# That $|a|\leq|b|$ implies existence of complex $z$ satisfying $|z-a|+|z+a|=2|b|$?

I'm looking at the equation $|z-a|+|z+a|=2|b|$. If there are complex values $z$ satisfying this equation, then $$2|b|=|z-a|+|z+a|=|a-z|+|z+a|\geq|(a-z)+(z+a)|=|2a|=2|a|$$ so $|a|\leq |b|$.

However, how is the converse true, that if $|a|\leq |b|$, then there is some complex $z$ such that $|z-a|+|z+a|=2|b|$? If such $z$ exists, is there a way to figure out the maximum and minimum values for $|z|$? Thank you.

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Existence is easy. Suppose that $z=ra$, where $1\le r\in\mathbb{R}$; then $$|z-a|+|z+a|=(r-1)|a|+(r+1)|a|=2r|a|\;,$$ so just take $r=|b|/|a|$. (Of course if $a=0$, take $z=b$.) –  Brian M. Scott Jan 18 '12 at 18:09

The converse is true. The equation $|z-a|+|z+a|=2|b|$ is the equation of an ellipse with foci at $\pm a$. The major axis will be on the line through the foci. The major radius will be $|b|$. The maximum modulus obtained by $z$ will be $|b|$ and it will occur when $z=\pm a|b|/|a|$.

In more detail: the equation describes the locus of points $z$ in the complex plane such that the distances from $z$ to the two points $a, -a$ add up to a constant, $2|b|$. This is one definition of an ellipse, with foci $a, -a$. As you proved, if $|b|<|a|$ there can't be any points like this, but if $|b|\geq |a|$ there definitely can. Imagine a string of length $2|b|$, with its endpoints at $a,-a$. Pull the string off to the side till it's tight, and you have found a point $z$ satisfying the given equation.

If $|b|=|a|$ exactly, the string is already tight and the locus of points is exactly the segment from $-a$ to $a$.

If $|b|>|a|$, then you get a true ellipse. Since it is centered at the origin (because $-a,a$ are symmetric with respect to the origin), the maximum modulus of a point on the ellipse occurs at the vertices. The vertices occur on the line through $-a,a$, thus they are real multiples of $a$. Also, being on the line with $a,-a$, their distance from the origin is their average distance from $-a,a$, which is always $2|b|/2=|b|$, since the sum of the distances must be $2|b|$. Thus the vertices are at distance $|b|$ from the origin, on the line through $-a,a$. Thus, normalize $\pm a$ in length: $\pm a/|a|$; then multiply by $|b|$: $z=\pm |b| \cdot a/|a|$ are the vertices.

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Ah nice, thank you. –  goguman Jan 18 '12 at 18:37
No prob! Glad useful. –  Ben Blum-Smith Jan 20 '12 at 22:56

Let's restrict $z$ to real numbers, and consider the left side as a function of $z$. Let's call it $f(z)$. $f$ is continuous.

$f(0)=2|a|$.

Applying the triangle inequality to $z-a$ and $z+a$ gives us $|z-a|+|z+a|\ge|2z|$, from which we see that $f(z)$ is unbounded above, so there must exist some $w$ such that $f(w)\ge2|b|$.

By the intermediate value theorem, there must be some $z$ in $\left[0,w\right]$ such that $f(z)=2|b|$.

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