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Finding the locus of points for complex analysis

  1. Sep 22, 2008 #1
    I think this should probably be easy, but I am stuck. My book is of no help.

    Find and describe the locus of points z satisfying the given equations:

    1. |z-i|=Re z

    2. |z-1|^2 =|z+1|^2 +6

    I am thinking for the 1st one that I have to square both sides, but then what? What happens to Re z?

    For the second one, I am totally lost. My first instinct would be to take the square root of both sides...and then...?

    Thanks to all.
  2. jcsd
  3. Sep 22, 2008 #2


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    Welcome to PF!

    Hi calculuskatie! Welcome to PF! :smile:

    Hint: |z - a| is just the distance from z to a.

    So just use geometry (including the cosine formula for triangles).

    How would you describe those two equations in geometry? :smile:
  4. Sep 22, 2008 #3


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    Have you tried replacing z by x+ iy so that each equation becomes an equation in x, y?

    That should make the curves easier to recognize.

    For example, if z= x+ iy, then z- i= x+iy- i= x+ (y-1)i and so [itex]|z-i|= |x+ (y-1)i|= \sqrt{(x+ (y-1)i)(x- (y-1)i)}[/itex] (because [itex]|z|= \sqrt{z \overline{z}}[/itex])
    [itex]= \sqrt{x^2- (y-1)^2}[/itex] and, of course, Re(z)= x. What kind of graph does [itex]\sqrt{x^2- (y-1)^2}= x[/itex] give? (Try squaring both sides.)

    Another, more geometric way to do this is to note that |z- i| is the distance from the point z, in the complex plane, to the point i. Re(z), on the other hand, is the distance from z to the y-axis. What kind of figure has distance from a point on the graph to a point equal to the distance from that point to a line?
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