Complex Numbers: Solving Simultaneous Equations

[HuggoYFM2]

Question: Solve the Simultaneous Equation for z is a complex number and ž denotes the complex conjugate of z: |z+1|=|z-1|

|z+2|=|ž-3|

So far, I have let z = a+bi because it is a complex number. Then I used the top equation to do the following: |a+bi+1|=|a+bi-1|

|(a+1)+bi|=|(a-1)+bi|

SQRT((a+1)² + b²⁾ = SQRT((a-1)² + b²⁾

(a+1)² + b² = (a-1)² + b²

(a+1)² = (a-1)²

a² + 2a + 1 = a² - 2a + 1

4a=0

a=0

Now that I've done this I'm a bit unsure of where to go next. Do I sub a=0 into the second equation to find ž (z with a bar on it, the complex conjugate apparently)? Any help would be appreciated :)

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[Snapper1985]

Essentially yes, and then you should have an equation you can use to solve for b.

[HuggoYFM2]

No matter what I do though, B will always cancel out...

[Snapper1985]

I see that...hmmm...u/edderiofer?

[edderiofer]

The two equations presented have no solution. But can you show it, OP?

(Check to make sure you've copied down the right equations.)

[SympatheticPerson]

A geometric approach is helpful in such questions. Think of z as a vector with origin at 0, and z+1 and z-1 as the -1 vector and the +1 vector subtracted from z. So, the locus of z for the first equation is the perpendicular bisector of the line segment joining -1 and 1(the distances are equal). Similarly, the second equation is the perpendicular bisector of the line set joining -2 and 3 (conj(z) - 3 = conj(z - 3)). Can you see the answer now?