proving that |sinz|^2 = sin^2(x) + sinh^2(y)?

Question

I have to prove |sinz|^2 = sin^2(x) + sinh^2(y)

(and the corresponding |cosz|^2= cos^2(x) + sinh^2(y))

I've been given

sinz = sin(x) * cosh(y) + i cos(x) * sinh(y)
cosz = cos(x)*cosh(y) - i sin(x)sinh(y)

so i just did sin(z) * sin(z) and was hoping to get the answer

I got

sin^2(x)*cosh(y) + 2i sin(x)cos(x)cosh(y)sinh(y) -cos^2(x)sinh^y

now I know these 3 formulas and can't quite see how to do it -i've tried completing the square for cos^2xsinh(y) but ended up with more mess

i think i need to use the fact that
sin^2(x) + cos^2(x) = 1
cosh^2(y) - sinh^2(y)=1
and 2sin(x)cos(x) = sin2x

I can see how they all might fit in, but just can't get to the next step... any ideas would be much appreciated

Answer 1

sinz = sin(x) * cosh(y) + i cos(x) * sinh(y)

take mod on both sides

| sin z| ^2 = sin^2 x cosh ^2 y + cos^2 x sinh^2 y
= sin ^2 x (1+sin h^2 y) + cos^2 x sin h^2 y
= sin^2 x + sinh^2 y ( sin ^ 2x + cos^2 x)
= sin ^ 2 x + sinh ^2 y

similiarly for cos z

Answer 2

You don't want to square sin z. You want to multiply it by its conjugate. That will give
|sin z|^2 =sin^2 x cosh^2 y +cos^2 x sinh^2 y.

Now play a bit with your identities.

Answer 3

I got a head ach from looking at it!!!!
I need a advil!
Sorry I was not very helpfull!

Answer 4

I just had to drop in and ask you, "What the ?!?" Nothing personal.