Three problems on Complex Analysis

[carlosbgois]

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

 

[SammyS]

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]

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.