Let S be the set of all $(\alpha, \beta), \pi<\alpha, \beta<2 \pi$, for which the complex number $\frac{1-i \sin \alpha}{1+2 i \sin \alpha}$ is purely imaginary and $\frac{1+i \cos \beta}{1-2 i \cos \beta}$ is purely real. Let $\mathrm{Z}_{\alpha \beta}=\sin 2 \alpha+\mathrm{i} \cos 2 \beta,(\alpha, \beta) \in \mathrm{S}$. Then $\sum_{(\alpha, \beta) \in \mathrm{S}}\left(i \mathrm{Z}_{\alpha \beta}+\frac{1}{i \overline{\mathrm{Z}}_{\alpha \beta}}\right)$ is equal to :

Complex Numbers — JEE Main · 2022 · 27 Jul 2022 · Shift 2 Question with Solution

Let S be the set of all

(α, β), π < α, β < 2 π

, for which the complex number

\frac{1 - i s i n α}{1 + 2 i s i n α}

is purely imaginary and

\frac{1 + i c o s β}{1 - 2 i c o s β}

is purely real. Let

Z_{α β} = sin 2 α + i cos 2 β, (α, β) \in S

.
Then

\sum_{(α, β) \in S} (i Z_{α β} + \frac{1}{i Z _{α β}})

is equal to :

Answer: (C)

For

\frac{1 - i s i n α}{1 + 2 i s i n α}

to be purely imaginary, the real part must be zero. Rationalizing:

\frac{( 1 - i s i n α ) ( 1 - 2 i s i n α )}{1 + 4 s i n ^{2} α} = \frac{1 - 2 s i n ^{2} α - 3 i s i n α}{1 + 4 s i n ^{2} α}

. Real part = 0 gives

sin^{2} α = \frac{1}{2}

, so

α \in {\frac{5 π}{4}, \frac{7 π}{4}}

(π, 2 π)

.

For

\frac{1 + i c o s β}{1 - 2 i c o s β}

to be purely real, the imaginary part must be zero. This gives

3 cos β = 0

, so

β = \frac{3 π}{2}

.

For

α = \frac{5 π}{4}

Z_{α β} = sin \frac{5 π}{2} + i cos 3 π = 1 - i

.
For

α = \frac{7 π}{4}

Z_{α β} = sin \frac{7 π}{2} + i cos 3 π = - 1 - i

.

The sum evaluates to

[i (1 - i) + \frac{1}{i ( 1 + i )}] + [i (- 1 - i) + \frac{1}{i ( - 1 + i )}] = 2 - 1 = 1

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Single correct option