Prove that $\cos^2(z)+\sin^2(z)=1$ Can anybody show me how to prove that $\cos^2(z)+\sin^2(z)=1$ for $z\in \mathbb{C}$? I can prove it for the case where $z$ is real, but can't seem to find a way to prove it for complex numbers in general.
 A: Ok so here is a conceptual proof: the function $f(z)=\cos^2 z+\sin^2 z-1$ is analytic and is zero on a set with a limit point (ie $\mathbb{R}$) and thus $f(z)=0$ identically. The comment above is incorrect. A more computational proof can be given from
$$\cos z=\frac{e^{iz}+e^{-iz}}{2}$$
$$\sin z=\frac{e^{iz}-e^{-iz}}{2i}$$
A: The very same proof that works for real values works for complex values: both $\sin z$ and $\cos z$ admit convergent powerseries on the whole complex plane, which you can differentiate term by term, giving that
$$(\cos z) ' = -\sin z$$
$$(\sin z)' = \cos z$$
If you consider $f(z) = \cos^2 z +\sin^2 z-1$ the above gives $f'(z) =0$:
$$f'(z) = -2\cos z\sin z+2\sin z\cos z=0$$ Thus $f$ must be constant, and because $f(0)=0$, it is constantly zero. 
A: For real trigonometric functions have interpretation in terms of ratios in right-angled triangle.  Then we have calculus (in one variable) that helps us calculate the Taylor  series for $\sin x$ and $\cos x$. For complex values we define trigonometric function by this infinite series and so function values will also be complex numbers.
These series happen to converge throughout the complex plane and define an entire (analytic everywhere) function.
Now consider the complex function $f(z)$ defined by $f(z)=\cos^2 z +sin ^2 z-1$. This function is also entire. And is identically zero on the real line (by the identity known for  highschool trig functions). 
Now invoke the Principle of Permanence of Identities in complex analysis.  This essentially means the set of points where an analytic function takes the value zero (or any other complex number for that matter) is an isolated set: i.e., a set with no limit point, unless the function is a constant function.
Here our $f(z)$ takes value zero on the real line, a set with limit points. SO it should be constantly zero.  SO the identity is valid for complex values of $z$ too.
A: Let $x,y\in\mathbb{R}$ a real number, so :
$$
\cos(y+ix)=\frac{e^{i(y+ix)}+e^{-i(y+ix)}}{2}=\frac{e^{iy}e^{-x}+e^{-iy}e^{x}}{2}
$$
and the same for $\sin$ :
$$
\sin(y+ix)=\frac{e^{i(y+ix)}-e^{-i(y+ix)}}{2i}=\frac{e^{iy}e^{-x}-e^{-iy}e^{x}}{2i}
$$
if we put $z=y+ix$ :
\begin{eqnarray}
4(\cos^2(z)+\sin^2(z))&=&(e^{iy}e^{-x}+e^{-iy}e^{x})^2 -(e^{iy}e^{-x}-e^{-iy}e^{x})^2\\
&=& e^{-2x+2iy}+e^{2x-2iy}+2e^{-x+iy+x-iy}-(e^{-2x+2iy}+e^{2x-2iy}-2e^{-x+iy+x-iy})\\
&=&4
\end{eqnarray}
so :
$$
\cos(z)^2+\sin(z)^2=1 \qquad \forall z \in \mathbb{C}
$$
