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I have played around with this a bit and keep getting something that doesn't seem right. Perhaps I'm overlooking something. Using the definition of distance in the complex plane I transform my equation into Cartesian form and with a little bit of college algebra end up with:

y$^2$ = 0

Is that right? I noticed that if I replace the 2 by 1 or 3 in the original equation I get something different, so I think I'm just misinterpreting or failing to see something.

Martin Sleziak
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Edi Madi
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    Let's see:the distance from one fixed point plus the distance to another fixed point is a constant, hmmm ... describes ... hmmm ... one of those funny things the Greeks, Kepler, Newton were interested in? – zhw. Apr 15 '16 at 04:27
  • You can find several similar posts on this site. For example: http://math.stackexchange.com/questions/626554/find-z-such-that-z1-z-1-4 and http://math.stackexchange.com/questions/1674177/describe-the-set-of-all-complex-numbers-z-such-that-z-a-z-b-c – Martin Sleziak Apr 15 '16 at 06:03

4 Answers4

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It is an ellipse, where the foci are at $1$ and $-1$ and the sum of the distances to the foci is $2$.

Since the foci are $2$ apart, all the points are on the x-axis between $-1$ and $1$, and these all have $y = 0$.

marty cohen
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Since equality holds in $$ 2 = \left|{1-z}\right|+\left|{z+1}\right| \ge \text{Re}(1-z) + \text{Re}(1+z) = 2 \, , $$ it follows that both $1-z$ and $1+z$ are non-negative real numbers, i.e. $z \in [-1, 1]$.

Martin R
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$\lvert z-1\rvert + \lvert z+1\rvert =2$

This sais: The sum of the lengths of the two line segments, from $(1,0)$ to a point $z$ and from $z$ to $(-1,0)$, is $2$.

Since the distance between $(1,0)$ and $(-1,0)$ is $2$ then the sum two line segments is exactly the length of the line segment between the end points.   This is only permissible if the point $z$ actually lies on the line segment between $(1,0)$ and $(-1,0)$.

$$z\in\big\{(x+iy): (x,y)\in \Bbb R^2, -1\leq x\leq 1, y=0\big\}$$

Graham Kemp
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Suppose $|w+z|=|w|+|z|$, then $w,z$ must lie on the same ray.

If $w=0$ then this is obvious, otherwise divide through by $w$ to get $|1+s|=1 + |s|$, where $s={z\over w}$.

Squaring and expanding gives: $(1+ \operatorname{re} s)^2 + (\operatorname{im} s)^2 = 1 + 2 |s| + |s|^2$, which reduces to $\operatorname{re}s = |s|$, and hence $s = |s|$, and hence $w,z$ lie on the same ray.

Returning to the question:

Let $a=1-z,b=1+z$, then $|a+b| = 2 = |a|+|b|$, hence $a,b$ lie on the same ray. If $b=0$, then $z=-1$, otherwise $a=tb$ for some $t \ge 0$, and solving this gives $z={1-t \over 1+t}$. The range of the latter expression for $t \ge 0$ is $(-1,1]$, hence the set of $z$ solving the original equation is $[-1,1]$.

copper.hat
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