Homework+1

1. Find the real part of 1 + ((1+i)/√2) + ((1+i)/√2) 2 +...+ ((1+i)/√2) 2008

One possible solution: let z=(1+i)/√2= cos(pi/4) + i sin(pi/4). Then math 1+z+...+z^{2008}=\frac{1-z^2009}{1-z}=\frac{1-\cos(2009 \pi/4)-i \sin (2009 \pi/4)}{1-\cos(\pi/4) - i \sin(\pi/4)}=1 math

2. Compute (1+i) 2013 /(1-i) 2011

3. There are six complex numbers satisfying z 6 +1=0. Find the sum of all of them.