1
$\begingroup$

I'm preparing for a complex analysis prelim which I'll take next summer by consulting Ahlfor's Complex Analysis: An Intro to the Theory of Analytic Functions of One Variable (3rd edition). My question pertains to Exercise 4 of Chapter 1, Section 1.5, and it is as follows:

Show that there are complex numbers $z$ satisfying $$|z-a| + |z+a|= 2|c|$$ if and only if $|a| \leq |c|$. If this condition is fulfilled, what are the smallest and largest values of |z|?

Well, for the first part of the question I was able to show only one direction, i.e. if there are complex numbers $z$ such that $|z-a| + |z+a|= 2|c|$, then $|a| \leq |c|$ by applying the triangle inequality. Specifically, I did the following: $$ \begin{array} {lcl} 2|a| &=& |2a| \\ &=& |a-z+z+a| \\&\leq& |a-z| + |z+a| \\&=& |z-a| + |z+a| \\&=& 2|c|. \end{array}$$ Dividing both sides of the inequality by $2$ yields the result in one direction. The other direction is giving me trouble, and so is the second part of the exercise. Some helpful hints and advice would be much appreciated.

$\endgroup$
1

2 Answers 2

2
$\begingroup$

Draw it, and you'll see that's euclidean geometry, not complex analysis. Anyway, something like $z=ia\sqrt{\frac{|c|^2}{|a|^2}-1}$ should work.

To be more specific: $$|z-a|=|a|\cdot|i\sqrt{\frac{|c|^2}{|a|^2}-1}-1|=|a|\cdot\sqrt{\frac{|c|^2}{|a|^2}-1+1}=|c|$$ $$|z+a|=|a|\cdot|i\sqrt{\frac{|c|^2}{|a|^2}-1}+1|=|a|\cdot\sqrt{\frac{|c|^2}{|a|^2}-1+1}=|c|$$

Edit: I didn't see the part about max/min. The points that satisfy the equation draw an ellipse with foci in $a$ and $-a$. Clearly the minimum and maximum values of $|z|$ are on the axes. The first one is the one I've already written, the second one is realised by $a|\frac ca|$.

$\endgroup$
4
  • $\begingroup$ Wouldn't this suggestion you gave have to consider the restriction $a \neq 0$? Drawing it is fine, but I think the question asks to be a proof. $\endgroup$
    – Libertron
    Dec 28, 2013 at 19:18
  • $\begingroup$ Well, if $a=0$, the exercise is not so difficult... Take $z=c$. Otherwise, you can check that the formula works: this is a correct proof. If you want to know where the formula comes from, you can draw a picture: the formula finds a point on the axis of the segment between $a$ and $-a$. $\endgroup$ Dec 28, 2013 at 19:21
  • $\begingroup$ I think showing the converse may require $z= \frac{|c|}{|a|} a$. Then I can use the fact that $\frac{|c|}{|a|} \geq 1$ from there. $\endgroup$
    – Libertron
    Dec 29, 2013 at 16:05
  • $\begingroup$ My answer is not clear? When $a\neq 0$, you have to use $z=ia\sqrt{\frac{|c|^2}{|a|^2}-1}$, and the fact that $|c|\geq|a|$ ensures that the square root is real. $\endgroup$ Dec 29, 2013 at 16:09
1
$\begingroup$

You can rotate the complex plane so that $a$ is real.

Let $z= x+ i \,y$. Then the left hand side is $$ \sqrt{(x+a)^2+y^2} + \sqrt{(x-a)^2+y^2}$$ is minimum when $y=0$, i.e when $z$ is real. Clearly all $z$ in the interval $[-a,a]$ satisfies the condition.

Hence $z=0$ is the minimum, $z=\pm a$ is maximum in magnitude

Note: Added as an after thought.

If you do not want to make the initial rotation, then just consider $z$ of the form $$ z = \lambda \, (-a) + (1-\lambda) a = a \,(1 - 2 \lambda),~~~ 0 \le \lambda \le 1$$ and show that all these $z$ meet the requirement.

$\endgroup$
1
  • $\begingroup$ ?? $z=0$ implies $|z-a|+|z+a|=2|a|\neq 2|c|$. $\endgroup$ Jan 8, 2014 at 17:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .