0

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 $ln$?

razercreed
  • 109
  • You can't multiply $\ln$ with anything. $\ln$ is a function, and it has an input. So for example, $\ln(2)$ is read "LN of 2", just like $f(2)$ is "F of 2". With $f(2)$, you understand that $f$ is a function and $2$ is the input. So for $\ln(2)$, $\ln$ is a function and $2$ is the input. – layman Dec 30 '14 at 14:03
  • BUT, the idea is, if $e^{zi} = \sqrt{1}$, then these two numbers are equal. So taking the $\ln$ of both sides should give you the same thing (if you plug the same number/input into a function twice, you should get the same output). So that means $\ln(e^{zi}) = \ln(\sqrt{1})$. Now, this isn't necessarily the right way to finish the problem, but I just wanted to clarify that you don't multiply $\ln$, you "Take the $\ln$ of both sides" because $\ln$ is a function. – layman Dec 30 '14 at 14:05
  • ok thank you and if multiply with $ln$ i will get zero – razercreed Dec 30 '14 at 14:09
  • There is no multiplying with $\ln$. This is unfortunately incorrect math language. Also, please check out some of the answers to your question because they are all helpful and solve the problem in a better (more complete) way. – layman Dec 30 '14 at 14:12

3 Answers3

1

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?

egreg
  • 238,574
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.

Yuval Filmus
  • 57,157
1

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.

kummerer94
  • 356