 widdonod1t

2022-01-16

How many solutions does $\mathrm{sin}\left(z\right)=10$ have in the complex plane? Kirsten Davis

Expert

- periodic, which means if you have a solution $z\in C$ such as $\mathrm{sin}z=10$, for any $k\in Z,\mathrm{sin}\left(z+2k\pi \right)=10$
Also, for any $x\in R$
$\mathrm{sin}\left(ix+\frac{\pi }{2}\right)=\frac{{e}^{i\left(ix+\frac{\pi }{2}\right)}-{e}^{-i\left(ix+\frac{\pi }{2}\right)}}{2i}$
$=\frac{{e}^{-x}{e}^{i\frac{\pi }{2}}-{e}^{x}{e}^{i\frac{\pi }{2}}}{2i}$
$=\frac{i{e}^{-x}+i{e}^{x}}{2i}$

We know that the range of , which implies there is at least one possible solution $x\in R$ such as  and at least one $z=ix+\frac{\pi }{2}\in C$ such as $\mathrm{sin}z=10$
As a result, the equation has an infinite number of solutions $\mathrm{sin}z=10\in C$. Janet Young

Expert

Just like with $\mathrm{sin}x=\frac{1}{2}$ there will be two supplementary angles that are solutions, that is, a pair of solutions that adds up to $\pi$. Then to each of those we can add $2\pi k$ for integer k to handle the periodicity. That’s a countably infinite number of solutions.
These aren’t too hard to solve exactly. I’ll solve
The expression for sine from Euler’s Formula is:
$\mathrm{sin}z=\frac{1}{2i}\left({e}^{iz}-{e}^{-iz}\right)$
Let's let $x={e}^{iz}$, then this becomes
$a=\frac{1}{2i}\left(x-\frac{1}{x}\right)$
${x}^{2}-2aix-1=0$
$x=ai±\sqrt{-{a}^{2}+1}$
Since $|a|>1$ the radicand is negative, so x is purely imaginary.
$x=ai±\sqrt{-\left({a}^{2}-1\right)}=i\left(a±\sqrt{{a}^{2}-1}\right)={e}^{i\frac{\pi }{2}}\left(a±\sqrt{{a}^{2}-1}\right)$
$\mathrm{ln}x=iz$
$z=-i\mathrm{ln}x=-i\left(i\frac{\pi }{2}+2\pi ki+\mathrm{ln}\left(a±\sqrt{{a}^{2}-1}\right)\right)$
$z=\left(\frac{1}{2}+2k\right)\pi +i\mathrm{ln}\left(a±\sqrt{{a}^{2}-1}\right)$
The cool thing about this is that the two expressions $\mathrm{ln}\left(a±\sqrt{{a}^{2}-1}\right)$ are negations of each other, as the arguments to the log are reciprocals. (I’ll leave that to you to check. It follows from the product of the x is being -1, because of the -1 in the quadratic equation.) So the z s come in conjugate pairs.
$z=\left(\frac{1}{2}+2k\right)\pi ±i\mathrm{ln}\left(a+\sqrt{{a}^{2}-1}\right)$
Considering $k=0$ for a moment, the real part of both conjugates is $\frac{\pi }{2}$, so when they’re summed they add to $\pi$ as the imaginary parts cancel. They’re supplementary angles.
Here we have $a=10$, so
$\mathrm{arcsin}10=\left(\frac{1}{2}+2k\right)\pi ±i\mathrm{ln}\left(10+\sqrt{99}\right)$
So there are countably infinite solutions to $\mathrm{sin}z=10$, a conjugate pair of two solutions for each integer k.

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