Question

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

• A. $\sqrt{3}+1$
• B. $\sqrt{5}+1$
• C. $2$
• D. $2+\sqrt{2}$

Answer 1

The correct option is B

$\sqrt{5}+1$

Explanation for the correct option.

Step 1. Form a compound inequality.

For any two complex number $a$ and $b$ $\left|\left|a\right|-\left|b\right|\right|\le \left|a-b\right|$.

So for complex number $z$ and $\frac{4}{z}$

$$\begin{gathered} \left|\left|z\right|-\frac{4}{\left|z\right|}\right|\le \left|z-\frac{4}{z}\right| \\ \Rightarrow \left|\left|z\right|-\frac{4}{\left|z\right|}\right|\le 2\left[\left|z-\frac{4}{z}\right|=2\right] \end{gathered}$$

Now, let $\left|z\right|=r>0$, then $\left|r-\frac{4}{r}\right|\le 2$ and so the compound inequality becomes

$-2\le r-\frac{4}{r}\le 2$

Step 2. Solve the left inequality.

From the compound inequality $-2\le r-\frac{4}{r}\le 2$, the left inequality is $r-\frac{4}{r}\ge -2$, the inequality can be written as:

$$\begin{gathered} r^2-4\ge -2r \\ \Rightarrow r^2+2r-4\ge 0 \end{gathered}$$

The roots of the quadratic equation $r^2+2r-4=0$ is given as:

$\begin{array}{rcl}r& =& \frac{-2\pm \sqrt{2^2-4\times 1\times \left(-4\right)}}{2\times 1}\\ & =& \frac{-2\pm \sqrt{4+16}}{2}\\ & =& \frac{-2\pm 2\sqrt{5}}{2}\\ & =& -1\pm \sqrt{5}\end{array}$

So, the solution of the inequality $r^2+2r-4\ge 0$ is $r\le -1-\sqrt{5}$ and $r\ge -1+\sqrt{5}$.

But as $r>0$, so $r\le -1-\sqrt{5}$ is rejected.

Thus the solution from the ldeft inequality is $r\ge -1+\sqrt{5}...\left(1\right)$

Step 3. Solve the right inequality.

From the compound inequality $-2\le r-\frac{4}{r}\le 2$, the right inequality is $r-\frac{4}{r}\le 2$, the inequality can be written as:

$$\begin{gathered} r^2-4\le 2r \\ \Rightarrow r^2-2r-4\le 0 \end{gathered}$$

The roots of the quadratic equation $r^2-2r-4=0$ is given as:

$\begin{array}{rcl}r& =& \frac{-(-2)\pm \sqrt{{(-2)}^2-4\times 1\times \left(-4\right)}}{2\times 1}\\ & =& \frac{2\pm \sqrt{4+16}}{2}\\ & =& \frac{2\pm 2\sqrt{5}}{2}\\ & =& 1\pm \sqrt{5}\end{array}$

So, the solution of the inequality $r^2-2r-4\le 0$ is $1-\sqrt{5}\le r\le 1+\sqrt{5}$ .

But as $r>0$, so the solution from the right inequality is $0<r\le 1+\sqrt{5}...\left(2\right)$.

Step 4. Find the combined solution and find the maximum value of $\left|z\right|$.

The solution from the left inequality is $r\ge -1+\sqrt{5}$ and from the right inequality is $0<r\le 1+\sqrt{5}$. So the combined solution is:

$-1+\sqrt{5}\le r\le 1+\sqrt{5}$.

As $r=\left|z\right|$, so $-1+\sqrt{5}\le \left|z\right|\le 1+\sqrt{5}$.

Thus the maximum value of $\left|z\right|$ is $\sqrt{5}+1$.

Hence, the correct option is B.