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Question
sinx + icos2x and cosx – isin2x are conjugate to each other for ______.
Options
x = nπ
x = `(n + 1/2) pi/2`
x = 0
No value of x
Solution
sinx + icos2x and cosx – isin2x are conjugate to each other for x = 0.
Explanation:
Let z = sinx + icos2x
`barz` = sinx – icos2x
But we are given that `barz` = cosx – isin2x
∴ sinx – icos2x = cosx – isin2x
Comparing the real and imaginary parts, we get
sinx = cosx and cos2x = sin2x
⇒ tanx = 1 and tan2x = 1
⇒ tanx = `tan pi/4` and tan2x = `pi/4`
∴ x = `npi + pi/4`, n ∈ I and 2x = `npi + pi/4`
⇒ x = 2x
⇒ 2x – x = 0
⇒ x = 0
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