Possible integral values of $|z|$

Question

Consider the complex number $z$ such that $|z-3|+|z|+|z+3|=12$. Then number of possible integral values of $|z|$ is/are

Try: Using Triangle Inequality $$\bigg|z_{1}+z_{2}+z_{3}\bigg|\leq |z_{1}|+|z_{2}|+|z_{3}|$$

So $$\bigg|(z-3)+z+(3+z)\bigg|\leq |z-3|+|z|+|z+3|=12$$

So $$|3z|\leq 12\Rightarrow |z|\leq 4$$.

i have seems that only $z=4$ satisfy

I did not understand how can i calculate that only $z=4$ satusfy the given eqution.

could some help me , Thanks

Answer 1

By triangle inequality $12 = |z -3| + |z| + |z+3| \ge |z-3 + z + z+3| = 3|z|$

So $|z| \le 4$

And $12= |z-3| + |z| + |z+3| \le |z| + |3| + |z| + |z| + 3 = 3|z| + 6$

So $|z| \ge 2$

so $|z| = 2,3,4$.

If $z = \pm 4$ then $|z-3| + |z| +|z+3| = 1+4 + 7 = 12$.

If $z = \pm 2i$ then $|z-3| + |z| + |z+3| = \sqrt{2^2 + 3^2} + 2 + \sqrt {2^2 + 3^2} = 12$.

To solve for $|z| = 3$ we need $z = a + bi$ and $\sqrt{(a+3)^2 + b^2} + 3 + \sqrt{(a-3)^2 + b^2}=12$ or $\sqrt{(a+3)^2 + b^2} + \sqrt{(a-3)^2 + b^2}=9$ and there are infinitely many solutions to that.

Answer 2

Continue your work a bit more with the triangle inequality: $12 \le |z|+3 + |z|+3 + |z|\implies 6 \le 3|z| \implies |z| \ge 2$ . Thus $2 \le |z| \le 4$ or $|z| \in [2,4]$ . And if you are looking for the integer values of $|z|$, then $|z| = 2, 3, 4$ are the only possible values.