Three problems on Complex Analysis

Why is it incorrect?In summary, in the first problem, (1-i)^{2} was solved to be -2i using polar form, but the final result was incorrect. In the second problem, it can be shown that any complex number with modulus 1, except z=1, can be written in the form \frac{a+i}{a-i}. Finally, in the third problem, the correct equation is z^{m/n}=(z^{1/n})^{m}, not z^{m/n}=(z
  • #1
carlosbgois
68
0

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
 
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  • #2
carlosbgois said:

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
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.
 

What is Complex Analysis?

Complex Analysis is a branch of mathematics that deals with the study of complex numbers and functions. It involves the application of calculus to complex numbers and the study of properties of complex functions.

What are the three main problems in Complex Analysis?

The three main problems in Complex Analysis are the Riemann Hypothesis, the Prime Number Theorem, and the Hadamard Conjecture. These problems have been open for centuries and are still actively studied by mathematicians.

What are some real-world applications of Complex Analysis?

Complex Analysis has many real-world applications, such as in physics, engineering, and finance. It is used to model and analyze complex systems, and to solve problems involving electric circuits, fluid dynamics, and signal processing.

What are the tools and techniques used in Complex Analysis?

Some of the tools and techniques used in Complex Analysis include contour integration, Cauchy's theorem, and the Cauchy-Riemann equations. Other important concepts include power series, Laurent series, and the Residue Theorem.

What are some common misconceptions about Complex Analysis?

One common misconception about Complex Analysis is that it is only applicable to theoretical mathematics and has no practical applications. Another misconception is that it is just a more complicated version of real analysis. In reality, Complex Analysis has many practical applications and is a distinct and important branch of mathematics.

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