Solving e^{iz}-e^{-iz}=4i: Why Is 2nd Way Better?

[fargoth]

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?

 

[HallsofIvy]

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

 

[fargoth]

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