{y}+e^{-y})=10\implies e^{y}+e^{-y}-10=0$$ Multiplying both sides by $e^{y}$ we get: $$e^{2y}-10e^{y}+1=0$$ and by quadratic formula we will get: $$e^{y}=5 \pm 2\sqrt{6}\implies y=\ln(5 \pm 2\sqrt{6})$$ So we now have $x=\frac{\pi}{2}+n\pi$ where $n$ is an even integer and $y=\ln(5 \pm 2\sqrt{6})$. Therefore, \begin{align} e^{iz} &= e^{i(x+iy)} \\ &= e^{i(\frac{\pi}{2}+n)-\ln(5 \pm 2\sqrt{6})}\\ &=-(5 \pm 2\sqrt{6})[\cos(\frac{\pi}{2}+n\pi)+i(\sin(\frac{\pi}{2}+n\pi)] \\ &=-(5 \pm 2\sqrt{6})i \end{align} since $n$ is even.

Is this correct or did I miss anything? The solution is quite long and I might have overlooked something. Is there perhaps a shorter way of doing this?

I would like to hear from you guys. Thank you! :)

UPDATE

Thank you for the answer guys. :) I'm just wondering where in my reasoning above is wrong.

The answer is supposed to be $=(5 \pm 2\sqrt{6})i$ but I got mine as negative.

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These are all great solutions. Kaster's is really nice, because of the pythagorean theorem. But in the general situation sinz = a+bi where both a,b are nonzero, I have not been able to use the pythagorean theorem to make it work. Mrf's method is the more general method. Also, why don't we care about multiples of these angles? (2pi) – imranfat Sep 9 '13 at 16:55
@imranfat, just in case, mrf's solution is not more general but completely equivalent. – Kaster Sep 9 '13 at 17:02
Yes, but I never understood how to use the pythagorean theorem conveniently in case of sinz = a + bi when both a and b are nonzero. I tried just one and it becomes a disaster...Working with complex e-powers goes straight forward... – imranfat Sep 9 '13 at 18:10

Shorter solution: Put $w=e^{iz}$ and note that $1/w=e^{-iz}$. From the given information, $$5=\sin z = \frac{w-\frac1w}{2i}$$ or $$10iw = w^2-1$$ which is a quadratic equation in $w$. Solve it.

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$$\sin z = 5 \implies \cos z = \pm \sqrt{24}i = \pm 2 \sqrt 6 i \\ e^{iz} = \cos z + i \sin z = \pm 2 \sqrt 6 i + 5i = (5 \pm 2 \sqrt 6)i$$

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+1. Nice solution. – Did Sep 9 '13 at 16:46
So the answer should be $(5\pm 2 \sqrt{6})i$. I wonder where I did wrong. thank you for the answer though. – chowching Sep 9 '13 at 23:55
How did you arrive at $cos z = \pm\sqrt{24}i$? Thanks! – chowching Sep 10 '13 at 0:08
@chowching from the fact that $\cos^2 z + \sin^2 z = 1$. – Kaster Sep 10 '13 at 0:23
@chowching you also have too many mistakes/typos in your text. I tried to correct them, but after each correction some new typos showed up. Please, correct those, you probably did some minor typo kind of mistake which led to the sign mismatch. – Kaster Sep 10 '13 at 0:28

Another way :

$$5=\sin(x+iy)=\sin x\cos(iy)+\cos x\sin(iy)=\sin x\cosh y+i\cos x\sinh y$$

Comparing the imaginary parts, $\cos x\sinh y =0$

Comparing the real parts, $\sin x\cosh y=5\implies \sin x>0$(why?)

Case $1:$ If $\cos x=0,\sin x=1\implies x=2n\pi+\frac\pi2$

$\implies \cosh y=5,$ hope you can solve it

Case $2:$ $\sinh y=0\implies \cosh y=\sqrt{1+\sinh^2y}=1$

$\sin x=5$ which is impossible, you know why

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