Is $\sin^2(z) + \cos^2(z)=1$ still true for $z \in \Bbb{C}$? Is $$\sin^2(z) + \cos^2(z)=1$$ still true for all $z \in \Bbb{C}$? I've tried rewriting it using complex definitions of $\sin$ and $\cos$, and I don't see why it wouldn't, but the text I'm reading asks this question as if it shouldn't hold?
 A: You can derive this identity from nothing more than the fundamental differential identities that define $\sin$ and $\cos$. For suppose you have functions $s(x)$ and $c(x)$ (that play the obvious roles) with $s'(x) = c(x)$ and $c'(x) = -s(x)$, where $s(0) = c'(0) = 0$ and $c(0)=s'(0)=1$. (If you are only given the more common requirement that $s'' = -s$ and $c''=-c$ then you usually obtain the above derivatives as a result of knowing that certain ODEs with specified initial conditions have unique solutions; I don't want to invoke this here.)
This is just because $(s^2+c^2)' = 2ss' + 2cc' = 2sc-2sc = 0$, and so it's constant; but you also know that $s^2(0)+c^2(0)=1$ by the definition of $s$ and $c$. In this proof I never actually used that these are real functions; just the formulas their derivatives satisfy.
There are other ways to conclude, too. One standard way is Euler's formula; another is the fact that, because $\sin$ and $\cos$ are both holomorphic, $\sin^2+\cos^2-1$ is holomorphic, and we know it's zero on the real line. The identity theorem says that any holomorphic function that vanishes on a set with a limit point (which the real line, of course, is) vanishes everywhere, so $\sin^2(z)+\cos^2(z)=1$ everywhere. It's not hard to prove the identity theorem, but the identity $\sin^2+\cos^2=1$ is even more fundamental than it is.
A: You can also prove this with inversions of Euler's identity:$$ 
\begin{align}
\operatorname{e}^{iz} &= \cos z + i \sin z \Rightarrow \\
\cos z & = \frac{\operatorname{e}^{iz} + \operatorname{e}^{-iz}}{2},\ \mathrm{and} \\
\sin z & = \frac{\operatorname{e}^{iz} - \operatorname{e}^{-iz}}{2i}.
\end{align}$$
These lead to:
$$\begin{align}
\cos^2 z & = \frac{\operatorname{e}^{2iz} + \operatorname{e}^{-2iz} + 2}{4},\ \mathrm{and} \\
\sin^2 z & = -\frac{\operatorname{e}^{2iz} + \operatorname{e}^{-2iz} - 2}{4}.
\end{align}$$
Add these together, and you get $1$. You can make the same proof with the power series definitions, trig substitutions with the power reduction formulae, etc.
