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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

$e^{iz_1}-e^{-iz_1}=e^{iz_2}-e^{-iz_2}$

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!

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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 $R$.

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  • $\begingroup$ 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
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$$e^{iz_1}-e^{-iz_1}=e^{iz_2}-e^{-iz_2}$$

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 $z_1 = z_2$.

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

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