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Complex Analysis - Values of Real and Imaginary parts

  1. Mar 12, 2012 #1
    1. The problem statement, all variables and given/known data
    Simplify in terms of real and imaginary parts of x and y and sketch them.

    1) Re [itex]\frac{z}{z-1}[/itex] = 0
    2) Im [itex]\frac{1}{z}[/itex] ≥ 1


    3. The attempt at a solution

    1)

    [itex]\frac{x + iy}{x + iy -1}[/itex] = 0

    Am I allowed to just vanish the imaginary components here and have [itex]\frac{x}{x -1}[/itex]?

    If not, I was thinking split up the fraction, and have [itex]\frac{x}{x + iy -1}[/itex] = [itex]\frac{-iy}{x + iy -1}[/itex]

    Hence, x = -iy, or x + iy = 0, and for the real component: x = 0

    2)

    [itex]\frac{1}{x + iy}[/itex] ≥ 1

    1 ≤ x + iy where y ≥ 1 for the imaginary component.

    I'm not very confident of these answers at all.
     
  2. jcsd
  3. Mar 12, 2012 #2

    tiny-tim

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    Hi NewtonianAlch! :smile:

    The trick, which your professor should have told you, is to multiply top and bottom by the complex conjugate of the bottom …

    that makes the bottom real! :wink:
     
  4. Mar 12, 2012 #3
    Hmm, I actually tried that before and got a whole jumble that I couldn't realistically do much with. Perhaps I missed something, I'll try it again.

    Thanks!
     
  5. Mar 13, 2012 #4
    Not quite getting anything I expected.

    For c) I get [itex]\frac{x^{2}+y^{2}-x-iy}{x^{2}+y^{2}-2x+1}[/itex] = 0

    Which does not look like anything I can simplify to something meaningful. What have I done wrong?
     
  6. Mar 13, 2012 #5
    Your answer is fine except for the -iy term on the top, there shouldn't be any imaginary terms in this fraction.

    So now you have that equation equal to 0. Your goal is to find out what kind of equation this is. You need to do a little bit of algebra to solve for it, the answer actually comes out to a nice looking formula. Be warned that you need to throw out one of the solutions to the final equation when you finally solve for it (why?).
     
  7. Mar 13, 2012 #6
    Well I multiplied through by (x - iy - 1), which would mean there would be -iy hanging around at the top.

    (x + iy)*(x- iy -1)

    Unless, the complex conjugate is (x - iy), but in that case, there would be no -x term at the top either.
     
  8. Mar 13, 2012 #7
    Remember your equation is [itex]Re(\frac{z}{z-1}) = 0[/itex]. You will have two imaginary terms and two real terms in the numerator. Split the fraction up and only consider the real part!
     
  9. Mar 13, 2012 #8
    Damn...the obvious isn't obvious when the questions change.
     
  10. Mar 13, 2012 #9

    tiny-tim

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    Hi NewtonianAlch! :smile:

    (just got up :zzz:)
    But that's fine! :smile:

    You've lost the plot :rolleyes:

    the plot was to prove that the Re was 0 …

    so we made the bottom real, so now all we need to prove is that Re(the top) = 0

    ie… ? :smile:
     
  11. Mar 13, 2012 #10
    So, x^2 + y^2 - x = 0

    Which is clearly a circle of some sort, but...it's shifted onto the positive x-axis...?
     
  12. Mar 13, 2012 #11
    Almost correct. You are right that that is the form a circle, now which circle exactly? (Hint: complete the square on the x-terms). Also, This circle is missing a single point from it, what point is that? (Hint: this is the point that will set the denominator of [itex]Re(\frac{z}{z-1})[/itex]) equal to 0.

    You're almost there!
     
  13. Mar 13, 2012 #12
    (x-1/2)^2 + y^2 = 1/4 ?

    How can the denominator be zero, that would make it infinity. Although I do remember something in the lectures about some circles going through the origin excluding that point in the origin, so I guess it would be that point. Though I'm not sure.

    So it means the circle is centred at x = 1/2.
     
  14. Mar 13, 2012 #13
    Yep!

    Right. So that is why you need to exclude whatever points make the bottom equal to zero. What is/are the point/points?
     
  15. Mar 13, 2012 #14
    The bottom was x^2 - 2x + y^2 + 1

    So for that to equal zero; x^2 - 2x + y^2 should equal -1

    Which results in a quadratic equation, and there are numerous values that could now go in here.

    I plotted this thing in Maple, and it doesn't look too suspicious, and I cannot see a broken circumference in the circle.

    I should have asked really. What made you realise that there is an exclusion point here? I cannot see from the equation any clue as to how one can recognise it.
     
  16. Mar 13, 2012 #15

    tiny-tim

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    complete the square again!

    but you needn't solve that equation …

    just go back to the original question, which has 0 on the bottom only fo z = 1 :wink:

    (btw you multiplied by 0/0 when you multiplied top and bottom by x + iy - 1 if x + iy = 1 ! :biggrin:)
     
  17. Mar 13, 2012 #16
    Hmm thanks, yea I did consider that initially, but thought that to find the exact point I'd have to look through the fuller equation.

    How does one know by looking at the numerator equation that there is going to be an exclusion point on the circle?
     
  18. Mar 13, 2012 #17

    tiny-tim

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    You don't!

    As scurty :smile: says, you look at the denominator! :wink:
     
  19. Mar 13, 2012 #18
    Only if one of those points that satisfy the numerator makes the denominator 0. :)

    Edit: Didn't see Tim's post on the new page. What he said!
     
  20. Mar 13, 2012 #19
    Didn't think such a tiny problem went so in-depth. Thanks so much for your help! I still have a long way to go with Complex Analysis, and then there's Vector Calculus still...*sigh*
     
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