Solving the complex equation $\sin(z) = \cos(z)$ To find the complex numbers z satisfying $\sin(z) = \cos(z)$, can I say:
$$\sin(z) = \frac{(e^{iz}-e^{-iz})}{2i}=\frac{(e^{iz}+e^{-iz})}{2}$$ 
and solve for z? So we then reduce this to $$-e^{-iz} = e^{-iz}$$ but this doesn't look right
 A: Use angle addition identities and the relations between trig and hyperbolic trig functions.
$$\begin{align}
&&\sin(z)&=\cos(z)\\
\implies&&\sin(a+bi)&=\cos(a+bi)\\
\implies&&\sin(a)\cos(bi)+\cos(a)\sin(bi)&=\cos(a)\cos(bi)-\sin(a)\sin(bi)\\
\implies&&\sin(a)\cosh(b)+i\cos(a)\sinh(b)&=\cos(a)\cosh(b)-i\sin(a)\sinh(b)\\
\end{align}$$
Now equating real parts (and keeping in mind that $a$ and $b$ are real), we see that $\sin(a)=\cos(a)$. (This also use the fact that $\cosh(b)$ cannot be $0$.) So $a=\frac{\pi}{4}+k\pi$ for some $k\in\mathbb{Z}$. 
Now we can divide across by $\sin(a)$ or $\cos(a)$. (Remember, we have deduced they are equal.)
$$\begin{align}
\implies&&\cosh(b)+i\sinh(b)&=\cosh(b)-i\sinh(b)\\
\implies&&i\sinh(b)&=-i\sinh(b)\\
\implies&&\sinh(b)&=-\sinh(b)\\
\implies&&\sinh(b)&=0\\
\end{align}$$
There is only one real solution for $b$: $b=0$.
So in conclusion, $z=\frac{\pi}{4}+k\pi$ for some $k\in\mathbb{Z}$.
A: The definition of $\sin(z)$ is
$$\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$$
A: $$\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$$ 
Then, $$\sin(z)=\cos(z)\implies \frac{e^{iz}-e^{-iz}}{2i}=\frac{e^{iz}+e^{-iz}}{2}$$ Now let $e^{iz}=t$ ,then $e^{-iz}=1/t$ and you will get a quadratic equation,solve for it and back substitute it to get $z$.
The equation becomes $t^2=\frac{1+i}{1-i}=i=e^{i(\pi/2+2k\pi)}\implies t=e^{i(\pi/4+k\pi)},e^{i(5\pi/4+k\pi)}$. Equating it with $e^{iz}$ we get $z=\pi/4+k\pi$
