Given z is a complex number satisfying z 2- z - z -2-64- z -5-0 and Re z -1-2 where Re z denotes real part of z - then - z - is less thanA- 1B- 2C- 3D- 4

The correct options are C 3 D 4 ∵ z^2 z=|z^2| 64/|z|^5 = a purely real number ⇒ z^2 z = z̅^2 z̅ ⇒ (z^2 z̅^2) (z z̅)=0 if z z̅=0 ⇒ z=z̅ ⇒ z is purely ...

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Byju's Answer

Standard XII

Mathematics

Algebra of Complex Numbers

Given z is a ...

Question

Given z is a complex number satisfying $z^{2} - z - | z |^{2} + \frac{64}{| z |^{5}} = 0$ and Re(z) $\neq \frac{1}{2}$ (where Re(z) denotes real part of z), then |z| is less than

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Solution

The correct options are
C
3

D
4

$∵ z^{2} - z = | z^{2} | - \frac{64}{| z |^{5}}$ = a purely real number
$\Rightarrow z^{2} - z = {¯ z}^{2} - ¯ z \Rightarrow (z^{2} - {¯ z}^{2}) - (z - ¯ z) = 0$
if $z - ¯ z = 0 \Rightarrow z = ¯ z$
$\Rightarrow$ z is purely real number
So put z = x in given equation.
$x^{2} - x - | x^{2} | + \frac{64}{x^{5}} = 0 (i f x > 0)$
( $∵$ x < 0 not possible)

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Given z is a complex number satisfying z2−z−|z|2+64|z|5=0 and Re(z) ≠12 (where Re(z) denotes real part of z), then |z| is less than

The correct options are C 3 D 4 ∵ z2−z=|z2|−64|z|5 = a purely real number ⇒z2−z=¯z2−¯z ⇒(z2−¯z2)−(z−¯z)=0 if z−¯z=0 ⇒z=¯z ⇒ z is purely real number So put z = x in given equation. x2−x−|x2|+64x5=0 (if x>0) (∵ x < 0 not possible)

Given z is a complex number satisfying $z^{2} - z - | z |^{2} + \frac{64}{| z |^{5}} = 0$ and Re(z) $\neq \frac{1}{2}$ (where Re(z) denotes real part of z), then |z| is less than