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Homework Help: Complex Analysis question

  1. Dec 20, 2005 #1
    Prove that if wz = 0, then w = 0 or z = 0. w and z are two complex numbers.
    I said that w = a + bi and z = c + di and set wz = 0. I got down to c(a+b) = d(b-a), but don't know where to go from here.

    I'm trying to teach myself complex analysis, anyone know any good sources? I took the 3 calculus classes and an elementary differential equations class. I'm a mechanical engineering major though and I don't think I'll have time to take a complex analysis class unless I want to stay in college after I get my BS in ME.
  2. jcsd
  3. Dec 20, 2005 #2


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    You can solve it like a system of variables:
    so (case i)
    since [itex]a[/itex] and [itex]b[/itex] are real numbers this gives that [itex]z=0[/itex]
    otherwise (case ii) c=0
    so, [itex]d=0[/itex] so we have [itex]w=0[/itex] or [itex]a=0 \rm{and} b=0[/itex] and we have [itex]z=0[/itex].
  4. Dec 20, 2005 #3
    Got it. Thanks.
    Know any good sources to learn complex analysis from?

    EDIT: I guess it would be smart to know how big of a problem I'm trying to tackle here. Is self-teaching complex analysis going to be a hard task? Is it comparable to...say, someone who only knows algebra trying to teach himself calculus?
    Last edited: Dec 20, 2005
  5. Dec 20, 2005 #4
    I like the approach used in Tristan Needham's Visual Complex Analysis, but as you've not yet built up any intuition for complex numbers, you may want to accompany it with a more traditional text.
  6. Dec 20, 2005 #5

    matt grime

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    Surely it is far easier to simply use |vw|=|v||w|? This isn't complex analysis.
  7. Dec 22, 2005 #6
    More questions

    I attached a file with three more questions in it. I did the first two using:
    z = x + yi and w = a + bi and solved them down in the same manner like the first question I posted. Is there an easier way to do these?

    I'm not really sure what to do for #6 though. Do I just set z = x + yi and find an equation f(y) = f(x) then graph it on an x-y plane? I did that for part (a) and got a circle with center (2, -3) and radius 2, but is that what I'm supposed to do?


    Attached Files:

  8. Dec 22, 2005 #7
    Try thinking about what the equations are saying geometrically. In the case of the first one that you solved, it is saying that the distance between the points z and the point (2,-3) is a constant 2. That's just a circle of radius 2 centered at (2,-3) with no algebra necessary.
    The second equation says that the distance between the points z and the point (0,2) is less than or equal to 1, which describes a closed disc around that point of radius 1. What about the geometry described by (d) ?
    Use the algebraic method only if the geometry is hopelessly obscured by the equation. Even then, you can now use the geometric result you found algebraically for later questions.
  9. Dec 22, 2005 #8
    Ahh thanks a lot...that makes things a bit easier. See if I got those right:

    b) The distance between z and -2i is less than or equal to 1. Which means it's a disk centered at -2i with a radius of 1.
    c) The real part of z is equal to 4. Straight vertical line at 4 because the imaginary part can take on any value.
    d) The distance between z and (1, -2i) is equal to the distance between z and (-3, -i). z again can be any value on a straight line that's a perpendicular bisector to the line connecting (1, -2i) and (-3, -i).
    e) Ellipse with focii (-1, 0i) and (1, 0i).
    f) Hyperbola with the same focii as part e.

    By the way, when you write (2, -3) are you talking about the point a = 2 - 3i in the complex plane? When I graph those lines and ellipses and circles, do I graph them all in a complex plane with an imaginary axis and a real one?
    Last edited: Dec 22, 2005
  10. Dec 22, 2005 #9
    Yep, all those interpretations are spot on. :smile:
    If you take the complex plane to be a 2-dimensional real vector space with basis vectors 1 and i, then you can write the element a+bi as the (position) vector (a,b) and preserve mathematical rigour.
    Last edited: Dec 22, 2005
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