# Finding the locus of points for complex analysis

1. Sep 22, 2008

### calculuskatie

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. Sep 22, 2008

### tiny-tim

Welcome to PF!

Hi calculuskatie! Welcome to PF!

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?

3. Sep 22, 2008

### HallsofIvy

Staff Emeritus
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 $|z-i|= |x+ (y-1)i|= \sqrt{(x+ (y-1)i)(x- (y-1)i)}$ (because $|z|= \sqrt{z \overline{z}}$)
$= \sqrt{x^2- (y-1)^2}$ and, of course, Re(z)= x. What kind of graph does $\sqrt{x^2- (y-1)^2}= x$ 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?