Geometrical interpretation $|z-1| = Re(z+1)$ 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?

 A: 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).
A: 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!
