COMPLEX ANALYSIS please help - cos(z)=3i

Jason Bourne · #1 $Old$ November 14th, 2007, 07:38 AM

Hi

I'm not 100% sure of the solution to the following problem:

Find all the solutions of:

cos(z) = 3i

Please could someone share their wisdom with me. (Also, could someone provide me with a link on how to do LATEX writing on this forum as I used to be able to do it but have since forgotten.)

Thankyou!

Jhevon · #2 $Old$ November 14th, 2007, 07:49 AM

Quote:

Originally Posted by Jason Bourne $View Post$

Hi

I'm not 100% sure of the solution to the following problem:

Find all the solutions of:

cos(z) = 3i

Please could someone share their wisdom with me. (Also, could someone provide me with a link on how to do LATEX writing on this forum as I used to be able to do it but have since forgotten.)

Thankyou!

i'm in kind of a rush, so i won't be able to finish this, but i'll start you off and get you through, what i think, is the hard part

$\cos z = 3~i$

Let $z = a + i~b$

$\Rightarrow \cos (a + i~b) = 3~i$

by the addition formula for cosine, we have

$\cos a \cos i~b - \sin a \sin i~b = 3~i$

$\Rightarrow \cos a \cosh b - i~ \sin a \sinh b = 3~i$

now equate the real and imaginary parts

$\Rightarrow \cos a \cosh b = 0$ and $- \sin a \sinh b = 3$

now continue

See here for LaTex help

Jason Bourne · #3 $Old$ November 14th, 2007, 09:59 AM

Thanks for your fast reply. I thought about doing something like that but I'm not sure how the answer would come out. I attempted doing this question the way my lecture notes seem to indicate, as follows:

$\cos z = 3~i$ , $\frac {e^{iz} +e^{-iz}}{2}=3~i$

$\Rightarrow e^{2iz} -6ie^{iz} +1=0$

$\Rightarrow e^{iz} = i(3\pm{\sqrt{10}})e^{2n\pi~i}$

$\Rightarrow iz=\ln({i(3\pm\sqrt{10})}) + 2n\pi~i$

$\Rightarrow z=2n\pi - i\ln({i(3\pm\sqrt{10})})$ , where $n=0,\pm1,\pm2,...$

But i'm really not sure this makes any sense?! Can anyone help me out of confusion?