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Complex analysis

  • Thread starter DotKite
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  • #1
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Homework Statement



Let f(z) = sqrt(z) be the branch of the square root function with sqrt(z) = (r^1/2) (e^iΘ/2),
0≤Θ<2[itex]\pi[/itex], r > 0

(a) for what values of z is sqrt(z^2) = z?

(b) Which part of the complex plane stretches, and which part shrinks under this transformation?

Homework Equations





The Attempt at a Solution



Ok so for this branch i believe the function will map all points within 0≤Θ<2[itex]\pi[/itex] to the upper half plane (ie Im(f) > 0).

I do not really understand what part a is asking and for part b it seems everything is shrunk.
 

Answers and Replies

  • #2
Simon Bridge
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For part (a), it is asking for ##z:f(z^2)=z##
For part (b) please show your reasoning. What does it mean to say that the complex plane has shrunk or stretched? How would you tell?
 
  • #3
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For part (a), it is asking for ##z:f(z^2)=z##
For part (b) please show your reasoning. What does it mean to say that the complex plane has shrunk or stretched? How would you tell?
For part a do we have to consider the principal nth root? Or particular branches of the square root function?

In otherwords where this function is not multivalued?
 
  • #4
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For part a do we have to consider the principal nth root? Or particular branches of the square root function?

In otherwords where this function is not multivalued?
No, the function is not multivalued. Your OP has given the branch which you should consider:

[tex]f(re^{i\theta}) = \sqrt{r} e^{i\theta/2}[/tex]

where ##0\leq \theta< 2\pi##. That last restriction on ##\theta## makes sure it's not multivalued.
 
  • #5
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No, the function is not multivalued. Your OP has given the branch which you should consider:

[tex]f(re^{i\theta}) = \sqrt{r} e^{i\theta/2}[/tex]

where ##0\leq \theta< 2\pi##. That last restriction on ##\theta## makes sure it's not multivalued.
then wouldnt part a be true for all z then? It seems obvious that sqrt(z^2) = z for all z. Why would that not be the case?
 
  • #6
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then wouldnt part a be true for all z then? It seems obvious that sqrt(z^2) = z for all z. Why would that not be the case?
Can you prove it?
 
  • #7
pasmith
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then wouldnt part a be true for all z then? It seems obvious that sqrt(z^2) = z for all z. Why would that not be the case?
Let [itex]z = e^{i3\pi/2}[/itex]. Then
[tex]z^2 = (e^{i3\pi/2})^2 = e^{i3\pi} = e^{i\pi}[/tex] since we need [itex]0 \leq \arg(z^2) < 2\pi[/itex]. But then
[tex]f(z^2) = (e^{i\pi})^{1/2} = e^{i\pi/2} \neq z.[/tex]
That's one [itex]z[/itex] for which [itex]f(z^2) \neq z[/itex]. Are there others?
 
  • #8
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It seems the equation fails for values of z where when you square them the argument is outside of 0 to 2pi
 
  • #9
lurflurf
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^Good which z are those? What can you say about their real and imaginary parts?

For (b) Suppose we have two nearby points so that d(P1,P2)=h
what can we say about d(sqrt(P1),sqrt(P2))?
which is bigger? What is the formula for distance? (you could also consider areas)
 
  • #10
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^Good which z are those? What can you say about their real and imaginary parts?

For (b) Suppose we have two nearby points so that d(P1,P2)=h
what can we say about d(sqrt(P1),sqrt(P2))?
which is bigger? What is the formula for distance? (you could also consider areas)
For part a it would be all z such that 0≤arg(z)≤pi. Therefore im(z) > 0.

Is the distance formula just the standard Euklidian distance formula?
 
  • #11
lurflurf
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^yes consider a short segment in the complex plane
If the length is h say what is the length after taking square root
Where in the plane does a segment stretch and where does it shrink?
Hint find where the length does not change. That region will separate the other two.
 

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