So the problem is : let $R$ be region $[{z\in \mathbb{C} : -\cfrac{\pi}{2}<Re(z)<\cfrac{\pi}{2}}]$. Show that sine function is injective in R.

I could think about two starting points:

i) we have $\sin(z) = \cfrac{e^{iz}-e^{-iz}}{2i}$. So if, say for $z_1,z_2 \in R$, $\sin(z_1)=\sin(z_2)$, then we have


but I have no idea how to continue from this. Since region $R$ mentions real part of the complex number, I will need to refer to it at some point but I dont seem to find how. I tried to put $z_1$ and $z_2$ into the form $x+iy$ but it only seemed to complicate things.

ii) So after above attempt, maybe I thought it was better to consider imaginary part and real part separately:

$\sin(x+iy)=\sin(x) \cosh(y)+i\cos(x) \sinh(y)$

so I proceeded in exactly same way, but in the end I had to relate two equations arising from equating real/imaginary parts somehow, but that actually made attept too complicated.

I have a feeling that it shouldn't be this hard... any hints are appreciated!


Let $\sin(x+iy)=a+ib$. Then $$\sin x\cosh y=a\ ,\quad \cos x\sinh y=b\tag{$*$}$$ which easily gives $$(a\cos x)^2-(b\sin x)^2=\sin^2x\cos^2x\ .$$ We can write this in terms of $t=\sin^2x$ as $$t^2-(a^2+b^2+1)t+a^2=0\ .$$ Denote the LHS by $q(t)$. We consider two cases.

  • Case 1, $b=0$. Then $(*)$ has the unique solution $y=0$, $x=\sin^{-1}a$. (In fact, there will be no solution if $|a|>1$; but you were not asked to prove the function is surjective.)
  • Case 2, $b\ne0$. Then $q(0)\ge0$ and $q(1)<0$, so $q(t)$ has a unique root in $[0,1)$. Therefore equations $(*)$ are satisfied by a unique value of $\sin^2x$; but the sign of $\sin x$ is the same as the sign of $a$, so we get a unique value for $\sin x$; and if $x\in R$ we then get a unique value for $x$; and finally we get a unique value for $y$ from the second equation in $(*)$.

So $\sin z$ is injective on $R$.

  • $\begingroup$ That's great! I'd never figured it out alone. Very nice ideas and clear representations. Thanks very much! Now I'm convinced :) $\endgroup$ – user160738 Nov 11 '14 at 1:53


Denote by $u_l = e^{iz_l}$. Then $u_1 - \frac{1}{u_1} = u_2 - \frac{1}{u_2} = s$ and therefore $u_1$, $-\frac{1}{u_1}$ and $u_2$, $-\frac{1}{u_2}$ are roots of the equation $$\lambda ^2 - s \lambda -1=0$$ and therefore $$\{ u_1, -\frac{1}{u_1}\} = \{ u_2, -\frac{1}{u_2}\} $$

Therefore, $u_1 = u_2$ or $u_1 \cdot u_2 = -1$ which for $z_1$, $z_2$ means

$$z_1 - z_2 = 2 k \pi $$ or $$z_1 + z_2 = (2 k+1) \pi$$

which, given the restrictions on $\mathcal{Re}(z)$ implies $z_1 = z_2$.

  • $\begingroup$ Thanks for the answer! I've never thought of relating them to quadratic equations... very cool $\endgroup$ – user160738 Nov 15 '14 at 6:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.