Absolute value of trigonometric functions of a complex number

[agnimusayoti]

Homework Statement

Find absolute value of $\sin (x-iy)$.

Relevant Equations

$$\sin z=\frac {e^{iz}-e^{-iz}}{2i}$$
If $z=x+iy$ then, absolute value of this complex number is $|z|=\sqrt {x^2+y^2}$

So far I've got the real part and imaginary part of this complex number. Assume: $z=\sin (x+iy)$, then
1. Real part: $\sin x \cosh y$
2. Imaginary part: $\cos x \sinh y$
If I use the absolute value formula, I got $|z|=\sqrt{\sin^2 {x}.\cosh^2 {y}+\cos^2 {x}.\sinh^2 {y} }$
How to simplify that answer to $|z|=\sqrt{\sin^2 {x}+\sinh^2 {y}}$?

Thanks

 

[Orodruin]

$\sin^2 + \cos^2 = 1$

 

[agnimusayoti]

I do not understand how could $\cosh^2 \cos^2$ eliminated?

 

[Orodruin]

Please show your work. If not it is impossible to know where you have gotten $\cosh^2 \cos^2$ from.

 

[lurflurf]

You want to eliminate cos and cosh at some point. Do it right away.
$$\cos^2(x)=1-\sin^2(x)$$
$$\cosh^2(y)=1+\sinh^2(y)$$

 

[agnimusayoti]

Uh there it is. Thanks!

 

[agnimusayoti]

Please show your work. If not it is impossible to know where you have gotten $\cosh^2 \cos^2$ from.

I mean $cosh^2 y$ at $sin^x cosh^2y$

 

[Orodruin]

Please show your work. All of it.

 

[agnimusayoti]

 

[lurflurf]

That is some nice work and handwriting.
a nice identity in there
$$u^2(1+v^2)+(1-u^2)v^2=u^2+v^2$$