Solve in $\mathbb{C}$ : $|z-i| = |z-1|$

Question

I just had that question in my final exam

Solve in $\mathbb{C}$ : $|z-i| = |z-1|$

and I couldn't do it. I found a similar thread here : Showing that $\{z\in\mathbb{C}:|z-1|<|z+i|\}$ is an open set but it's too complicated for me to understand.

I tried 2 different ways to solve this

First, I developed the $z$ which gave me $$\begin{align} & (a+bi)-i=(a+bi)-1 \\ & = a-i+bi=a-1+bi \\ & = a-1(i+b)=a-1+bi \\ \end{align} $$ However we are looking at the module of this, so i have to do $\sqrt{((a^2)+(-1(i+b))^2} = \sqrt{((a-1)^2+(bi)^2)}$

Then I got some sort of $\sqrt{a^2-b^2-2ab}$ that equals to similar on the other side but couldn't find anything.

I tried the geometric way and the "logic" way (how can $i = 1$ when $i^2 = -1$), but I just don't understand how i'm suppose to solve this.

I also tried using the roots which say that for $z^n = w$

$w = r^{(1/n)} \text{Ei}((\angle m \cdot \frac{2k\pi}{n}))$

Every formula I knew failed

Answer 1

Hint: The geometric way is probably the simplest. The expression $|z-w|$ gives the distance between the points $z$ and $w$, so $z$ satisfying $|z-i|=|z-1|$ corresponds to a point which is equidistant from $1$ and $i$. (Or, in Cartesian coordinates, $(1,0)$ and $(0,1)$). But there's only one geometric figure corresponding to the set of points which are equidistant to two other points...

Answer 2

Write $z = a + bi$. Then you are trying to find all $z$ for which $$\sqrt{a^2 + (b - 1)^2} = \sqrt{(a - 1)^2 + b^2}$$ Squaring both sides and expanding the squares, this becomes $$a^2 + b^2 - 2b + 1 = a^2 - 2a + 1 + b^2$$ Subtracting $a^2 + b^2 - 1$ from both sides, this becomes $$-2b = -2a$$ So we are reduced to $a = b$. All steps here are reversible, so we conclude that all complex numbers where $b = a$ satisfy the equality. Geometrically this is the line $y = x$, the points equidistant from $(1,0)$ and $(0,1)$.

Answer 3

Your approach works, you just made a mistake in algebra. The condition $|z-i|=|z-1|$ is equivalent to $|z-i|^2=|z-1|^2$. Plugging in $z:=a+bi$, this is equivalent to $$ \begin{align} |a+bi-i|^2&=|a+bi-1|^2\\ \Longleftrightarrow\ |a+i(b-1)|^2&=|(a-1)+bi|^2.\\ \end{align} $$ Using the identity $|x+iy|^2=x^2+y^2$ we get $$ a^2+(b-1)^2=(a-1)^2+b^2.\tag1$$ Expanding (1) and simplifying gives you $$a=b$$

Answer 4

$$|z-i| = |z-1|$$

This says that the distance between $z$ and $i$ is equal to the distance between $z$ and $1$. That describes the perpendicular bisector of the segment with endpoints at $i$ and $1$.

The midpoint of that segment is $\frac 12(1+i)$.

The segment is parallel to the direction $i-1$. The direction perpendicular to that is $i(i-1) = -1-i$ since multiplying by $i$ is a rotation of $90^\circ$.

So the set of all points is described by $$\text{(initial point) + $t$(direction of line)}$$

$$z = \frac 12(1+i) + t(-1-i)$$

$$z = \left(\frac 12 - t\right) + i\left(\frac 12 -t\right)$$ $$z = s + is$$