- #1

- 315

- 6

[tex]e^{iz}-e^{-iz} = 4i[/tex]

[tex]e^{2iz}-4ie^{iz} = 1[/tex]

[tex]iz \ln (e^{iz}-4i) = 0[/tex]

[tex] z=0[/tex]

when solving it by

[tex]w = e^{iz}[/tex]

[tex]w^2-4wi-1 = 0[/tex]

i get one more solution, why is the first way not as good as the second way?