1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Complex analysis

  1. Jun 6, 2014 #1
    1. The problem statement, all variables and given/known data

    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?

    2. Relevant equations

    3. 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.
  2. jcsd
  3. Jun 7, 2014 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper

    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?
  4. Jun 7, 2014 #3
    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?
  5. Jun 7, 2014 #4
    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.
  6. Jun 7, 2014 #5
    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?
  7. Jun 7, 2014 #6
    Can you prove it?
  8. Jun 7, 2014 #7


    User Avatar
    Homework Helper

    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?
  9. Jun 7, 2014 #8
    It seems the equation fails for values of z where when you square them the argument is outside of 0 to 2pi
  10. Jun 7, 2014 #9


    User Avatar
    Homework Helper

    ^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)
  11. Jun 7, 2014 #10
    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?
  12. Jun 8, 2014 #11


    User Avatar
    Homework Helper

    ^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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted