Question

Show that Log(-1) = (2n +1)i.​

Answer 1

Step-by-step explanation:

According to Euler's identity,

$e^{i\theta} = \cos\theta + i\sin\theta$

$e^{i(2n+1)\pi} = \cos(2n+1)\pi + i\sin(2n+1)\pi$

$e^{i(2n+1)\pi} = -1 + i(0)= -1$

Take natural logarithm on both sides,

$i(2n+1)\pi =\log_e(-1)$

Q.E.D