Find all complex numbers z such that $|z-1|=|z+3|=|z-i|$.

Question

Find all complex numbers $z$ such that $|z-1|=|z+3|=|z-i|$.

I am not sure how to do this at all. I tried graphing $|z-1| = |z+3|$ and I figured out that that was a straight line. Is the easiest way to graph them and find their intersection/intersections, or is there another was because I am not sure how to graph $|z-1| = |z-i|$ or $|z+3|=|z-i|$. Thank you very much

Answer 1

Hint: $\;\require{cancel}|z-1|=|z+3|$ $\iff (z-1)(\bar z -1) = (z+3)(\bar z + 3)$ $\iff \bcancel{|z|^2} - (z+ \bar z)+1 = \bcancel{|z|^2} + 3(z+\bar z) + 9$ $\iff \operatorname{Re}(z) = \frac{1}{2}(z+\bar z) = -1\,$.

Repeat for $|z-1|=|z-i| \iff (z-1)(\bar z - 1)=(z-i)(\bar z +i) \iff\;\ldots$

Answer 2

The dummies solution:

$|z-1|=|a+ib-1|=\sqrt{(a-1)^2+b^2}$

$|z+3|=|a+3+ib|=\sqrt{(a+3)^2+b^2}$

$|z-i|=|a+3+ib|=\sqrt{a^2+(b-1)^2}$

We have 1

$\sqrt{(a-1)^2+b^2}=\sqrt{(a+3)^2+b^2}$

from where

$a^2-2a+1+b^2=a^2+6a+9+b^2$

or

$a=-1.$

We have 2

$\sqrt{(a-1)^2+b^2}=\sqrt{a^2+(b-1)^2}$

from where

$a^2-2a+1+b^2=a^2+b^2-2b+1$

or

$a=b.$

So far:

$$a=b=-1\tag 1$$

We have 3

$\sqrt{(a+3)^2+b^2}=\sqrt{a^2+(b-1)^2}$

from where

$a^2+6a+9+b^2=a^2+b^2-2b+1$

or

$3a=-2b-4$

contradicting $(1)$.

Answer 3

Let $z=a+bi$ then we have $(a-1)^2+b^2=(a+3)^2+b^2=a^2+(b-1)^2$. Thus, $(a-1)^2=(a+3)^2$, $a=-1$. Let's substitute $a$ with it's value: $(-1-1)^2+b^2=(-1)^2+(b-1)^2$; $4+b^2=1+b^2-2b+1$; $b=-1$. The answer is $z=-1-i$

Answer 4

A bit of geometry in the Gaussian plane:

1)$ |z-1|=|z-i| $.

A straight line

$z = t +it $, passing through the origin, slope $=1.$

2)$ |z-1|= |z-(-3)|$.

A straight line

$z = -1 +is$, parallel to the $i-$axis passing through $(-1,0).$

Intersection : $t =-1$, $s=t.$

Hence: $z = -1 - i$.

Comments welcome.