5

If the complex number $z$ satisfies the equation $\left\lvert z- {4\over z}\right\rvert = 2$ then the least and the greatest values of $|z|$ are ?

My try $\left\lvert z- {4\over z}\right\rvert = 2$

$\left| |z| - \left\lvert z- {4\over z}\right\rvert \right| \le 2$.

David K
  • 98,388
Nitro phenol
  • 245

2 Answers2

8

Given $$\left|z-\frac{4}{z}\right| = 2$$ and here we have to find $\max$ and $\min$ of $|z|$

So $$|z| = \left|\left(z-\frac{4}{z}\right)+\frac{4}{z}\right|\leq \left|z-\frac{4}{z}\right|+\frac{4}{|z|}$$

Above we have used $\bf{\triangle \; Inequality}$

So $$|z|\leq 2+\frac{4}{|z|}\Rightarrow |z|^2-2|z|\leq 4$$

So $$\left(|z|-1\right)^2 \leq 5$$

Now after that You can solve it.

juantheron
  • 53,015
  • 1
    Hey. (excuse me for no mathjax. I will update this). If you consider the identity |z1(+-)z2| <= |z1| + |z2|, where z1=z and z2=4/z, then wont the inequality become 2<= |z|+|4/z|, which is totally the different thing? – Shashank Holla Dec 05 '16 at 13:07
6

If you know $\left|\frac zw\right|=\frac{|z|}{|w|}$, then you can solve your inequality for real $|z|$. Then see if real $z$ attains those max and min.
$$|r-4/r|\leq2\\-2r\leq r^2-4\leq2r\\5\leq(r+1)^2,(r-1)^2\leq5\\-1+\sqrt{5}\leq r\leq1+\sqrt{5}$$

Empy2
  • 50,853