Dumb question alert!
I'm not too sure if I recall this question, but it's been baffling me:
If z=-4i what is the modulus-argument of Iz^3I
Wouldn't the modulus be 64 and the argument be 0?
However the teacher said that the argument was pi/2
Please help this poor soul
Interesting - not sure you've gone and calculated the argument for this, but there's two ways to look at it. The first is "what is the modulus and argument of the number |z^3| (the modulus of z^3)", in which case |z^3| is (by definition) a real number, and the argument MUST be 0, as you've initially suggested.
The second (and I imagine this is where the teacher was going), is where you ignore the modulus signs, and she actually intended you to find the modulus and argument of z^3 - the complex number. The modulus is, of course, done the same, but now for the argument, you'd need to use the formula arg(z^n)=n*arg(z), and then normalise it so it's in the domain (-pi,pi].