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A question about complex numbers

  1. Oct 13, 2015 #1
    1. The problem statement, all variables and given/known data
    upload_2015-10-13_18-24-33.png
    I don't understand example 2. For part a, I got a slightly different answer.

    2. Relevant equations
    see picture

    3. The attempt at a solution
    ##|z-1|=2=\sqrt{x^2+y^2-1}##
    ##4=x^2+y^2-1 \neq (x-1)^2+y^2##
     

    Attached Files:

  2. jcsd
  3. Oct 13, 2015 #2

    berkeman

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    Staff: Mentor

    But the -1 is real, isn't it? Why would it affect the y (imaginary) axis? I may not understand the question well enough, though.

    EDIT -- Well, that math quote didn't work so well, eh?
     
  4. Oct 14, 2015 #3

    SteamKing

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    Staff Emeritus
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    Homework Helper

    Expressing z into its real and imaginary parts:
    z = x + iy

    We can also create another complex number, w, such that:

    w = u + iv

    We can also specify that

    u = x - 1
    v = y

    So that

    w = u + iv = (x-1) + iy = z - 1

    If we let the modulus of (z - 1) = 2, then

    |w| = |z - 1| = 2

    then by the definition of the modulus of a complex number, we get

    |w| = (u2 + v2)1/2 = 2

    Squaring both sides and substituting for u and v,

    u2 + v2 = 4 or

    (x - 1)2 + y2 = 4

    which is the equation of a circle of radius = 2 centered at (1, 0).
     
  5. Oct 14, 2015 #4

    Mark44

    Staff: Mentor

    Geometrically, the equation |z - 1| = 2 represents all of the points z in the complex plane that are 2 units away from 1; i.e., 1 + 0i. The geometry here should suggest that we're dealing with a circle of radius 2, centered at (1, 0).

    For the algebra, a slightly different tack from the one SteamKing took...
    ##|z - 1| = 2##
    ##\Rightarrow |x + yi - 1| = 2##
    ##\Rightarrow \sqrt{ [(x - 1) + yi] \cdot [(x - 1) - yi]} = 2## (using the fact that ##|z| = \sqrt{z \cdot \bar{z}}##)
    ##\Rightarrow \sqrt{(x - 1)^2 + y^2} = 2##
    ##\Rightarrow (x - 1)^2 + y^2 = 4##
     
  6. Oct 14, 2015 #5

    Given ##|z-1|## how did you decide to make the substitution of ##u=x-1##? Why not ##y-1##? So my method was incorrect? While it is true that ##|z| = \sqrt{x^2 +y^2}##, ##|z-1| \neq \sqrt{x^2 +y^2-1}##? Thank you.
     
    Last edited by a moderator: Oct 14, 2015
  7. Oct 14, 2015 #6
    Ok, this method makes a lot of sense. But just to clarify, was it wrong that I approached this question with ##|z-1|=\sqrt{x^2+y^2-1}##
     
  8. Oct 14, 2015 #7

    Mark44

    Staff: Mentor

    Yes, because ##\sqrt{x^2+y^2-1} \ne \sqrt{(x - 1)^2 + y^2}##
     
  9. Oct 14, 2015 #8
    Okay thank you
     
  10. Oct 14, 2015 #9
    It is easier for me to follow Mark44's method, but I still thank you for your response.
     
  11. Oct 14, 2015 #10

    Mark44

    Staff: Mentor

    Not even close. Subtracting 1 from z doesn't translate into subtracting 1 from the stuff inside the radical. The operations involved are very non-linear.

    Forget complex numbers for a moment, and consider the hypotenuse on a 3-4-5 right triangle. We have ##5 = \sqrt{3^2 + 4^2}##. If we subtract 1 from 5, does that correspond to a subtraction of 1 on the quantity in the radical? IOW, does ##5 - 1 = \sqrt{3^2 + 4^2 - 1}##? I.e., is ##4 = \sqrt{24}##? That's essentially what you're trying to do above.
     
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