Test Index

Complex Numbers Part 2

Section: Mathematics

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Question : 94 of 106

Marks: +1, -0

Let

S

be the set of all

(\alpha, \beta), \pi < \alpha, \beta < 2\pi

, for which the complex number

\; \frac{1-i\sin\alpha}{1+2i\sin\alpha}

is purely imaginary and

\; \frac{1+i\cos\beta}{1-2i\cos\beta}

is purely real. Let

Z_{\alpha \beta} = \sin 2\alpha + i \cos 2\beta, (\alpha, \beta) \in S

\sum\limits_{(\alpha, \beta) \in S} \left( i Z_{\alpha \beta} + \; \frac{1}{i Z_{\alpha \beta}} \right)

[27-Jul-2022-Shift-2]

3
$3 i$
1
$2 - i$

Solution:

\because \; \frac{1-i\sin\alpha}{1+2i\sin\alpha}

is purely imaginary

\therefore \; \frac{1-i\sin\alpha}{1+2i\sin\alpha} + \; \frac{1+i\sin\alpha}{1-2i\sin\alpha} = 0

\Rightarrow 1-2\sin^2\alpha = 0

\therefore \alpha = \; \frac{5\pi}{4}, \; \frac{7\pi}{4}

and

\; \frac{1+i\cos\beta}{1-2i\cos\beta}

is purely real

\; \frac{1+i\cos\beta}{1-2i\cos\beta} - \; \frac{1-i\cos\beta}{1+2i\cos\beta} = 0

\Rightarrow \cos\beta = 0

\therefore \beta = \; \frac{3\pi}{2}

\therefore S=\{ \left(\; \frac{5\pi}{2}, \; \frac{3\pi}{2}\right), \left(\; \frac{7\pi}{4}, \; \frac{3\pi}{2}\right) \}

Z_{\alpha \beta} = 1 - i

and

Z_{\alpha \beta} = -1 - i

\therefore \sum\limits_{(\alpha, \beta) \in S} \left( i Z_{\alpha \beta} + \; \frac{1}{i \overline{Z}_{\alpha \beta}} \right) = i(-2 i) + \; \frac{1}{i} \left[ \; \frac{1}{1+i} + \; \frac{1}{-1+i} \right]

=2+\; \frac{1}{i} \; \frac{2i}{-2}=1

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