# Three problems on Complex Analysis

## Homework Statement

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 $z^{m/n}=(z^{1/n})^{m}$

## 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

Related Calculus and Beyond Homework Help News on Phys.org
SammyS
Staff Emeritus
Homework Helper
Gold Member

## Homework Statement

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 $z^{m/n}=(z^{1/n})^{m}$

## 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?
...
Many Thanks
Yes, something is wrong.

cos(±π/2) = 0

Bacle2