If $\left|z-\frac{4}{z}\right|=2$, then the maximum value of $|z|$ is equal to
If $\left|z-\frac{4}{z}\right|=2$, then the maximum value of $|z|$ is equal to
-
$\sqrt{3}+1$
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$\sqrt{5}+1$
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2
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$2+\sqrt{2}$
Solution
$
\begin{aligned}
& |z|=\left|\left(z-\frac{4}{z}\right)+\frac{4}{z}\right| \Rightarrow|z|=\left|z-\frac{4}{z}+\frac{4}{z}\right| \\
& \Rightarrow|z| \leq\left|z-\frac{4}{z}\right|+\frac{4}{|z|} \Rightarrow|z| \leq 2+\frac{4}{|z|}
\end{aligned}
$
$
\begin{aligned}
& \Rightarrow|z|^2-2|z|-4 \leq 0 \\
& (|z|-(\sqrt{5}+1))(|z|-(1-\sqrt{5})) \leq 0 \Rightarrow 1-\sqrt{5} \leq|z| \leq \sqrt{5}+1
\end{aligned}
$
Asked in: JEE Main 2009
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