Solve $\tan z = 1$ for $z = x + yi$
$$\tan z = 1 \iff \sin z = \cos z$$
$$\iff i(e^{iz}+e^{-iz}) = e^{iz}-e^{-iz}$$
$$\iff ie^{-y}(\cos x + i \sin x) + i e^y (\cos x – i \sin x) = e^{-y}(\cos x + i \sin x) + e^y(\cos x -i \sin x)$$
And comparing the imaginary and real parts I obtain: $$\cos x = \sin x, \cos x = – \sin x$$
which doesn't yield the correct answer.
Where did I go wrong?
Best Answer
I don't see how you deduced what you deduced by comparing the real and the imaginary parts. Anyway, there is no need to write $z$ as $x+yi$. Note that\begin{align}\tan z=1&\iff\sin z=\cos z\\&\iff\sin(z)-\cos(z)=0\\&\iff\sin(z)\cos\left(\frac\pi4\right)-\cos(z)\sin\left(\frac\pi4\right)=0\\&\iff\sin\left(z-\frac\pi4\right)=0\\&\iff z=\frac\pi4+k\pi,\end{align}for some $k\in\mathbb Z$.