Sketch set of complex numbers $z$ such that $\Re\left(\frac{z-2}{z-2i}\right) = 0$

Question

I have this problem on my maths exam and I got stuck.

Sketch all of numbers that satisfy $\Re\left(\frac{z-2}{z-2i}\right) = 0$.

What I did:

$$\frac{z-2}{z-2i} = \frac{(x-2) + iy}{x + i(y-2)}\\ \frac{(x-2) + iy}{x + i(y-2)} × \frac{x - i(y-2)}{x - i(y-2)}\\ \cdots$$ Finally getting this $$\Re\left(\frac{z-2}{z-2i}\right) = 0 \implies x(x-2) + y(y-2) = 0$$

Idk what to do next.

EDIT: Thanks for all the answers. I got it figured out. I just had to "complete the square" to get an equation of the circle and after was just matter of plotting it.

Answer 1

You can also use that $Re\left(\frac{z-2}{z-2i}\right) = 0\iff \frac{z-2}{z-2i}$ is purely imaginary, therefore

$$\frac{z-2}{z-2i}=-\frac{\bar z-2}{\bar z+2i} \iff z\bar z+2iz-2\bar z-4i=-z\bar z+2z+2i\bar z-4i \\\iff (z+\bar z)-i(z+z\bar)=z\bar z \iff 2Re(z)+2Im(z)=|z|^2 \\\iff 2x+2y=x^2+y^2\iff x^2-2x+y^2-2y=0\iff (x-1)^2+(y-1)^2=2$$

Answer 2

This is a Möbius transform, hence the image as well as pre-image of a line-or-circle is a line-or-circle (in the extended complex plane, i.e., with a point at infinity added). Here, we are looking for the pre-image of the imaginary axis (i.e., a line), so it must be a line or a circle.

Let's see if we can guess enough points to determine the line or circle:

• $z=2$ maps to $0$, so is on our curve.
• $z=0$ maps to $\frac{-2}{-2i}=-i$, so is on our curve.
• If $z=2i$ makes the denominator zero, hence maps to $\infty$ (which is a point on every line in the extended plane), so $2i$ is on our curve. (To be precise, whether or not the point $2i$ belongs to the answer curve, may depend on your local customs, see below)

As these three points are not collinear, we are looking for a circle, not for a line. It must be the circle passing through $0$, $2$, and $2i$. Depending on whether or not you routinely work with the extended complex numbers, you may have to remove the single point $2i$ from this circle and indicate that gap accordingly in your sketch.

Answer 3

Argument of the given quotient is 90 or 270 degeee since it's real part is zero. So angle between segment joining $z$ and $2$ and $z$ and $2i$ is either 90 or 270. This implies locus is a circle with diametric end points $2$ and $2i$.

But $z$ cannot be $2i$ since then quotient is I'll defined.

Answer 4

1) Decompose the number $z$ in real ane imaginary part: $$z = a + bi$$, your equation will be then:

$$\mathrm{Re}\Bigl(\frac{a+ib - 2}{a+ib-2i}\Bigr) = 0$$

2) be aware that the denominator should be not zero: your solution cannot be $a+ib -2i = 0$ which implies that $(a=0, b=2)$ could not be a solution of our equation

3) now you could rename your fraction as

$$\frac{a+ib - 2}{a+ib-2i} = c + di$$

and since you want the real part $c$ to be 0, then you can quickly write:

$$\frac{a+ib - 2}{a+ib-2i} = di$$

which implies

$$a+ib - 2 = di(a+ib-2i)$$

which implies

$$a+ib - 2 -dai +db - 2d = 0.$$

4) Solve the equation using a system, where you ensure both real part and imaginary part to be equal to zero

$$b-da = 0$$ and

$$a - 2 +db - 2d = 0.$$. First equation brings you to say $d = \frac{b}{a}$. We substitute to the second equation, and we get

$$a - 2 +\frac{b}{a} b - 2\frac{b}{a} = 0.$$.

thus, $a^2 - 2a + b^2 - 2b = 0 $ provided both $a\neq 0$ and not $(a=0, b=2)$, is the space of your solution .