Define: $(e^{iz}+e^{-iz})/2= cos z$ where $z \in \Bbb C $, i.e, the cosine function is defined for complex $z$. Now, is it true that for each $w \in \Bbb C $ there is $z \in \Bbb C $ such that $cos z = w$?
I claim yes and prove it in the following manner but I have some issues with certain steps in the proof:
Proof:
Let $u=e^{iz}$
then, $cos z = w \implies u^{2}- 2wu+ 1 = 0$
$\implies u=[w \pm \sqrt (w^{2}-1)] \implies e^{iz}= [w \pm \sqrt (w^{2}-1)]$
$\implies e^{iz}= [w \pm \sqrt (w^{2}-1)]e^{i2\pi k}, k=0,\pm 1, \pm 2,... $
$\implies log (e^{iz})= log ([w \pm \sqrt (w^{2}-1)]e^{i2\pi k}) $
$\implies z=-i log (w \pm \sqrt (w^{2}-1)]) + 2\pi k$
Now I have two fundamental questions.
1) Is the logarithm of complex numbers defined and allowed as I have done in the proof?
2) Does the quadratic formula for calculating roots still hold good for complex coefficients ($w$ in this case)?
Finally, please suggest a modified proof if something is wrong.