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If k + |k + z^2| = |z|^2 (k R^-), then possible argument of z

Question

If $k + ∣ ∣ k + z^{2} ∣ ∣ = ∣ z ∣^{2} (k \in R^{-}),$ then possible argument of $z$ is

A

0

B

π

C

\frac{π}{2}

D

none of these

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If

k + ∣ ∣ k + z^{2} ∣ ∣ = ∣ z ∣^{2} (k \in R^{-}),

then possible argument of

z

is

12 mins ago

Discuss this question LIVE

12 mins ago

Text solutionVerified

k + ∣ ∣ k + z^{2} ∣ ∣ = ∣ z ∣^{2} (k \in R^{-})

...(1)

Let

z = r (cos θ + i sin θ)

Substitute

z

in equation (1)

k + ∣ ∣ k + r^{2} (cos θ + i sin θ)^{2} ∣ ∣ = r^{2} \Rightarrow (k + r^{2} cos 2 θ)^{2} + (r^{2} sin 2 θ)^{2} = (r^{2} - k)^{2} \Rightarrow k^{2} + r^{4} (sin^{2} 2 θ + cos^{2} 2 θ) + 2 k r^{2} cos 2 θ = k^{2} + r^{4} - 2 k r^{2} \Rightarrow cos 2 θ = - 1

One of many solution of above equation is

2 θ = π

.

∴ θ = \frac{π}{2}

Hence, option 'C' is correct.

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Question Text	If $k + ∣ ∣ k + z^{2} ∣ ∣ = ∣ z ∣^{2} (k \in R^{-}),$ then possible argument of $z$ is
Answer Type	Text solution:1
Upvotes	150