How Does the Complex Logarithm Function Differ Across Branches?

[neginf]

Homework Statement

Show that

(a) log(i^2) = 2*log(i) when log z=ln r + i * theta (r>0 and pi/4 < theta < 9*pi/4)
(b) log(i^2) <> 2*log(i) when log z=ln r + i * theta (r>0 and 3*pi/4 < theta < 11*pi/4)

Homework Equations

log z = ln r + i * theta

The Attempt at a Solution

Got log(i) = i * pi/2
log(i^2)= i * pi.
Are those right ?

Do not know what to do with ranges given for theta.

 

[vela]

You're assuming i = e^iπ/2, but you also have i = e^i(π/2+2π) = e^i(π/2+4π) and so on. Now do you see what the problem is getting at?