First slide

De-moivre's theorem

Question

$If a complex number z satisfies | z |^{2} + \frac{4}{| z |^{2}} - 2 (\frac{z}{\bar{z}} + \frac{\bar{z}}{z}) - 16 = 0, then the maximum value of |Z| is$

Moderate

A

$\sqrt{6} + 1$
B

2
C

6
D

$2 + \sqrt{6}$

Solution

$\begin{array}{l} Given that {|z|}^{2} + \frac{4}{| z |^{2}} - 2 (\frac{z}{\bar{z}} + \frac{\bar{z}}{z}) - 16 = 0 \underset{}{\underset{}{\dots \dots}} (1) \\ Put z = r (\cos θ + i \sin θ) \\ \Rightarrow \bar{z} = r (\cos θ - i \sin θ) \\ And \frac{z}{\bar{z}} = \cos 2 θ + i \sin 2 θ, \frac{\bar{z}}{z} = \cos 2 θ - i \sin 2 θ \\ | \bar{z} |^{2} = r^{2} \\ Substitute all these in equation (1),we get \\ r^{2} + \frac{4}{r^{2}} - 4 \cos 2 θ - 16 = 0 \end{array}$
$\begin{array}{l} \Rightarrow r^{4} + 4 - r^{2} (4 \cos 2 θ + 16) = 0 \\ Put r^{2} = t \\ \Rightarrow t^{2} - t (4 \cos 2 θ + 16) + 4 = 0 \\ \Rightarrow t = \frac{4 \cos 2 θ + 16 \pm \sqrt{(4 \cos 2 θ + 16)^{2} - 16}}{2} \\ \Rightarrow r^{2} = 2 (\cos 2 θ + 4) \pm 2 \sqrt{(\cos 2 θ + 4)^{2} - 1} \\ ∴ Maximum of r^{2} = 10 + 2 \sqrt{24}, w h e n \cos 2 θ = 1 \\ = 6 + 4 + 2 \sqrt{6 \times 4} \\ = (\sqrt{6} + \sqrt{4})^{2} \\ ∴ Maximum of = 2 + \sqrt{6} \end{array}$

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