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Find the number of complex numbers satisfying $|z|=\text{max}\{|z-1|,|z+1|\}$

My Attempt: Let $z=x+iy$, so, $$\sqrt{x^2+y^2}=\text{max}\{\sqrt{(x-1)^2+y^2},\sqrt{(x+1)^2+y^2}\}$$

Case I: $\sqrt{x^2+y^2}=\sqrt{(x-1)^2+y^2}\implies \pm x=x-1\implies x=\frac12$

Case II: $\sqrt{x^2+y^2}=\sqrt{(x+1)^2+y^2}\implies \pm x=x+1\implies x=-\frac12$

Does that mean if the complex numbers are lying on two lines, either $x=\frac12$ or $x=-\frac12$ then they would satisfy the required equation?

Therefore, the required number of complex numbers is infinite?

But the answer given is zero.

Can we solve this graphically?

aarbee
  • 8,246

2 Answers2

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It follows from $$ |z-1|^2 = (x-1)^2 + y^2 = x^2 + y^2 - 2x + 1 \\ |z+1|^2 = (x-1)^2 + y^2 = x^2 + y^2 + 2x + 1 $$ that $$ \max(|z-1|, |z+1|)^2 = x^2 + y^2 + 2|x| + 1 > x^2 + y^2 = |z|^2 $$ which shows that the equation has no solution.

Geometrically: $$ \max(|z-1|, |z+1|) = |z+1| $$ means that the distance from $z$ to $1$ is less than or equal to the distance from $z$ to $-1$, that are exactly the points in the closed right half-plane. But points in the right half-plane are closer to the origin than to the point $-1$, so that $$ \max(|z-1|, |z+1|) = |z+1| > |z| $$

In the same way you can show that the equation has no solution in the (closed) left half-plane.

With respect to your approach: It is correct that $|z| = |z-1|$ if and only if $x=1/2$ (case I). But for those $z$ is $$ |z-1| < |z+1| $$ so that does not give any solution. Similar in case II.

CiaPan
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Martin R
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$|z| = max \{ |z-1|, |z+1|\} \implies |z| \ge |z-1|$ and $|z| \ge |z+1|$ . Summing the two equations, we have $|2z| \ge |z-1| + |z+1|$. But triangle inequality states that $|z-1| + |z+1| \ge |2z|$ so $|z-1| + |z+1| = 2|z|$. Because, $|z|$ is one of $|z-1|$ and $|z+1|$, we then deduce that $|z-1| = |z| = |z+1|$. $z$, $z-1$, and $z+1$ are distinct and belong to the horizontal line that pass by $z$. They cannot then be part of the circle of centre 0 and radius $|z|$ because a circle can't have more than 2 intersection points with a line. We conclude then that there's no z that satisfy our last equality.

glk0
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