Results 1 to 12 of 12

Math Help - oh boy do i need help - complex analysis

  1. #1
    Junior Member
    Joined
    Sep 2008
    From
    Edmonton
    Posts
    67

    oh boy do i need help - complex analysis

    http://www.math.ualberta.ca/~runde/files/ass411-1.pdf
    I can't do #3 or #4
    I have a feeling I can do 3 but I'm not quite sure where to go from
    and for 4 I know that to show it's not injective show that exp(0,0) = exp(0,2pi)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    3)The problem here is that how is really defined what \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} means? I assume it means that we think of f: D\to \mathbb{C} as f(x,y) = (u(x,y),v(x,y)) (since the complex numbers are simply points in \mathbb{R}^2 formally).
    Now define \frac{\partial f}{\partial x} = \left( \frac{\partial u}{\partial x}, \frac{\partial v}{\partial x} \right) and \frac{\partial f}{\partial y} = \left( \frac{\partial u}{\partial y}, \frac{\partial v}{\partial y} \right).
    This means \left( \frac{\partial u}{\partial x}, \frac{\partial v}{\partial x} \right) + i \left( \frac{\partial u}{\partial y}, \frac{\partial v}{\partial y} \right) = \left( \frac{\partial u}{\partial x}, \frac{\partial v}{\partial x} \right) + \left( - \frac{\partial v}{\partial y} , \frac{\partial u}{\partial y} \right).

    However, \frac{\partial u}{\partial x} - \frac{\partial u}{\partial y} = 0 and \frac{\partial v}{\partial x} + \frac{\partial u}{\partial y} = 0 by Cauchy-Riemann equations.

    4)Note \exp(0) = \exp(2\pi i) thus it is not injective.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2008
    From
    Edmonton
    Posts
    67
    thank you
    now i'll try 5 and 6 on my own for a while before asking help for those
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Aug 2008
    Posts
    903
    May I offer a different perspective?

    If x=\frac{z+\overline{z}}{2} and y=\frac{z-\overline{z}}{2i} then:

    <br />
\frac{\partial f}{\partial \overline{z}}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial \overline{z}}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial \overline{z}}+i\left(\frac{\partial v}{\partial x}\frac{\partial x}{\partial \overline{z}}+\frac{\partial v}{\partial y}\frac{\partial y}{\partial \overline{z}}\right)
    ..... =\frac{1}{2}\left(\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y}\right)+\frac{i}{2}\left(\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}\right)<br />

    Well there you go: If f is analytic then the partials satisfy the Cauchy-Riemann equations so \frac{\partial f}{\partial \overline{z}}=0 and the only way for
    \frac{1}{2}\left(\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y}\right)+\frac{i}{2}\left(\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}\right)=0
    is for the partials to satisfy the Cauchy-Riemann equations, that is, f must be analytic.
    Last edited by shawsend; September 14th 2008 at 03:43 PM. Reason: corrected formulas
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Sep 2008
    From
    Edmonton
    Posts
    67
    thank you but i got that one already
    also for #4
    where it says show that exp(C) = C/{0}
    what does that mean? the notation is unfamiliar to me and hasn't been explained in course lecture or notes
    also there is no textbook for the class which makes this particularly frustrating
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Aug 2008
    Posts
    903
    Hey jb. It's a function at a domain. Suppose I define the unit circle as the domain D=\left\{z : |z|<=1\right\}

    Then I could say, what is the function at that domain? or what is f(D) meaning how does the function map the domain D. Same with e^{\mathbb{C}}. How does the exponential function map all of \mathbb{C}? It does so by the expression:

    e^{\mathbb{C}}=\mathbb{C}\backslash\{0\} meaning it maps all of \mathbb{C} to the deleted neighborhood: \mathbb{C}\backslash\{0\} which is \mathbb{C} minus the origin.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Sep 2008
    From
    Edmonton
    Posts
    67
    thank you
    the complex plane not including the origin <- thats what I needed to know
    I'm still at a loss to show how the exp maps to that but I'm sure I'll figure it out by tonight
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    If \alpha \in \mathbb{C}^{\times} then you should know the fact that is is possible to write \alpha = re^{i\theta} where r>0. Now let z = \log r + i\theta. Then e^z = e^{\log r + i \theta} = re^{i\theta} = \alpha. This shows that the equation e^z = \alpha always has a solution for any \alpha\in \mathbb{C}^{\times}. Thus, the function \exp : \mathbb{C} \to \mathbb{C}^{\times} is onto.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Sep 2008
    From
    Edmonton
    Posts
    67
    umm I got #5 a and b d
    but I can't get c
    also if it's not too much of a bother could someone give hints at 6
    I don't want them entirely answered for me because I still want to be learning this stuff
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    It may be late (since it's past 3pm...), but I don't care
    I'm also learning how to do this stuff !

    For #5 c), use Cauchy-Hadamard theorem : Cauchy-Hadamard theorem - Wikipedia, the free encyclopedia
    My teacher says the best way above all to solve this problem is to use Abel's theorem :
    R=\sup \{ ~ r \in \mathbb{R} / a_n r^n \text{ is bounded } \}
    (R is the radius of convergence)
    Here, Hadamard's theorem is more straightforward.


    #6 is above my capacities...
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by Moo View Post
    #6 is above my capacities...
    You can find it here.

    And root test works wonderful for 5c.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by ThePerfectHacker View Post
    You can find it here.

    And root test works wonderful for 5c.
    Cauchy-Hadamard's theorem is the name for this "root test".
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: October 4th 2011, 05:30 AM
  2. Replies: 6
    Last Post: September 13th 2011, 07:16 AM
  3. Replies: 1
    Last Post: October 2nd 2010, 01:54 PM
  4. Replies: 12
    Last Post: June 2nd 2010, 02:30 PM
  5. Replies: 1
    Last Post: March 3rd 2008, 07:17 AM

Search Tags


/mathhelpforum @mathhelpforum