Let z be a complex number such that -2 z -1- z -1 and z - then minimum value of 8 sin 2- isA- 0B- 8C- 7D- 5

Given, |2z + 1z| = 1 ⋯ (i) We know, z = r(cosθ + isinθ) ⋯ (ii) Putting (ii) in (i) |2(r(cosθ + isinθ)) + 1r(cosθ + isinθ)| = 1 ⇒(2r + 1r)^2cos^2θ + (2r 1r)^2s ...

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Standard XII

Mathematics

Geometrical Representation of Argument and Modulus

Let z be a co...

Question

Let $z$ be a complex number such that $∣ ∣ ∣ 2 z + \frac{1}{z} ∣ ∣ ∣ = 1$ and $arg (z) = θ$ , then minimum value of $8 {sin}^{2} θ$ is

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Solution

The correct option is C $7$
Given, $∣ ∣ ∣ 2 z + \frac{1}{z} ∣ ∣ ∣ = 1 \dots (i)$
We know, $z = r (cos θ + i sin θ) \dots (i i)$
Putting $(i i)$ in $(i)$
$∣ ∣ ∣ 2 (r (cos θ + i sin θ)) + \frac{1}{r (cos θ + i sin θ)} ∣ ∣ ∣ = 1$
$\Rightarrow {(2 r + \frac{1}{r})}^{2} {cos}^{2} θ + {(2 r - \frac{1}{r})}^{2} {sin}^{2} θ = 1$
$\Rightarrow {(2 r + \frac{1}{r})}^{2} - 1 = ({(2 r + \frac{1}{r})}^{2} - {(2 r - \frac{1}{r})}^{2}) {sin}^{2} θ$
$\Rightarrow 4 r^{2} + \frac{1}{r^{2}} + 3 = 8 {sin}^{2} θ$
$\Rightarrow 4 r^{2} + \frac{1}{r^{2}} = 8 {sin}^{2} θ - 3$
By using $A . M . \geq G . M .$ , we have
$\frac{4 r^{2} + \frac{1}{r^{2}}}{2} \geq \sqrt{4 r^{2} \times \frac{1}{r^{2}}}$
$\Rightarrow 4 r^{2} + \frac{1}{r^{2}} \geq 4$

$∴ 8 {sin}^{2} θ - 3 \geq 4$
$\Rightarrow 8 {sin}^{2} θ \geq 7$

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Geometrical Representation of Argument and Modulus

Standard XII Mathematics

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Let $z$ be a complex number such that $∣ ∣ ∣ 2 z + \frac{1}{z} ∣ ∣ ∣ = 1$ and $arg (z) = θ$ , then minimum value of $8 {sin}^{2} θ$ is