widdonod1t

Answered

2022-01-16

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

Answer & Explanation

Kirsten Davis

Expert

2022-01-17Added 27 answers

$\mathrm{sin}\text{is}\text{}2\pi$- periodic, which means if you have a solution $z\in C$ such as $\mathrm{sin}z=10$, for any $k\in Z,\mathrm{sin}(z+2k\pi )=10$.

Also, for any $x\in R$:

$\mathrm{sin}(ix+\frac{\pi}{2})=\frac{{e}^{i(ix+\frac{\pi}{2})}-{e}^{-i(ix+\frac{\pi}{2})}}{2i}$

$=\frac{{e}^{-x}{e}^{i\frac{\pi}{2}}-{e}^{x}{e}^{i\frac{\pi}{2}}}{2i}$

$=\frac{i{e}^{-x}+i{e}^{x}}{2i}$

$=\frac{{e}^{x}+{e}^{-x}}{2}=\mathrm{cos}h\text{}x$

We know that the range of $\mathrm{cos}h\text{}\text{is}\text{}[1,+\mathrm{\infty})$, which implies there is at least one possible solution $x\in R$ such as $\mathrm{cos}h\text{}x=10$ 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

2022-01-18Added 32 answers

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$\mathrm{sin}z=a\text{}\text{for real}\text{}a,\left|a\right|1$

The expression for sine from Euler’s Formula is:

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

Let's let$x={e}^{iz}$ , then this becomes

$a=\frac{1}{2i}(x-\frac{1}{x})$

${x}^{2}-2aix-1=0$

$x=ai\pm \sqrt{-{a}^{2}+1}$

Since$\left|a\right|>1$ the radicand is negative, so x is purely imaginary.

$x=ai\pm \sqrt{-({a}^{2}-1)}=i(a\pm \sqrt{{a}^{2}-1})={e}^{i\frac{\pi}{2}}(a\pm \sqrt{{a}^{2}-1})$

$\mathrm{ln}x=iz$

$z=-i\mathrm{ln}x=-i(i\frac{\pi}{2}+2\pi ki+\mathrm{ln}(a\pm \sqrt{{a}^{2}-1}))$

$z=(\frac{1}{2}+2k)\pi +i\mathrm{ln}(a\pm \sqrt{{a}^{2}-1})$

The cool thing about this is that the two expressions$\mathrm{ln}(a\pm \sqrt{{a}^{2}-1})$ 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.

$z=(\frac{1}{2}+2k)\pi \pm i\mathrm{ln}(a+\sqrt{{a}^{2}-1})$

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

$\mathrm{arcsin}10=(\frac{1}{2}+2k)\pi \pm i\mathrm{ln}(10+\sqrt{99})$

So there are countably infinite solutions to$\mathrm{sin}z=10$ , a conjugate pair of two solutions for each integer k.

These aren’t too hard to solve exactly. I’ll solve

The expression for sine from Euler’s Formula is:

Let's let

Since

The cool thing about this is that the two expressions

Considering

Here we have

So there are countably infinite solutions to

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