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
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