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Three problems on Complex Analysis

  • #1

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
 

Answers and Replies

  • #2
SammyS
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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?
...
Many Thanks
Yes, something is wrong.

cos(±π/2) = 0
 
  • #3
Bacle2
Science Advisor
1,089
10
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.
 

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