Complex Analysis: Solving z^2 = sqrt(z) & Mapping of Plane

In summary: For part a it would be all z such that 0≤arg(z)≤pi. Therefore im(z) > 0. For part b it seems that if the distance between the points is h then sqrt(P1) > sqrt(P2).
  • #1
DotKite
81
1

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.
 
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  • #2
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
Simon Bridge said:
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
DotKite said:
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
micromass said:
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 wouldn't 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
DotKite said:
then wouldn't 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
DotKite said:
then wouldn't 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
It seems the equation fails for values of z where when you square them the argument is outside of 0 to 2pi
 
  • #9
^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
lurflurf said:
^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
^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.
 

1. What is complex analysis?

Complex analysis is a branch of mathematics that deals with the study of complex numbers and their functions. It involves the use of complex numbers, which are numbers that have both a real and imaginary component, to solve mathematical problems.

2. What is the equation z^2 = sqrt(z)?

The equation z^2 = sqrt(z) is a complex equation that involves a variable, z, raised to the power of 2 and also under a square root. It is a common equation in complex analysis and can be solved using various techniques such as mapping of the complex plane.

3. How do you solve z^2 = sqrt(z)?

To solve z^2 = sqrt(z), one approach is to rewrite the equation as z^2 - sqrt(z) = 0 and then use the quadratic formula. Another approach is to use the polar form of complex numbers and equate the real and imaginary parts to solve for z.

4. What is the mapping of the complex plane?

The mapping of the complex plane is a technique used in complex analysis to transform complex numbers onto a different coordinate system. It involves using functions or transformations to map points from the complex plane onto another plane, such as the Cartesian plane.

5. How is the mapping of the complex plane used in solving equations like z^2 = sqrt(z)?

The mapping of the complex plane can be used to transform the equation z^2 = sqrt(z) into a simpler form that is easier to solve. By mapping the complex numbers onto the Cartesian plane, the equation can be visualized and solved using geometric techniques such as finding the intersection of curves.

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