Find the values of $\sin z=0$

Question

Hello can someone check my work and tell me what to do next.

Find the values of $\sin z = 0$

we know that; $$\sin z = {{{e^{zi}} - {e^{ - zi}}} \over {2i}}$$

So; $${{{e^{zi}} - {e^{ - zi}}} \over {2i}} = 0$$ $${e^{zi}} - {e^{ - zi}} = 0$$ Then I multiply with $e^{jz}$; $${({e^{zi}})^2} - 1 = 0$$ $${({e^{zi}})^2} = 1$$ $$({e^{zi}}) = \sqrt 1 $$

I stuck at this part. Did I need to multiply with $ln$?

Answer 1

Instead of taking a square root, put $z = x+yi$ and try $$ 1 = (e^{zi})^2 = e^{2zi} = e^{2xi-2y} = e^{-2y}(\cos 2x + i\sin 2x). $$ Identifying the real and imaginary parts, you need $\sin 2x = 0$ and $e^{-2y} \cos 2x = 1$. You take it from here.

Answer 2

You have $e^{iz}=1$ or $e^{iz}=-1$. Now write $z=x+iy$, so $iz=-y+ix$ and $$ e^{iz}=e^{-y}e^{ix} $$ so $e^{-y}=|e^{iz}|$.

In the first case you have $e^{-y}=1$ and $e^{ix}=1$, in the second case $e^{-y}=1$ and $e^{ix}=-1$.

Can you go on?

Answer 3

Make sure your last step is right, it should say:

$e^{i z} = \pm 1 $

You can now write $ z = a + bi $ and solve for $ iz $. Use that to evaluate the $e^x$ expression.