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Complex equation

  1. Aug 31, 2011 #1

    I am trying to solve arcsin(z)=w. Where z is a complex number z=x+iy.

    I got to this.


    where p= exp(i*w)

    I dont know how to solve this quadratic equation.

    This is not homework, I got this problem out of my head. I am curious.

    Thank you.
    Last edited: Aug 31, 2011
  2. jcsd
  3. Aug 31, 2011 #2
    You can solve that quadratic I bet. It's just:


    so using the quadratic formula for ap^2+bp+c=0:


    now just substitute what a, b and c are. That is a=1, b=-2iz, and c=-1 right?

    then let p=e^{iw}, take (multi) logs of both sides to get:


    bet you can get that and note I didn't use plus minus since in the big leagues, that square root symbol really means the multi-valued square root function and the log symbol means the multi-valued log function but don't worry about that for now.
  4. Aug 31, 2011 #3
    yup its clear, just correction "ln" instead of "log"(im meticulous). Again why not a minus sign too? I didn't understand why we dont include that
  5. Aug 31, 2011 #4
    Ok, I can tell you exactly why. Cus' you're using z and w and not x and y. You're implying the complex-variable version of the inverse sine function and in Complex Analysis, the square root, log, and inverse trig functions are multi-valued and not just the single-valued real-version you use in high-school. It's ok to use plus, minus for the square root in that expression but then you know, that's baby-looking. Also, the log function in that expression is actually infinitely-valued. Best then to treat these functions as a single multivalued entity and deal with them appropriately. That is, when you actually use it to compute something, you're smart enough to realize the expression is multivalued and if necessary, manually extract the plus and minus version of the root object for each value of the log object. That is, the function is actually doubly infinitely valued, one set of infinite values for each root although the purist in here may object to calling it twice infinity.
  6. Aug 31, 2011 #5
    But this gave me another puzzle. How does graph of complex square root look like?

    I've seen some color graphs but i don't know what they mean.

    But then again

    consider equation

    the solutions are:


    This isn't one and the same complex number, how come we include +/- here then?
    Last edited: Aug 31, 2011
  7. Aug 31, 2011 #6


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    Homework Helper

    The reason is that jackmell is using the convention that in complex analysis, [itex]\sqrt{\dots}[/itex] is understood to be multi-valued, so the "plus/minus" sign is implict. Alternatively, because the two roots of [itex]z^{1/2}[/itex] need not be purely real, one prefers, I suppose, to think of the two roots as being [itex]\pi[/itex] radians apart in the complex plane rather than negatives of one another (even though it does amount to a minus sign difference), so the plus/minus isn't used.

    Also, in complex analysis, only log base e is ever used, so the notation "log" refers only to the log base e. Even in ordinary math, "log" is typically understood to mean base e. The only time I really see "ln" when dealing with complex numbers is when one rights the complex logarithm in terms of its real and imaginary parts:

    [tex]\log z = \ln |z| + i\mbox{arg}(z).[/tex]
  8. Aug 31, 2011 #7
    but log z still has base "e" here? I found that confusing when i saw it in books.
  9. Aug 31, 2011 #8


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    Homework Helper

    Yes. The complex logarithm is always base e. For this reason, and because mathematicians tend to use "log(x)" to mean [itex]\log_e(x)[/itex] even when working only with real numbers, the complex logarithm is just written log(z), because there is no other base it could be.
  10. Sep 1, 2011 #9
    Got Mathematica? Really, I think it's essential to get good with it if you're in a technical field. You know, why use a hand saw when you can just buy a skill saw? I mean really. Anyway, here's the code to generate a simple plot of the imaginary component of the complex square root function:

    Code (Text):
    p1 = Plot3D[Im[Sqrt[x + I y]], {x, -1, 1}, {y, -1, 1}]
    p2 = Plot3D[Im[-Sqrt[x + I y]], {x, -1, 1}, {y, -1, 1}]
    Show[{p1, p2}, PlotRange -> All]
    nothing fancy, we could do better. So that if you take [itex]\sqrt{z}[/itex] letting z=x+iy in the x-y coordinate plane, and put a point for each of the (double) value of the imaginary part of square root of x+iy, connect them all together and make surfaces, you get that plot.
  11. Sep 1, 2011 #10

    I am fond of matlab. But even in matlab im not that crafty. I will google something up. Thank you.
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