Using the definitions of the corresponding functions, prove…
Using the definitions of the corresponding functions, prove: $\ \sin^2z+\cos^2z=1$.
Explanation of Method
Start from the exponential definitions $\sin z=\dfrac{e^{iz}-e^{-iz}}{2i}$, $\cos z=\dfrac{e^{iz}+e^{-iz}}2$, square each, and add.
Step 1 · Square $\sin z$
$$ \sin^2z=\left(\frac{e^{iz}-e^{-iz}}{2i}\right)^2=\frac{(e^{iz}-e^{-iz})^2}{-4}=\frac{2-e^{2iz}-e^{-2iz}}4 $$
(using $(e^{iz}-e^{-iz})^2=e^{2iz}-2+e^{-2iz}$ and $\dfrac1{(2i)^2}=\dfrac1{-4}$.)
Step 2 · Square $\cos z$
$$ \cos^2z=\left(\frac{e^{iz}+e^{-iz}}2\right)^2=\frac{e^{2iz}+2+e^{-2iz}}4 $$
Step 3 · Add
$$ \sin^2z+\cos^2z=\frac{(2-e^{2iz}-e^{-2iz})+(e^{2iz}+2+e^{-2iz})}4=\frac44=1 $$
Alternative proof
Multiply $e^{iz}=\cos z+i\sin z$ by $e^{-iz}=\cos z-i\sin z$:
$$ e^{iz}e^{-iz}=e^0=1=(\cos z+i\sin z)(\cos z-i\sin z)=\cos^2z-i^2\sin^2z=\cos^2z+\sin^2z $$
Result
The Pythagorean identity extends unchanged from real to complex arguments — it is an algebraic consequence of the exponential definitions, not a geometric statement about a unit circle (indeed $\sin z,\cos z$ are unbounded on $\mathbb C$).