- #1

carlosbgois

- 68

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## Homework Statement

**1)**Show that [itex](1-i)^{2}=-2i[/itex] then evaluate [itex](1-i)^{2004}+(1-i)^{2005}[/itex]

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

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

## The Attempt at a Solution

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

**2)**Let z=x+yi. Letting [itex]x+yi=\frac{a+i}{a-i}[/itex] I can obtain the following equations: [itex]xa+y=a; ya-x=1[/itex], which, solved simultaneously, yields [itex]a=\sqrt{\frac{1+x}{1-x}}[/itex], 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