# Whats wrong here?

$$sin(z)=2$$
$$e^{iz}-e^{-iz} = 4i$$
$$e^{2iz}-4ie^{iz} = 1$$
$$iz \ln (e^{iz}-4i) = 0$$
$$z=0$$

when solving it by
$$w = e^{iz}$$
$$w^2-4wi-1 = 0$$
i get one more solution, why is the first way not as good as the second way?

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HallsofIvy
Homework Helper
ln(ab)= ln(a)+ ln(b), not ln(a)*ln(b).

If $$e^{2iz}-4ie^{iz} = 1$$
then $$e^{iz}(e^{iz}- 4i)= 1$$
so $$iz+ ln(e^{iz}- 4i)= 0$$
NOT $$iz \ln (e^{iz}-4i) = 0$$

hehe, right, that was a dumb mistake
thank you for pointing it out.