Comparing log(i^2) and 2*log(i)

[neginf]

Homework Statement

Show log(i^2) and 2*log(i) have different sets of values.

Homework Equations

log z=ln|z|+i*arg z

The Attempt at a Solution

log(i^2) = ln|i^2| + i*arg(i^2)
= ln|-1| + i*(pi + 2*n*pi)
= 0 + i*pi*(1 + 2*n)
= i*pi*(1+2*n)

2*log(i) = 2*(ln|i| + i*arg(i))
= 2*(ln|1| + i*(pi/2 + 2*n*pi))
= 2*(0 + i*(pi/2 + 2*n*pi ))
= i*pi*(1+4*n)

If this isn't right, what did I miss ?

 

[praharmitra]

It's correct.

 

[neginf]

Thank you very much.