Deragz

2022-01-16

What's the value of x when $\mathrm{sin}x=2$ and x is a complex number?

peterpan7117i

I like these. I think I did $\mathrm{sin}x=8$ in a previous answer.
${e}^{ix}=\mathrm{cos}x+i\mathrm{sin}x$
${e}^{ix}=\mathrm{cos}x-i\mathrm{sin}x$
${e}^{ix}-{e}^{-ix}=2i\mathrm{sin}x$
$\mathrm{sin}x=\frac{{e}^{ix}-{e}^{-ix}}{2i}$
Let $y={e}^{ix}$. Then ${e}^{-ix}=\frac{1}{y}$.
$2\left(2i\right)=y-\frac{1}{y}$
${y}^{2}-4iy-1=0$
$y=2i±\sqrt{-3}=i\left(2±\sqrt{3}\right)$
Convert to polar and get everything in the exponent. We note for integer k.
${e}^{ix}=y={e}^{\frac{i\pi }{2}}{e}^{\mathrm{ln}\left(2±\sqrt{3}\right)}{e}^{2\pi ki}$
$ix=i\frac{\pi }{2}+\mathrm{ln}\left(2±\sqrt{3}\right)+2\pi ki$
$x=\frac{\pi }{2}+2\pi k-i\mathrm{ln}\left(2±\sqrt{3}\right)$ for integer k
Check: Let’s check one, plus root, $k=0$.
$x=\frac{\pi }{2}-i\mathrm{ln}\left(2+\sqrt{3}\right)$
${e}^{ix}={e}^{\frac{i\pi }{2}}\left(2+\sqrt{3}\right)=i\left(2-\sqrt{3}\right)$
${e}^{-ix}=\frac{1}{i\left(2-\sqrt{3}\right)}=-i\left(2+\sqrt{3}\right)=i\left(-2-\sqrt{3}\right)$
${e}^{ix}-{e}^{-ix}=4i$
$\mathrm{sin}x=\frac{{e}^{ix}-{e}^{-ix}}{2i}=2$

Esther Phillips

$\mathrm{sin}x=2$
We use the fact that
$\mathrm{sin}x=\frac{{e}^{ix}-{e}^{-ix}}{2i}$
So our equation is
$\frac{{e}^{ix}-{e}^{-ix}}{2i}=2$
Now, set $z={e}^{ix}$. So we have
$\frac{z-\frac{1}{z}}{2i}=2$
which can be rearranged as
${z}^{2}-4iz-1=0$
Now, solve this
${\left(z-2i\right)}^{2}+4-1=0$
in other words
${\left(z-2i\right)}^{2}-{\left(i\sqrt{3}\right)}^{2}=0$
which is also
$\left(z-\left(2+\sqrt{3}\right)i\right)\left(z-\left(2-\sqrt{3}\right)i\right)=0$
So we have two solutions:

Now, let’s get back to our initial variable x.
${e}^{ix}=\left(2+\sqrt{3}\right)i$
implies that
$x=-i\mathrm{ln}i-i\mathrm{ln}\left(2+\sqrt{3}\right)+2k\pi ,\left(k\in Z\right)$
$=2k\pi +\frac{\pi }{2}-i\mathrm{ln}\left(2+\sqrt{3}\right)$
Likewise, we find that ${x}^{\prime }=2k\pi +\frac{\pi }{2}-i\mathrm{ln}\left(2-\sqrt{3}\right)$ are the other class of solutions.
Hence, the solutions are $\left(2k+\frac{1}{2}\right)\pi -i\mathrm{ln}\left(2±\sqrt{3}\right)$ where $k\in Z$.

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