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


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...

  • $\begingroup$ I'm not sure I understand, sorry. I tried to draw a cart to represent what you said : imgur.com/ZvXgX9l But I don't quite get your point...Do you mean I need to calculate the hypothenuse of a right triangle? $\endgroup$ – Maude Jun 15 '16 at 21:45
  • $\begingroup$ @Maude I really don't understand what's in that diagram. And this really doesn't have much to do with right triangles (though I can see how it might appear so). The idea is simpler: If I pick two points, then in general any other point in the plane will be closer to one than the other. How would you represent that idea in a drawing? $\endgroup$ – Semiclassical Jun 15 '16 at 21:49
  • $\begingroup$ I'm sorry i'm very bad in geometry. If you pick 2 points, then the third one will make a triangle, no? $\endgroup$ – Maude Jun 15 '16 at 21:57
  • 1
    $\begingroup$ Given that this guessing game doesn't seem to be helpful, I'll just give say what I'm getting at: If I pick any two points, the set of points which is equidistant from both of them is the straight line which lies halfway between them. If I had the points $(1,0)$ and $(-1,0)$, this would be the $y$-axis. In the complex plane, with points $z=1$ and $z=i$, this corresponds to the condition $\text{Re }z=\text{Im }z$ i.e. the line through the points $z=0$ and $z=1+i$ (both of which satisfy the equation under consideration.) $\endgroup$ – Semiclassical Jun 15 '16 at 22:05

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 $(0,1)$.

  • $\begingroup$ Oh this is exactly what I did but I was stuck at square rooting a^2+b^2-2b+1..I didn't know how to remove the root without messing up the equation. I guess I was on the right track! $\endgroup$ – Maude Jun 15 '16 at 21:54

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 $$a=b$$

  • $\begingroup$ Thanks! i'm really happy to know I wasn't wrong all along =) $\endgroup$ – Maude Jun 15 '16 at 22:02

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