What is the Complex Tangent Formula Proof for Homework?

[allamid06]

Homework Statement

This is an easy one, but keep in mind I'm kind of a newbie, anyway I can't figure out how to get the next formula...
tan(z) = (tan(a)+i tanh(b))/(1 - i tan(a)tan(b))

Homework Equations

This is the third part of an excercise, previous I proof the follow, -all using the definitions, of complex sin,cos, and tan, and definitions of real sinh,cosh, and tanh-...

cos(z)=cos(a)cosh(b)-i sin(a)sinh(b);
sin(z)=sin(a)cosh(b)+icos(a)sinh(b)

The Attempt at a Solution

I tried a lot of things, but couldn't get any way, maybe I'm missing something important, and that's what I fear.
I tried to replace tan(z) = sin(z)/cos(z) with the other equations but, I'm getting nothing.

Sorry for my english, lot of thanks for the help!

 

[Svein]

$\tan(z)=\frac{\sin(z)}{\cos(z)}=\frac{\frac{e^{z}-e^{-z}}{2i}}{\frac{e^{z}+e^{-z}}{2}}=\frac{1}{i}\frac{e^{z}-e^{-z}}{e^{z}+e^{-z}}=\frac{1}{i}\frac{e^{a+ib}-e^{-(a+ib)}}{e^{a+ib}+e^{-(a+ib)}}$
Now write out the expression for $e^{a+ib}$...

 

[allamid06]

$\tan(z)=\frac{\sin(z)}{\cos(z)}=\frac{\frac{e^{z}-e^{-z}}{2i}}{\frac{e^{z}+e^{-z}}{2}}=\frac{1}{i}\frac{e^{z}-e^{-z}}{e^{z}+e^{-z}}=\frac{1}{i}\frac{e^{a+ib}-e^{-(a+ib)}}{e^{a+ib}+e^{-(a+ib)}}$
Now write out the expression for $e^{a+ib}$...

Thanks for that!, but, I will have to ask for even more help. Because, I already knew that formula, and I can't find how to pass to the one I'm asking for...

$tan(z) = (tan(a)+i tanh(b))/(1 - i tan(a)tan(b))$

 

[allamid06]

I've solved it! and I wanted to tell you. The biggest problem was that the damn book have a cute problem tan(z)=(tan(a)+itanh(b))/(1−itan(a)tan(b)) is really tan(z)=(tan(a)+itanh(b))/(1−itan(a)tanh(b)) having that in mind and with some equations it's really simple to get. I mean, using cosh2(x)- sinh2(x)=1,cos2(x)+sin2(x)=1, and tan(x)=sin(x)/cos(x), ...and the most important, after the two other equations I posted at first... (and the simplest one) sin(x)=cos(x)tan(x)