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How to solve $\sin(z) = 2$,where $z$ is a complex number.

Another approach I tried:

$$\sin z=\frac{1}{2i}\left(e^{iz}-e^{-iz}\right)=2$$

$$e^{iz}-e^{-iz}=4i$$

Let $x=e^{iz}$ so that $x^2-4ix-1=0$. Then, I would use the quadratic formula to solve would this be correct?

Quanto
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robotgirl21
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  • It's a good start. Now note that $a$ is real, so you must have $\cos(x)\sinh(y) = 0$. Which options for $x$ and $y$ does that give you? – Daniel Fischer Dec 12 '20 at 15:17
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    @robotgirl21 Please do not change radically your question like this, after edition answers do not correspond. If you need to show two approaches then just append to the existing text. – zwim Dec 12 '20 at 17:18

3 Answers3

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Continuing your answer, we have now $\sin x\cosh y=a$ and $\cos x\sinh y=0.$ From the second equation $\cos x=0$ or $\sinh y=0.$ If $\cos x=0$ then $x=\pi/2(2n+1)$. But the first equation then gives that $\cosh y=a$ which is not possible because $\cosh y\geq1.$ Hence $\sinh y=0$ and therefore $y=0$.

Chetan Vuppulury
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user159888
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An alternative: write $w:=\exp(iz)$ so $w^2-2iaw-1=0$, with roots $w=ia\pm\sqrt{1-a^2}$. Since the surd is real, $|w|^2=a^2+1-a^2=1$, so $z\in\Bbb R$.

J.G.
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Note $$2=\sin z= \cos(2\pi k +\frac\pi2 -z) = \cosh[i(2\pi k +\frac\pi2 -z)] $$ which leads to $i(2\pi k +\frac\pi2 -z)=\pm \cosh^{-1} 2$ and the solutions $$z= 2\pi k +\frac\pi2\pm i \cosh^{-1} 2$$

Quanto
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