Question

If $|z-\frac{4}{z}|$ = 2, then the maximum value of $\left|z\right|$ is equal to

• A. -1
  
• B. +1
  
• C. 2
• D. 5

Answer 1

The correct option is B

+1

We are asked to find the maximum value of $\left|z\right|$. We will apply the triangle inequalities and try to find relations involving $\left|z\right|$.
$|z-\frac{4}{z}|\le \left|z\right|+\left|\frac{4}{z}\right|2\le \left|z\right|+\frac{4}{\left|z\right|}Let\left|z\right|=r2\le r+\frac{4}{r}{r}^2-2r+4\ge 0{(r-1)}^2+3\ge 0$
which is always true. We can't conclude anything about the maximum value from this relation. So we will consider
$\left|z\right|=|z-\frac{4}{z}+\frac{4}{z}|\le |z-\frac{4}{z}|+\left|\frac{4}{z}\right|\le 2+\frac{4}{\left|z\right|}Let\left|z\right|=r\Rightarrow r\le 2+\frac{4}{r}{r}^2-2r-4\le 0\Rightarrow If\alpha ,\beta aretherootsof{r}^2-2r-4=0,thenrϵ[\alpha ,\beta ].Alsor\ge 0Therootsare\frac{2\mp \sqrt{4+16}}{2}=1\mp \sqrt{5}r>0,rϵ[1-\sqrt{5},1+\sqrt{5}]$. Combining both the conditions, we get
$rϵ[0,1+\sqrt{5}]\Rightarrow Maximumvalueis1+\sqrt{5}$.