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Complex Number Question(s)

  1. Jan 31, 2016 #1
    1. The problem statement, all variables and given/known data
    a) Solve equation z + 2i z(with a line above it i.e. complex conjugate) = -9 +2i
    I want it in the form x + iy and I am solving for z.

    b)
    The equation |z-9+9i| = |z-6+3i| describes the straight line in the complex plane that is the perpendicular bisector of the line segment from 9-9i to 6-3i.
    Find the value of m and c

    2. Relevant equations
    Complex conjugate: z = x + iy, z(line above it) = x -iy


    3. The attempt at a solution
    a) I'm not exactly sure how to approach this but I have a few ideas.
    1. let z = -9 and 2i z(line above) = 2i and solve that way
    2. Let the entire equation = z, however when I do this what do I do about the complex conjugate z?
    3. Let entire equation = z complex conjugate and change the signs?
    I really am unsure which will work.

    b) If you write it in term of x +iy, you can write the equation like so: y = mx+c.
    I now need to find the values of m and c, my textbook literally jumps to insanely easy complex numbers to questions like these, I presume they are just difficultly worded questions and that they aren't actually difficult. Any help on how to approach these questions?
     
  2. jcsd
  3. Jan 31, 2016 #2

    HallsofIvy

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    These are only "difficultly worded" if you don't understand what the words mean. What points in the plane represent the complex numbers 9- 9i and 6- 3i? What is the equation of the line through those two points? What is the equation of the perpendicular bisector of that line segment? [itex]|x+ iy|= \sqrt{x^2+ y^2}[/itex] so how can you write |z- 9- 9i|= |z-6+3i| in terms of x and y? What do you get when you square both sides of that?
     
    Last edited: Jan 31, 2016
  4. Jan 31, 2016 #3

    SteamKing

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    Let's take a) which is relatively simple.

    You need to solve ##z + 2i\bar z = -9 +2i##

    Use the definition of ##z = x + iy## and ##\bar z = x - iy## and substitute these back into the original equation. Do the multiplication of ##\bar z## and ##2i## as indicated.

    You find z by equating the real part and the complex part on both sides of the equation.
     
  5. Jan 31, 2016 #4
    So rewrite it as... x+iy +2i x-iy = -9 +2i?

    I understand what you mean by equating the real part and the complex part on both sides of the equation i.e. reals and reals together, imaginary and imaginary together right? :)
    but what do you mean by "Do the multiplication of z(conjugate) +2i as indicated? :)
     
  6. Jan 31, 2016 #5

    SteamKing

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    Exactly that. Calculate the product ##2i ⋅ (x - iy)##. What do you get when you multiply everything out?
     
  7. Jan 31, 2016 #6

    HallsofIvy

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    No, you rewrite it as x+ iy- 2i(x- iy)= -9+ 2i. Do you see the difference?

    He means "multiply the two complex numbers"!
     
  8. Jan 31, 2016 #7
    when you multiply both those together you get -2i2y but because i2 = -1 you get 2ix + 2? :/
    Then you can rewrite it as follows right?
    x+ iy- 2ix +2= -9+ 2i
     
  9. Jan 31, 2016 #8

    SteamKing

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    What happened to the y when you multiplied 2i by -iy ?
     
  10. Jan 31, 2016 #9
    (2i)(x-iy)

    (2i)(x) + (2i)(-iy)

    2ix - 2i^2y (i^2 = -1)

    2ix +2y whoops sorry. Is that better?
     
  11. Jan 31, 2016 #10

    SteamKing

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    Yep.
     
  12. Jan 31, 2016 #11
    x + iy +2ix +2y = -9 + 2i

    Now I want to put Re numbers with Re numbers and the Im with the Im numbers so...
    x + 2y + 9 = -2ix -iy +2i

    Am I making progress here? :)
     
  13. Jan 31, 2016 #12

    SteamKing

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    A little, but it's all jumbled up.

    Remember, you want to equate the real parts on the left side of the equation with the real parts on the other, namely the -9 + 2i quantity. Same for the complex parts.
    You turn the original equation into two new equations in this process.
     
  14. Jan 31, 2016 #13
    So in other words, keep it like in this form: x + iy +2ix +2y = -9 + 2i
    Then equate Re with Re and Im with Im so for example, the Real parts in this case...

    x + 2y = -9 ?
     
  15. Jan 31, 2016 #14

    SteamKing

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    Yep. And do the same for the imaginary parts. You should wind up with two equations in two unknowns. Solve for x and y.
     
  16. Jan 31, 2016 #15
    x+2y = -9
    2x + y = 2

    Solve simultaneously and I should get the value of x and y to be fractions, then it will look something like x = a/b, y = c/di
    z = a/b + c/di where a,b,c,d represent numbers? :)
     
  17. Jan 31, 2016 #16

    SteamKing

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    Yes. And you can check your calculations by substituting the values of x and y back into the original equation.
     
  18. Jan 31, 2016 #17
    Woo :D I understand now! thanks!! huge help! now for Question b.
    As HallsofIvy said above, |z-9+9i| can be written in the form √x^2 + y^2. However in |z-9+9i| I have 3 variables, a z, a real number and an imaginary number. do I just replace the z with x+iy again in this case?
     
  19. Jan 31, 2016 #18

    SteamKing

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    Yes. The definition z = x + iy can always be used.
     
  20. Feb 1, 2016 #19
    Sorry for the late response I was studying.
    |z-9+9i| = |x+iy-9+9i|
    |z-6+3i| = |x+iy-6+3i|
    |x+iy-9+9i| = |x+iy-6+3i| or √(-9+x)^2 + (9i +iy)^2 = √(-6+x)^2 + (3i+iy)^2

    Is this correct?
     
  21. Feb 1, 2016 #20
     
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