If complex number z(z=2) satisfies the equation z2=4z+∣z∣2+∣z∣31, then the value of ∣z∣−4 is

z=reiθ z1=r1e−iθ Substituting in the given equationr2e2iθ =4reiθ +r2+r31Comparing real part and imaginary part r2cos2θ=4rcosθ+r2+r31 ...(1)and r2sin2θ=4rsinθ ... (2)Solving (2), we getr2sin2θ=4rsinθ⇒2r2sinθcosθ−4rsinθ=0⇒rsinθ=0 or rcosθ=2Taking casesCase I: rsinθ=0, ⇒cosθ=±1Use cosθ=1 in (1)There is no solution.Use cosθ=−1 in (1)r4=41Case II:rcosθ=2There is no solution.Therefore∣z∣−4=r−4=4

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If complex number $z (z \neq = 2)$ satisfies the equation $z^{2} = 4 z + ∣ z ∣^{2} + \frac{1}{∣ z ∣ ^{3}}$ , then the value of $∣ z ∣^{- 4}$ is

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If complex number

z (z \neq = 2)

satisfies the equation

z^{2} = 4 z + ∣ z ∣^{2} + \frac{1}{∣ z ∣ ^{3}}

, then the value of

∣ z ∣^{- 4}

7 mins ago

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Text solutionVerified

z = re^{i θ} \frac{1}{z} = \frac{1}{r} e^{- i θ}

Substituting in the given equation

r^{2} e^{2 i θ} = 4 r e^{i θ} + r^{2} + \frac{1}{r ^{3}}

Comparing real part and imaginary part

r^{2} cos 2 θ = 4 r cos θ + r^{2} + \frac{1}{r ^{3}}

...(1)

and

r^{2} sin 2 θ = 4 r sin θ

... (2)

Solving (2), we get

r^{2} sin 2 θ = 4 r sin θ

\Rightarrow 2 r^{2} sin θ cos θ - 4 r sin θ = 0

\Rightarrow r sin θ = 0

r cos θ = 2

Taking cases
Case I:

r sin θ = 0

\Rightarrow cos θ = \pm 1

Use

cos θ = 1

in (1)

There is no solution.

Use

cos θ = - 1

in (1)

r^{4} = \frac{1}{4}

Case II:

r cos θ = 2

There is no solution.

Therefore

∣ z ∣^{- 4} = r^{- 4} = 4

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Question Text	If complex number $z (z \neq = 2)$ satisfies the equation $z^{2} = 4 z + ∣ z ∣^{2} + \frac{1}{∣ z ∣ ^{3}}$ , then the value of $∣ z ∣^{- 4}$ is
Answer Type	Text solution:1
Upvotes	150

If complex number z(z=2) satisfies the equation z2=4z+∣z∣2+∣z∣31​, then the value of ∣z∣−4 is

Text solutionVerified

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If complex number $z (z \neq = 2)$ satisfies the equation $z^{2} = 4 z + ∣ z ∣^{2} + \frac{1}{∣ z ∣ ^{3}}$ , then the value of $∣ z ∣^{- 4}$ is