What does $ \lvert z-a \rvert = \mathit Re(z)+a $ look like?

Question

What does a loci with the equation look like?

$ \lvert z-a \rvert = \mathit Re(z)+a $

This is for the applying complex numbers topic of an advanced HSC maths course. I was asked to describe the loci.

I know that $ \lvert z-a \rvert $ would get me either a perpendicular bisector or a circle. I also know that $ \mathit Re(z) $ refers to the horizontal values on the complex plane. But I just can't imagine what it looks like.

Answer 1

We can treat complex numbers $z = x + iy$ as equations over $(x, y) \in \mathbb R^2$, and use a geometry plotter to plot them. In this case, the equation system is:

$$ \begin{align*} |x + iy - a| &= Re(x+iy) + a \\ |(x - a) + iy| &= x + a\\ \sqrt{(x-a)^2 + y^2} &= x + a \\ (x-a)^2 + y^2 &= (x + a)^2 \end{align*} $$

One can use a tool like Desmos for plotting curves like these. In this case, here is a playable version of the graph with $a$ as a parameter.

The image for one choice of $a$ is:

Answer 2

Given $|z-a|$= $\Re(z)+a$ $$|x+iy-a|=x+a\\|(x-a)+iy|=x+a\\ $$

$$\sqrt{(x-a)^2+y^2}=x+a\\$$

taking square on both side

$$(x-a)^2+y^2=(x+a)^2\\x^2+a^2-2ax+y^2=x^2+a^2+2ax$$

we get $$y^2=4ax$$

This is a right handed parabola with focus $(a,0)$

Answer 3

First note that $a$ must be real as it is the difference between a magnitude (real) and the real part of a complex number (also real).

So we can let $z = x+yi$ and proceed as follows:

$\sqrt{(x-a)^2 + y^2}= x+a$

$(x-a)^2 + y^2= (x+a)^2$

Rearrange, use the difference of squares identity,

$2a(2x) = y^2$

$y^2 = 4ax$

which is a parabola that's of the same shape as $y^2 = x$ with some scaling adjustments.

Answer 4

Geometrical $\vert {z-a}\vert$ is distance of Z from point a And distance of Z from line .$$ a\bar z + \bar a z + b = 0 $$ Is $$ \vert\frac {a\bar z + \bar a z + b }{2\vert a \vert}\vert$$ In given question Re(z)+a = $ \frac{z}{2}\,+\,\frac{\bar z}{2}+ a $ Which is distance of Z from line. Hence locus is a parabola.