I'd like to ask you about the example below (I have to draw a geometrical interpretation in an argand diagram).

$|z-1| = Re(z+1)$

I know that: Re(z+1) = $Re(x + yi + 1) = x + 1$

But what is the most efficient way to solve it?

What I did:

$|x + yi - 1| = x + 1$

$\sqrt{(x-1)^2 + y^2} = x + 1$

$(x-1)^2 + y^2 = (x+1)^2$

$x^2 - 2x + 1 + y^2 = x^2 + 2x + 1$

$ - 4x + y^2 = 0$

$y^2 = 4x$

Is the interpretation correct?

enter image description here

  • 4
    $\begingroup$ Yes, that looks correct. If you want to proceed through geometry alone, consider that $|z-1|$ is the distance between $z$ and the point $(1,0)$ and, for $z>-1$ we have $Re(z+1)$ is the distance between $z$ and the line $L=\{x=yi:\:x=-1,y\in\mathbb{R}\}$. So since the distances are equal, we have a parabola where the point $(1,0)$ is the focus, and the line is the directrix. $\endgroup$ – Marconius Oct 19 '15 at 19:06
  • $\begingroup$ @Marconius, thank you. Should I include that $x +1 \ge 0 -> x \ge -1$? $\endgroup$ – belford Oct 19 '15 at 19:35

Yeah, what you did is correct.

Another way of reaching that answer is noticing that $|x+yi−1|$ is the distance of a point P from a fixed point $(1,0)$ and is equal to $Re(z+1)$, which is the perpendicular distance of the point P from the lines x=-1.

From the definition of a parabola

The parabola is the locus of points in that plane that are equidistant from both the directrix and the focus

We see that the required locus is a parabola having its focus at (1,0) and the directrix as x=-1. Hence you get the required locus as $y^2 = 4x$ ( standard parabola).


Great work and great post. Your method is the way to do it. My answer is more of an elaboration about the geometric interpretation of this parabola. I hope it's useful.

$$|z-1| = Re(z+1)$$

We can read this as "the distance of $z$ from $1$ is equal to one more than the real part $x$ of $z$". Incidentally, this is directly associated with the geometric definition of a parabola.

A parabola is the set of all points that are equidistant from one point (the focus) and a line (the directrix). Let the parabola have vertex $h+ik$. Now define $c$ to be the distance from the vertex to the focus; then $c$ is also the perpendicular distance from the vertex to directrix. If the parabola opens to the right, the focus is $h+c+ik$. Then geometrically, $$|z-focus|=|z-directrix|$$ $$|z-(h+c+ik)| = |z - (h-c+iy)|$$

$$|z-h-c-ik)| = |z - h+c-iy)|$$

Now compare with your parabola. It's vertex in rectangular coordinates is $(0,0)$.

$$|z-c| = |z+c-iy|$$

It appears that $c=1$ and thus your focus is $(1,0)$. Likewise, the distance from the directrix is given by the real part of $z$. The directrix itself is the line $x=-1$. This information was encoded in the original equation all along!


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