Is it true that $ |\sin^2z+\cos^2z|=1, \forall z \in\Bbb C$? We know that equation $ \sin^2z+ \cos^2z=1$ which holds  $ \forall z \in\Bbb R$, actually holds $ \forall z \in\Bbb C$.
Is it true that $ |\sin^2z+\cos^2z|=1, \forall z \in\Bbb C$?
Thanks in advance.
 A: Here is a start
$$ \sin z = \sin(x+iy)=\sin(x)\cos(iy)+\sin(iy)\cos(x),\quad i=\sqrt{-1}. $$

$$ \implies \sin z = \sin(x) \cosh(y)+i\sinh(y)\cos(x). $$

Now, do the same with $\cos z$ and substitute back in your equation 

$$ |\sin^2 z +\cos^2 z | $$

and see what you get.
Added: If you work the problem further, then we have
$$ \sin^2 z +\cos^2 z $$
$$=  \sin^2  x   \cosh^2 y
 - \cos^2 x  \sinh^2  y +  \cos^2  x \cosh^2 y -
 \sin^2 x \sinh^2 y =1. $$
Note:

$$\cosh^2 t - \sinh^2 t =1. $$ 

A: In general, if an identity holds for a smooth function over real numbers, then the same identity also holds over the complex numbers. (Amazing!)
In this case: $$\cos^2(z)+\sin^2(z)=(\cos(z)+i\sin(z))(\cos(z)-i\sin(z))=e^{iz}e^{-iz}=1$$
where you need the exponential definition of trigonometric functions. There is no need for the absolute value in your question.
A: When you take the Absolute Value of $z$ you get a number in $\mathbb R$, and since $\left|z\right|\in \mathbb R$, $\sin^2|z|+\cos^2|z|=1$, so the answer is yes
