Complex Variables: |z-i|=Re(z) Locus of Points

[kingwinner]

Homework Statement

Let z=x+iy. Describe the locus of points z satisfying |z-i|=Re(z).

Homework Equations

N/A

The Attempt at a Solution

|z-i|=Re(z)
=> sqrt[x²+(y-1)²] = x
=> x²+(y-1)² = x²
=> (y-1)² = 0
=> y=1

But the answer says y=1 and x≥0. Why? I think maybe I'm not getting the x≥0 because my implcations above isn't "if and only if". But what exactly is the reason and how can we rigorously justify that x≥0?

Any help is appreciated!

 

[Office_Shredder]

If x<0, what sign is Re(z)? What sign does |z-i| have to be?

The place where you lost the if and only if is here:

=> sqrt[x²+(y-1)²] = x
=> x²+(y-1)² = x²

The if is perfectly fine, but when you try to go backwards, you're taking square roots so you don't know if the signs match up.

Whenever you have to divide by stuff or square things and aren't sure whether you've been doing if and only if statements the whole time, one way to check is to just take your solution and plug it back into the original problem. You know that every solution has to be of the form you have, but now you can see if there are any further restrictions involved. here we would spot it easily. If y=1 then

|z-i|=|x+i-i|=|x|

And Re(z)=x

So we need |x|=x