Showing a complicated mess is equal to cot(z)

[nathan12343]

Homework Statement

Find the function f(z) whose real part is

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

where z = x + iy

Homework 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)}

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

 

[HallsofIvy]

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

 

[nathan12343]

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

 

[tiny-tim]

Hi Nathan!

Hint:

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

 

[nathan12343]

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

I get,

<br /> \frac{1\ -\ i\,cot(z)}{1\ +\ i\,cot(z)} = -(Cos(2z)+Sin(2z))<br />

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

 

[tiny-tim]

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

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)} = … ?

 

[nathan12343]

Thanks for the help, tiny-tim

I think I dropped an i

<br /> \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}<br />

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

 

[tiny-tim]

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}

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}

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

You wanted to prove that cot(z) = \frac{-sin(2x)+isinh(2y)}{cos(2x)-cosh(2y)} …

just plug that into the above formula.

 

[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?

 

[tiny-tim]

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

erm … I'm familiar with (1 + tanhz)/(1 - tanhz) = e^2z …

(as you should be! )

so I just used iz instead of z.