# Complex solutions of $\sin z = i \alpha \cos z$

I'm trying to solve the following question:

Let $\alpha$ $\in [-1, 1]$ be a real number. Find all complex numbers $z$ that satisfy the equation:

$\sin z = i \alpha \cos z$

This is what I've done so far:

I let $z = x + iy$ which makes the the equation equal:

$$\sin (x+iy) = i \alpha \cos(x+iy)$$

I simplified it to:

$$\sin x + i\sin y = a(i\cos x-\cos y)$$

I don't know what to next because I have never encountered anything like this before.

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First of all, $\sin$ and $\cos$ are not linear, not even over the complex numbers. Instead try using that $\sin(z) = \frac{1}{2i}(e^{iz} - e^{-iz})$ and $\cos(s) = \frac{1}{2}(e^{iz} + e^{-iz})$. –  Neal Dec 3 '12 at 0:25
You might also gain insight by dividing through by $\cos z$ to get $\tan z=i\alpha$. –  Lubin Dec 3 '12 at 0:59
Ok if I do that method, I've managed to get to e$^{iz}$ - e$^{-iz}$ = -2a (e$^{iz}$ - e$^{-iz}$). If I expand and take it to one side I get 1+2a (e$^{iz}$ - e$^{-iz}$) =0. Solving this a=-1/2 but what would I do next with the e$^{iz}$ = e$^{-iz}$ to solve the complex roots ? –  methuselah Dec 3 '12 at 1:03

I'll try to give some insight ...

We know that $\frac {\sin z}{\cos z} = \tan z$. So our equation becomes.. $$\tan z = i\alpha$$

Now use the exponential form of $\sin$ and $\cos$ to find the exponential form of $\tan z$. And what is the exponential form of $i$? Now we can solve the question.

For much general result, you can see next (though I'll strongly recommend doing this yourself)

want solution for $\tan z = z_0 \ \ ;z_0,z \in \mathbb{C}$

Solution:

$$\tan(z) = \frac{e^{iz} - e^{-iz}}{i(e^{iz} + e^{-iz})} = z_0$$

Putting $p= e^{iz} ;\frac 1p = e^{-iz}$ our equation becomes,

$$\frac{p - 1/p}{i(p + 1/p)} = z_0$$ $$\frac{p^2 - 1}{p^2 + 1} = iz_0$$ $$p^2 - 1 = iz_0(p^2 + 1)$$ $$(1 - iz_0)p^2 = 1 + iz_0$$ $$p^2 = \frac{1 + iz_0}{1 - iz_0}$$

Now plugging in original value, we get, $$e^{2iz} = \frac{1 + iz_0}{1 - iz_0}$$

Giving the solution to be,

$$z = \frac 1{2i} \ln \frac{1 + iz_0}{1 - iz_0}$$

Disclaimer: Just typed quickly, might be errors in it but method will be same.

EDIT:

As asked in comment by the poster. Put $z_0=i\alpha$ in the solution.

Then we get, $$z=\frac 1{2i} \ln \frac{1 + i \cdot i \alpha}{1 - i \cdot i \alpha}=\frac 1{2i} \ln \frac{1 + i^2 \alpha}{1 - i^2 \alpha}=\frac 1{2i} \ln \frac{1 - \alpha}{1 + \alpha}$$.

I suppose it is clear now

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what happened to the tanz=iα ? –  methuselah Dec 3 '12 at 1:13
put $z_0=i\alpha$ in the final answer.. it also gives you a very important onservation, if $\alpha = 1$ there is no solution. –  TheJoker Dec 3 '12 at 1:15

Your equation is equivalent to $\tan(z)=i\alpha$. Even for complex $a$ and $b$, $$\tan(a+b)=\frac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}$$

So let $z=a+b\,i$, with $a,b$ real, and \begin{align}\tan(a+bi)&=\frac{\tan(a)+\tan(bi)}{1-\tan(a)\tan(bi)}\\ &=\frac{\tan(a)+i\tanh(b)}{1-i\tan(a)\tanh(b)}\\ &=\frac{(\tan(a)+i\tanh(b))(1+i\tan(a)\tanh(b))}{1+\tan^2(a)\tanh^2(b)} \end{align} Here we have used that $\tan(ix)=i\tanh(x)$.

You want the real part of this expression to be $0$, since you want it to equal $i\alpha$. So $\tan(a)-\tan(a)\tanh^2(b)=0$. That is, $\tan(a)(1-\tanh^2(b))=0$. Now, there are no real numbers $b$ with $\tanh(b)=\pm1$, so $\tan(a)=0$. Therefore $a$ must be $k\pi$ for some integer $k$.

Back substituting, \begin{align}\frac{(\tan(k\pi)+i\tanh(b))(1+i\tan(k\pi)\tanh(b))}{1+\tan^2(k\pi)\tanh^2(b)}&=\alpha i\\ {i\tanh(b)}&=\alpha i\\ {\tanh(b)}&=\alpha \\ \end{align}

So there is only a solution if $\alpha\in(-1,1)$, in which case $b=\tanh^{-1}(\alpha)$, and $z=k\pi+\tanh^{-1}(\alpha)$ for some integer $k$.

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