if

tan(x+i*y)=A+i*B

Show that

A/B=sin(2*x)/sinh(2*y)

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First you have to separate real and imaginary parts of Tan(x+iy)=Tan(z)=sin(z)/cos(z)

sinz=sin(x+iy)=sinxcos(iy)+cosxsin(iy)=sinxcoshy-icosx sinhy

cosz=cos(x+iy)=cosxcos(iy)-sinxsin(iy)=cosxcoshyâˆ’isinxsinhy

Now if you plug in Tan(z) and simplify (it is easy!) you get

Tan(z)=(sin(2x)+isinh(2y))/(cos(2x)+cosh(2y))= A+iB.

This means that

A=sin(2x)/(cos(2x)+cosh(2y)) and B= sinh(2y)/(cos(2x)+cosh(2y))

Now,

A/B=sin(2x)/sinh(2y)

If any questions, let me know.

Could you please explain more step by step

Ok

We have Tan(x+iy) and we need to separate real and imaginary parts: A and B (as you denoted). How?

We know that

1. Tan(x+iy)=sin(x+iy)/cos(x+iy).

Now we have to remind a few identities (formulas):

a) sin(a+b)=...

b) cos(a+b)=...

c) cosh(a)=cos(ia)

d) sinh(a)=-isin(ia)

2. Write down (expand) tan(x+iy)= .. and by using a)-d) simplify that such that you can see real and imaginary parts. That all.

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## Comments

^{iz}-e^{-iz})/(e^{iz}+e^{-iz}). From this expression you can separate real and imaginary parts.