Showing a complicated mess is equal to cot(z)

1. Sep 1, 2008

nathan12343

1. The problem statement, all variables and given/known data

Find the function f(z) whose real part is

$$\frac{sin(x)}{cosh(2y)-cos(2x)}$$

where z = x + iy

2. Relevant equations

I already know f(z)=cot(z), using FullSimplify[] in mathematica on the expression I get for u + iv when I apply the Cauchy-Reimann equations.

More explicitly,

f(x,y) = u(x,y)+iv(x,y) = $$\frac{-sin(2x)+isinh(2y)}{cos(2x)-cosh(2y)}$$

3. The attempt at a solution

I've tried to derive without mathematica that this is cot(z) about a thousand different ways. I can't seem to manipulate the expression into a usable form. Am I missing something obvious?

Help!!!

Thanks!!

-Nathan

2. Sep 1, 2008

HallsofIvy

Staff Emeritus
There can exist an infinite number of functions having that real part. Do you mean a function analytic on the entire complex plane?

3. Sep 1, 2008

nathan12343

Yes, a function analytic on a certain subset of the complex plane.

4. Sep 1, 2008

tiny-tim

Hi Nathan!

Hint:

$$\frac{1\ -\ i\,cot(z)}{1\ +\ i\,cot(z)}$$ = … ?

5. Sep 1, 2008

nathan12343

I'm sorry, Tiny Tim, I don't see how that helps me:

I get,

$$\frac{1\ -\ i\,cot(z)}{1\ +\ i\,cot(z)} = -(Cos(2z)+Sin(2z))$$

What does that have to do with reducing what I found above?

6. Sep 1, 2008

tiny-tim

Hi Nathan!

How did you get that?

I get $$\frac{1\ -\ i\,cot(z)}{1\ +\ i\,cot(z)}\ =\ \frac{sin(z)\ -\ i\,cos(z)}{sin(z)\ +\ i\,cos(z)}$$ = … ?

7. Sep 1, 2008

nathan12343

Thanks for the help, tiny-tim

I think I dropped an i

$$\frac{1\ -\ i\,cot(z)}{1\ +\ i\,cot(z)}\ =\ \frac{sin(z)\ -\ i\,cos(z)}{sin(z)\ +\ i\,cos(z)} = sin^2(z) - cos^2(z) - 2isin(z)cos(z) = -(cos(2z) + i sin(2z)) = -e^{2iz}$$

But I still don't see how this helps me with my original problem...

8. Sep 1, 2008

tiny-tim

hmm … quicker would be …
$$\frac{1\ -\ i\,cot(z)}{1\ +\ i\,cot(z)}\ =\ \frac{sin(z)\ -\ i\,cos(z)}{sin(z)\ +\ i\,cos(z)}\ =\ -\frac{i\,e^{iz}}{i\,e{-iz}}\ =\ -e^{2iz}$$
You wanted to prove that cot(z) = $$\frac{-sin(2x)+isinh(2y)}{cos(2x)-cosh(2y)}$$ …

just plug that into the above formula.

Last edited: Sep 1, 2008
9. Sep 1, 2008

nathan12343

Wow, that was a really hard problem...

Thanks, for the help, tiny-tim!

If you wouldn't mind my asking, how did you come up with that?

10. Sep 1, 2008

tiny-tim

erm … I'm familiar with (1 + tanhz)/(1 - tanhz) = e2z

(as you should be! )

so I just used iz instead of z.