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Complex Plane Regions

  1. Jan 18, 2014 #1
    1. The problem statement, all variables and given/known data

    Shade each region in the complex plane. Justify your solution.

    1) z - Conjugate[z] = 4

    2) 1 + z, where |z| = 1

    3. The attempt at a solution

    So for my attempt for 1 is:

    Let z = x + iy therefore Conjugate[z] = x - iy

    z - Conjugate[z] = 4
    x + iy - (x - iy) = 4
    2iy = 4
    iy = 2 ***multiply both sides by i
    -y = 2i
    y = -2i

    Now from my understanding y is supposed to be a real number, is it not? So what exactly does
    y = -2i represent? What region would it be in the complex plane?

    And my attempt for 2:

    |z| = 1 represents a circle of unit radius 1.
    Given 1+z and |z|=1, I changed this to into the equation of a disk in the complex plane (can I even do that?)

    Let z = x + iy
    |z+1| < 1 ***For this step would I use less than, less than or equal to, or just equal to?
    |x+iy+1| < 1
    (x+1)2 + y2 < 12

    Now this gives me an equation for a circle of radius 1 centred around the point (-1,0)
    Is this correct?

    Thanks
     
  2. jcsd
  3. Jan 18, 2014 #2

    Dick

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    z-conjugate(z)=4 has no solutions in the complex plane, as you've correctly deduced. So there is no region. It's just the empty set. For the second part, if |z|=1 is a circle around z=0 of radius 1 then isn't |z-c|=1 a circle around z=c of radius 1? |(z+1)-1|=|z|=1.
     
  4. Jan 18, 2014 #3
    I do not understand how z - conjugate(z)=4 has no solutions in the complex plane. Can you explain that?
     
  5. Jan 18, 2014 #4

    Dick

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    You explained it. z-conjugate(z) is a purely imaginary number, it's 2yi where y is real. 4 is real and not zero. They can't be equal.
     
    Last edited: Jan 18, 2014
  6. Jan 18, 2014 #5
    Okay that makes sense, thank you!!
     
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