# Thread: solve basic trigonometric equation in complex plane

1. ## solve basic trigonometric equation in complex plane

$\sin z = \sin c \hfill \\$
$z = x + iy \hfill \\$
$c = a + ib \hfill \\$
$a,b,x,y \in \mathbb{R} \hfill \\$
${\text{seperating the real and imaginary part I have the following system to solve}} \hfill \\$
$\left\{ \begin{gathered}
\sin x\cosh y = \sin a\cosh b \hfill \\
\cos x\sinh y = \cos a\sinh b \hfill \\
\end{gathered} \right. \hfill \\$

But I don't see how I can solve this. Any idea?

2. You need to know this identity.
$\sin (x + yi) = \sin (x)\cosh (y) + i\left[ {\cos (x)\sinh (y)} \right]$

3. I do know this formula; this is how I reached the system I need to solve.

This is where the problem begins, at least for me.

civodul

4. Originally Posted by civodul
I do know this formula; this is how I reached the system I need to solve.
Well in that case, I would square both; add them together; use identities to eliminate one variable.

5. If $\sin(z)=\sin(c)$ then why can't we just write:

$z=-i\log\bigg(iw+\sqrt{1-w^2}\bigg),\quad w=\sin(c)$

6. Why do you not stay with your original question?
Of course there are many ways to solve this question!
But you were adamant about the way you wanted it done.
Most mathematicians would have gone for $\sin (z) = \frac{{e^{iz} - e^{ - iz} }}{{2i}}$.

7. Hi Plato, Shawsend,

Thanks for your answers.
Plato, it was Shawsend who suggested this other way, not me!

I don't understand why the assignement directs us to seperate real and imaginary parts instead of going the 'mathematicians way'.
It could be just for training students in trigonometric calculous.

civodul

8. Sorry for causing trouble in your thread. I was just genuinely curious that's all. I'm sure Plato is much better at this than me.