# Three problems on Complex Analysis

1. Aug 31, 2012

### carlosbgois

1. The problem statement, all variables and given/known data

1)Show that $(1-i)^{2}=-2i$ then evaluate $(1-i)^{2004}+(1-i)^{2005}$

2)Prove that every complex number with moduli 1, except z=1, can be put in the form $\frac{a+i}{a-i}$

3)Let m and n be positive integers without a common factor. Define $z^{m/n}=(z^{m})^{1/n}$, and show that $z^{m/n}=(z^{1/n})^{m}$

3. The attempt at a solution

1) I solved $(1-i)^{2}=-2i$ by expanding, and then solved in the polar form, that is, $(1-i)^{2}=(\sqrt{2}(cos(-pi/4)+i*sin(-pi/4))^{2}=2(cos(-pi/2)+i*sin(-pi/2)=\sqrt{2}(1-i)$, hence, $\sqrt{2}(1-i)=-2i$. Then I stopped because something's wrong, isn't it?

2)Let z=x+yi. Letting $x+yi=\frac{a+i}{a-i}$ I can obtain the following equations: $xa+y=a; ya-x=1$, which, solved simultaneously, yields $a=\sqrt{\frac{1+x}{1-x}}$, then I got stuck.

3)I've just used DeMoivre's with z in polar form and particular properties of real numbers to invert the powers, but idk if that's correct.

Many Thanks

2. Aug 31, 2012

### SammyS

Staff Emeritus
Yes, something is wrong.

cos(±π/2) = 0

3. Aug 31, 2012

### Bacle2

For 1 , you can look at multiplication as multiplying the respective moduli and

adding the arguments (mod 2pi if necessary). For 2, it seems you can just solve

for a in a single equation, to have a in terms of x,y ; after all, z seems to be given

in terms of x and y.