# Math Help - Complex Analysis Help Please (not complicated :D)

1. ## Complex Analysis Help Please (not complicated :D)

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

Find all the solutions of:

, where z=x+iy

I attempted doing this question the way my lecture notes seem to indicate, as follows:

,

, where

But i'm really not sure this makes any sense?! Can anyone help me out of confusion? Is there another way of doing this or is this vaugely right?

2. Originally Posted by Jason Bourne
I'm not 100% sure of the solution to the following problem: Find all the solutions of:
, where z=x+iy
The standard way is to write cos(z) in the u+iv form.
$\cos (z) = \cos (x)\cosh (y) + i\left( { - \sin (x)\sinh (y)} \right)$.
So we get $\cos (x)\cosh (y) = 0\;\& \; - \sin (x)\sinh (y) = 3$.
Now note that $\cosh (y) \ne 0\; \Rightarrow \;\cos (x) = 0\; \Rightarrow \;x = \frac{{\left( {2k + 1} \right)\pi }}{2}$.
Now you can finish?

3. Originally Posted by Jason Bourne
I'm not 100% sure of the solution to the following problem:

Find all the solutions of:

, where z=x+iy

I attempted doing this question the way my lecture notes seem to indicate, as follows:

,

, where

But i'm really not sure this makes any sense?! Can anyone help me out of confusion? Is there another way of doing this or is this vaugely right?
Please don't double post. See rule #1 here.

-Dan

4. Originally Posted by Jason Bourne
Find all the solutions of:

, where z=x+iy

...

, where
This solution is completely correct. You could simplify it a bit by using properties of the logarithm: $\ln(i(3\pm\sqrt10)) = \ln i + \ln(3\pm\sqrt10)$, and $\ln i = \ln(e^{i\pi/2}) = i\pi/2$. So $z = 2n\pi - i\ln(i(3\pm\sqrt10)) = (2n+{\textstyle \frac12})\pi-i\ln(3\pm\sqrt10)$.

5. Thankyou all for your help and sorry for the double post. I'm not confused now but was wondering, the real part suggested by Plato is $
x = \frac{{\left( {2n + 1} \right)\pi }}{2}
$

but the real part calculated using the other method is $
x = (2n+{\textstyle \frac12})\pi$

am I missing something here? Thanks Anyway.

6. Originally Posted by Jason Bourne
the real part suggested by Plato is $
x = \frac{{\left( {2n + 1} \right)\pi }}{2}
$

but the real part calculated using the other method is $
x = (2n+{\textstyle \frac12})\pi$

am I missing something here?
That's a very interesting question. The answer is that there is no discrepancy between the two solutions, though it took me a while to figure out how they fit together.

Start with the solution in the form $z = (2n+{\textstyle \frac12})\pi-i\ln(3\pm\sqrt10)$. There are two solutions for each value of n, namely

$z = (2n+{\textstyle \frac12})\pi-i\ln(3+\sqrt10)$ [1] and $z = (2n+{\textstyle \frac12})\pi-i\ln(3-\sqrt10)$ [2].

Looking at the second of these, the first thing you should notice is that 3–√10 is negative. In fact, $3-\sqrt{10} = \frac{-1}{3+\sqrt{10}}$ (this follows from the fact that $(3+\sqrt{10})(3-\sqrt{10}) = 9-10=-1$). Therefore $\ln(3-\sqrt{10}) = \ln(-1) - \ln(3+\sqrt{10}) = i\pi - \ln(3+\sqrt{10})$, and we can rewrite [2] as $z = (2n+1+{\textstyle \frac12})\pi+i\ln(3+\sqrt10)$. This can be combined with [1] to form the single expression $z = (n+{\textstyle \frac12})\pi-i(-1)^n\ln(3+\sqrt10)$, which agrees with Plato's method for finding the solution.