Need help with complex equation: $|z−i|+|z+i|=2$

Question

I am trying to solve this equation: |z−i|+|z+i|=2 and don't know how to do it. This what I have: $$\sqrt{(x+1)^2+y^2}+ \sqrt{(x-1)^2+y^2} = 2 /^2$$ $$(x+1)^2+y^2+(x-1)^2+y^2 + 2\sqrt{[(x+1)^2+y^2][(x-1)^2+y^2]} = 4$$ $$x^2 +2x + 1+y^2+x^2-2x + 1+y^2 + 2\sqrt{[(x^2 +2x + 1)+y^2][(x^2-2x + 1)+y^2]}=4$$ $$2x^2+2y^2 + 2 + 2\sqrt{[(x^2 +2x + 1)+y^2][(x^2-2x + 1)+y^2]}=4$$ $$2x^2+2y^2 + 2 + 2\sqrt{(x^2 +2x + 1)(x^2-2x + 1)+x^2y^2-2xy^2+y^2+x^2y^2-2xy^2+y^2+y^4}=4$$ $$2x^2+2y^2 + 2 + 2\sqrt{(x^2 +2x + 1)(x^2-2x + 1)+2x^2y^2-4xy^2+2y^2+y^4}=4$$ $$2x^2+2y^2 + 2 + 2\sqrt{x^4+2x^3+x^2-2x^3+4x^2-2x+x^2+2x+1+2x^2y^2-4xy^2+2y^2+y^4}=4$$ $$2x^2+2y^2 + 2 + 2\sqrt{x^4+6x^2+1+2x^2y^2-4xy^2+2y^2+y^4}=4$$

And I have no idea what to do next. Any help would be much appreciated

Note that $\pm i$ satisfies the equation. A little consideration shows that there are lots more solutions. — copper.hat, Nov 01 '20 at 22:41
It's just 90 degrees rotated version of What region in $\mathbb{C}$ does $\left|{z-1}\right|+\left|{z+1}\right|$ = 2 describe? — Sil, Nov 01 '20 at 23:16

marty cohen · Answer 1 · 2020-11-01T22:47:15.923

The two points i and -i are distance 2 apart. The origin is distance one from one of them and one from the other so it satisfies the equation.

All the points on the line between those two points satisfy the equation.

Any other point got on the line between these two has the sum of the distances to those two points being greater than 2. Think of the theorem in euclidean geometry that the sum of the length of two sides of a triangle is greater than the length of the third side.

Therefore the only solutions are all the points between the two points.

In general the solutions to the equation $|z-a|+|z-b| = c$ where $c$ is greater than the distance between $a$ and $b$ is the eclipse with foci at a and b and the sum of the distances to a and b equal to c.

If c is equal to the distance between a and b, the solution is the points on the line segment joining a and b.

If c is less than the distance between a and b, there is no solution.

I fixed it. Should be correct now. I upvoted copper.hat's comment because it was correct about my first answer. My correct answer is there now. — marty cohen, Nov 01 '20 at 22:48

score 2 · Accepted Answer · edited Dec 15 '23 at 07:18

2

From the beginning, move one of the absolute values to the other side. It will simplify the calculations: $$|z+i|=2-|z-i|$$ Square this and you get $$|z+i|^2=(2-|z-i|)^2=4-4|z-i|+|z-i|^2$$ $$x^2+(y+1)^2=4-4|z-i|+x^2+(y-1)^2$$ $$x^2+y^2+2y+1=4-4|z-i|+y^2-2y+1$$ $$4y-4=-4|z-i|$$ $$y-1=-|z-1|$$ $$(y-1)^2=x^2+(y-1)^2$$ The only solution is $x=0$. This is independent of $y$. Next is to check if by squaring we add extra solutions. Not all $y$'s might be valid. $$|y-1|+|y+1|=2$$ If $y\ge1$, the equation is $$2y=2$$So $y=1$. If $y\le -1$ then $$-(y-1)-(y+1)=2,$$ meaning $y=-1$.In between, $$-(y-1)+y+1=2$$or $2=2$. The final answer is $z=\alpha i$, for $\alpha\in[-1,1]$

