# Homework Help: Complex Analysis (i need help,immediately)

1. Oct 13, 2009

### phykb

1) if z^3=1, show that (1-z)(1-z^2)(1-z^4)(1-z^5)=9, zEC

2) if cos(x)+cos(y)+cos(t)=0, sin(x)+sin(y)+sin(t)=0 show that cos(3x)+cos(3y)+cos(3t)=3cos(x+y+t)

3)show that, the roots the equations (1+z)^(2n) +(1-z)^(2n)=0, nEN, zEC are given by the relation z=itan((2κ+1)π)/4n), κ=0,1,...,n-1

2. Oct 13, 2009

### Staff: Mentor

You must show some effort of your own, before we can offer tutorial help. How would you start working on each of these?

3. Oct 13, 2009

### phykb

Problem (1), (3): i have no idea
Problem (2): question...cos(x)cos(y)cos(t)-cos(x)sin(y)sin(t)-sin(x)cos(y)sin(t)-sin(x)sin(y)cos(t)=cos(y) (cos(t) cos(x)-sin(t) sin(x))-sin(y) (sin(t) cos(x)+cos(t) sin(x))=cos(x+y+t)? if answer=yes, i'm ok with this problem, i solve this problem!

4. Oct 13, 2009

### Staff: Mentor

You mean problems 1 and 3.

For 1, if z3 = 1, then z must be one of the complex cube roots of 1. All three have magnitude 1, but different args (angles). One of them has an arg of 2pi/3. Maybe you can come up with the other two.

For each one, evaluate (1-z)(1-z2)(1-z4)(1-z5), and see what you get. That's how I would approach it.

For 3, I don't have any insights right now, but I would start playing with it. For example, I would check that z = i tan(1/4) is a solution of your equation. (This is the solution for K = 0.)

5. Oct 13, 2009

### Dick

For 1) there are three complex roots of z^3=1, as Mark44 said. But z=1 doesn't work since (1-z)=0. So you must mean one of the other ones. But there is a simple way to do it. Since z^3=1, z^4=z and z^5=z^2. So you've now got (1-z)^2*(1-z^2)^2. Since |z|=1 and z(z^2)=1 that means z^2=z* (*=complex conjugate). Now you've got ((1-z)(1-z*))^2. If you expand that one of the parts is z+z*. Can you show that's (-1) for z a complex root of z^3=1?