Looking for fully evaluated solution to: $\sin(z) = 3$

Question

I'm looking for someone to verify my solution to $\sin(z) = 3$. I can't seem to find a solution anywhere else.

My result is: $z = -i(\ln(i)+\ln(3±\sqrt{10})$

$=\pi /2 - i(\ln(3±\sqrt{10})$

Thanks in advance.

Answer 1

First, substitute $\frac{e^{iz}-e^{-iz}}{2i}$ for $\sin(z)$:

$\frac{e^{iz}-e^{-iz}}{2i}=3$

Then multiply by $2i$:

$e^{iz}-e^{-iz}=6i$

Multiply by $e^{iz}$:

$(e^{iz})^2-1=6ie^{iz}$

Set equal to zero and solve using quadratic formula:

$(e^{iz})^2-6ie^{iz}-1=0 \\ \frac{6i\pm\sqrt{-36-4(1)(-1)}}{2}=\frac{6i\pm4i\sqrt{2}}{2}=3i\pm2i\sqrt{2} \\ e^{iz}=i(3\pm2\sqrt2)$

$\ln$ on both sides:

$iz=\ln(i(3\pm2\sqrt2)=\ln(i)+\ln(3\pm2\sqrt2)$

By the formula $e^{iz}=\cos(z)+i\sin(z)$, $e^\frac{i\pi}{2}=i$.

$\ln(i)=\ln(e^{\frac{i\pi}{2}})=\frac{i\pi}{2}$

Therefore,

$z=\frac{\frac{i\pi}{2}+\ln(3\pm2\sqrt{2})}{i}=-i(\frac{i\pi}{2}+\ln(3\pm2\sqrt{2}))=\frac{\pi}{2}-i\ln(3\pm2\sqrt 2)$

Your answer was close, the only mistake is the $\sqrt{10}$.