edited Dec 15 '23 at 07:18

Martin Sleziak

53,687

answered Nov 01 '20 at 22:43

Andrei

37,370

1

$\pm i$ are solutions by inspection. – copper.hat Nov 01 '20 at 22:43
2

Yes, those correspond to $x=0$ – Andrei Nov 01 '20 at 22:44
1

You should phrase your answer a little more precisely. The only value of $x$ that satisfies the equations is $x=0$, but $x=0 $ is not a solution. – copper.hat Nov 01 '20 at 22:45
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Once you get $x=0$ the equation becomes $$|iy+i|+|iy-i|=2 \ |y+1|+|y-1|=2$$Now, since $y$ is real, the standard technique of expressing $|y+1|, |y-1|$ depending on when $y <-1, -1 \leq y \leq 1, y>1$ completes the solution. – N. S. Nov 01 '20 at 22:49
x=0 and $-1 \le y \le 1$ is a solution. Other values of y make this sum too a large. – marty cohen Nov 01 '20 at 22:51
@martycohen Yes, I've modified the answer already. – Andrei Nov 01 '20 at 22:52

score 2 · Answer 3 · answered Nov 01 '20 at 22:55

2

The set of solutions is the segment $\{\alpha i\mid-1\le\alpha\le1\}$.

This follows from the following geometrical argument. $|z-i|$ is the distance from $z$ to $i$; $|z+i|$ is the distance from $z$ to $-i$. Which point $z\in\mathbb C$ can have the sum of distances to $i$ and to $-i$ equal to $2$, knowing that the distance between $i$ and $-i$ is already $2$? It must be colinear with $i$ and $-i$ (or otherwise the LHS would be $\gt 2$ due to triangle inequality) and it must lie in the actual segment with endpoints $i,-i$, or otherwise one of the moduli $|z-i|,|z+i|$ would alone be $\gt 2$.

answered Nov 01 '20 at 22:55

1

Same answer I came up with. Upvoted. – marty cohen Nov 01 '20 at 22:58
This kind of geometrical approach should be one of the main tools in any student’s box! It’s a shame that so many introductions to complex algebra manage to leave it out almost entirely, and focus so strongly on algebraic approaches. (I suspect mostly because the latter are easier to design + grade homework problems on.) – Peter LeFanu Lumsdaine Nov 02 '20 at 07:10

score 0 · Answer 4 · answered Nov 02 '20 at 01:45

Let $C$ be the solutions. Since $C = -C$ so we can look for solutions with non negative real part.

Note that $2 = |2i| = |z+i - (z-i)| \le |z+i|+|z-i|$.

Let $f(x) = |x+iy+i| + |x+iy-i|$ and note that $f$ is continuous everywhere, differentiable for $ x > 0$ and $f'(x) >0$ for such $x$. It follows from the mean value theorem that $f(x) > f(0) \ge 2$ for $x>0$. Hence it follows that $x=0$.

It is not hard to see that $|iy+i| + |iy-i| = |y+1|+|y-1| = \max(2,2|y|)$, hence we must have $|y| \le 1$, that is $C \subset [-i,i]$.

It is straightforward to verify that $[-i,i] \subset C$.

fleablood · Answer 5 · 2020-11-02T02:49:39.753

Triangle inequality tells us $|z+i| + |z-i| \ge |(z+i)-(z-i)| = |2i| = 2$ with equality holding if and only if geometrically of $0$ is colinear and between $z+i$ and $z-i$.

The only values colinear between $z + i$ and $z-i$ must all have the same real value so the real part as $z$ has and if $0$ is one such number, the real part of $z$ must be $0$.

If $Im(z) = b$ then $z = bi$ and we have $|bi+i| + |bi - i| = |b+1|+|b-1|=2$. By cases we can verify then this will hold if and only if $b \in [-1,1]$.

(If $b > 1$ then $|b+1| > 0$ and $|b-1| > 2$ and if $b < -1$ then $|b+1| > 2$ and $|b-1| > 0$; but if $-1 < b < 1$ then $|b+1| = b+1$ and $|b-1|= 1-b$ and $|b+1|+|b-1| = (b+1)+(1-b)=2$.)

So $z$ can be any $bi; -1 \le b \le 1$.

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Alternatively (in hindsight) note:

If $z = a + bi$ then $|z\pm i| = \sqrt{a^2 + (b\pm 1)^2}$ so

$|z+i| +|z-i|$
$=\sqrt {a^2 + (b+ 1)^2} + \sqrt {a^2 + (b-1)^2}$
$[*]\ge\sqrt {(b+1)^2} +\sqrt {(b-1)^2}$
$ =|b+1|+|b-1|$
$=\begin{cases}(b+1)+(b-1)=2b>2& b> 1\\ (b+1)+(1-b)=2 &-1\le b \le 1\\(-b-1)+(1-b)=-2b>2 & b < -1\end{cases}$
$[**] \ge 2$

with equality holding if and only if $a = 0$ and $-1 \le b \le 1$.

[*] Equality holding if and only if $a = 0$.
[**] Equality holding if and only if $-1 \le b \le 1$.

Need help with complex equation: $|z−i|+|z+i|=2$

5 Answers5