Let $z=x+iy$ be a complex number.

I want to find when the solutions of the equation $$\sin z =u,u\in\mathbb{R}$$ are real.

I used this approach:

Since $\sin z=\sin x\cosh y+i \cos x\sinh y=u$, then $$\sin x\cosh y=u,\cos x \sinh y=0.$$ From the second equation we have $\cos x=0$ or $y=0$. If $y=0$, the solutions are real, (since $\cosh 0=1$ the solutions would be the solutions of $\sin x=u$). If $\cos x=0$ then $x=\frac{\pi}{2}+n\pi,n\in\mathbb{Z}$ and $\sin x=(-1)^n$.

So, if $-1<u<1$, the case $\cos x=0, y\neq 0$ is impossible since $|\pm\cosh y |\geq 1$ for all $y$, so in this case all the solutions are real.

For the case $|u|=1$ we get $\pm\cosh(y)=u$ and restricting $x$ appropriately we get $y=0$ too.

In the case that $u>1$ or $u<-1$ then it's clear that $z\notin\mathbb{R}$ so the case $y=0$ is impossible. But since $\cosh y$ doesn't change it's sign, if $u>1$ we must restrict $x=\frac{\pi}{2}+2k\pi$ and we get 2 solutions $y_1=-y_2$ of $\cosh y=u$. So the solutions are given by $y=\pm y_1$ and $x=\frac{\pi}{2}+2k\pi$.

In the case $u<-1$ it's similar, the solutions are $y=\pm y_1',x=-\frac{\pi}{2}+2k\pi$.

I might use another approach:

$$z=\arcsin u=-i\ln(iu\pm\sqrt{1-u^2})$$ If we consider $|u|<1$, then $\sqrt{1-u^2}$ is real, so $$|\pm\sqrt{1-u^2}+iu|=1$$ and $$-i\ln(iu\pm\sqrt{1-u^2})=\theta_u+2k\pi$$ where $\theta_u$ satisfies $\cos \theta_u =\pm\sqrt{1-u^2}, \sin \theta_u=u$. So the solutions are real.

In the case $|u|>1$, $\sqrt{1-u^2}=i\sqrt{u^2-1}$ so in this case $$|i\pm\sqrt{u^2-1}+iu|=|\pm\sqrt{u^2-1}+u|$$ and $$-i\ln(iu\pm\sqrt{1-u^2})=-i\ln|\pm\sqrt{u^2-1}+u|+2k\pi$$ in this case the solutions are not real, since $u^2-1\neq u^2$ for all $u$.

Is there any easier, shorter, funnier, or even a strange approach to the problem (It doesn't have to be shorter or easier, it may be a large and strange approach, or a geometric one, or something like that)?

Note that we're trying to find out when all the zeroes of $\sin z-u=0$ are real.

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    $\begingroup$ if the equation has a real solution then u must be between minus one to one.conversely, for any real number bwtee minus one to one, we must have a solution. $\endgroup$ – Ben Dec 1 '16 at 1:36
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    $\begingroup$ Yes, that's true. But how would you prove that for any real number between minus one to one, the only possible solutions are real? $\endgroup$ – David Molano Dec 1 '16 at 1:41
  • $\begingroup$ oh right. you have a point here. $\endgroup$ – Ben Dec 1 '16 at 1:48

Writing $$\sin z = \frac{e^{iz}-e^{-iz}}{2i} = -\frac{i}{2}\left(e^{-y}(\cos x + i\sin x)-e^y (\cos x -i\sin x)\right)\\ =\sin x + \frac{i}{2} \cos x \left( e^y-e^{-y} \right) $$ For the imaginary part of $\sin z$ to be zero, either:

  • $y=0$ so that $e^{y}-e^{-y}=0$, in which case $z$ is real and $\sin z \in [-1,1]$.

  • or $\cos x = 0$ in which case $z = \frac{2n+1}{2}\pi+i y$, and $\sin z = 0]$.

  • $\begingroup$ I got lost in the last part you say $\sin z=0$. In your formula for $\sin z$, when $\cos x=0$, $\sin z=\sin x=\pm 1$. And to say the truth, when I make the calculations i don't get that formula, but \begin{align*}-\frac{i}{2}\left( e^{-y}(\cos x+i\sin x)-e^{y}(\cos x-i\sin x)\right) &=\frac{e^{-y}+e^{y}}{2}\sin x+i\frac{e^{y}-e^{-y}}{2}\cos x\\&=\cosh y\sin x+i\sinh y \cos x\end{align*} $\endgroup$ – David Molano Dec 1 '16 at 2:06

Since you assume that $u$ is real, there is an alternative method: Use Pythagorean Theorem we get $\cos z=\pm\sqrt{1-u^2}$=$\pm{i}\sqrt{u^2-1}$. Now use $e^{iz}=\cos z+i\sin z$, take the $\ln$ on both sides and you are done. (assuming you know how to take the complex $\ln$). A numerical problem is worked out here, it might be helpful: Finding $z\in \mathbb C$ with $\sin z = 2$?

  • $\begingroup$ If we use $\cos z=\pm\sqrt{1-\sin^2z}$ instead $\pm\sqrt{1-\sin^2z}$, yes, that's a way. But in the end, it's a (an interesting one) way to derive the formula for the complex $\arcsin$, which I use in my second approach. $\endgroup$ – David Molano Dec 1 '16 at 2:35
  • $\begingroup$ Yes. My approach is quick if $u$ is real. In case you want to solve $sinz=a+bi$ as a general case, then you need to resort to stuff like you have posted, and the answer Mark has given to you. A general formula isn't "compact" in the sense that it is easy to remember, unfortunately... $\endgroup$ – imranfat Dec 1 '16 at 2:43

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