Let S be the set of all (\alpha, \beta), \pi\lt \alpha, \beta\lt 2 \pi , for which the complex number \frac1-i \sin \alpha1+2 i \sin \alpha is purely imaginary and \frac1+i \cos \beta1-2 i \cos \beta is purely real. Let Z_\alpha \beta=\sin 2 \alpha+i \cos 2 \beta,(\alpha, \beta) \in S . Then \sum\limits_(\alpha, \beta) \in S(i Z_\alpha \beta+\frac1i \barZ_\alpha \beta) is equal to :

Complex Numbers PYQ — JEE Main 2022 · 27 Jul · Shift 2 · Q23 with Solution

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Correct answer: C

Find all admissible $\alpha$ from $\frac{1-i\sin\alpha}{1+2i\sin\alpha}$ being purely imaginary.

Let $x=\sin\alpha$ . Then $\frac{1-ix}{1+2ix}.$ Multiply numerator and denominator by the conjugate of the denominator:

=\frac{(1-ix)(1-2ix)}{1+4x^2}.$$ Now, $$ (1-ix)(1-2ix)=1-3ix-2x^2. $$ So $$\frac{1-ix}{1+2ix}=\frac{1-2x^2}{1+4x^2}-i\frac{3x}{1+4x^2}.$$ For this to be purely imaginary, its real part must be zero: $$1-2x^2=0 \implies x^2=\frac12.$$ Hence $$\sin\alpha=\pm \frac{1}{\sqrt2}.$$ Given $\pi<\alpha<2\pi$, we get $$\alpha=\frac{5\pi}{4},\ \frac{7\pi}{4}.$$ --- 2. **Find all admissible $\beta$** from $$\frac{1+i\cos\beta}{1-2i\cos\beta}$$ being purely real. Let $y=\cos\beta$. Then $$\frac{1+iy}{1-2iy}.$$ Multiply numerator and denominator by $1+2iy$: $$\frac{1+iy}{1-2iy}\cdot\frac{1+2iy}{1+2iy} =\frac{(1+iy)(1+2iy)}{1+4y^2}.$$ Now, $$ (1+iy)(1+2iy)=1+3iy-2y^2. $$ So $$\frac{1+i\cos\beta}{1-2i\cos\beta}=\frac{1-2y^2}{1+4y^2}+i\frac{3y}{1+4y^2}.$$ For this to be purely real, imaginary part must be zero: $$3y=0 \implies y=0.$$ Thus $$\cos\beta=0.$$ Given $\pi<\beta<2\pi$, we get $$\beta=\frac{3\pi}{2}.$$ --- 3. Therefore the set $S$ consists of two pairs: $$S=\left\{\left(\frac{5\pi}{4},\frac{3\pi}{2}\right),\left(\frac{7\pi}{4},\frac{3\pi}{2}\right)\right\}.$$ --- 4. Compute $Z_{\alpha\beta}=\sin2\alpha+i\cos2\beta$ for each pair. Since $$\beta=\frac{3\pi}{2}\implies 2\beta=3\pi \implies \cos2\beta=\cos3\pi=-1,$$ we have $$Z_{\alpha\beta}=\sin2\alpha-i.$$ - For $\alpha=\frac{5\pi}{4}$: $$2\alpha=\frac{5\pi}{2},\quad \sin\frac{5\pi}{2}=1,$$ so $$Z=1-i.$$ - For $\alpha=\frac{7\pi}{4}$: $$2\alpha=\frac{7\pi}{2},\quad \sin\frac{7\pi}{2}=-1,$$ so $$Z=-1-i.$$ --- 5. Evaluate $$iZ+\frac{1}{i\bar Z}$$ for each value. ## Case 1: $Z=1-i$ Then $$\bar Z=1+i,$$ so $$iZ=i(1-i)=1+i.$$ Also, $$i\bar Z=i(1+i)=i+i^2=-1+i,$$ thus $$\frac{1}{i\bar Z}=\frac{1}{-1+i}=\frac{-1-i}{(-1)^2+1^2}=\frac{-1-i}{2}.$$ Hence $$iZ+\frac{1}{i\bar Z}=(1+i)+\frac{-1-i}{2}=\frac{1+i}{2}.$$ ## Case 2: $Z=-1-i$ Then $$\bar Z=-1+i,$$ so $$iZ=i(-1-i)=1-i.$$ Also, $$i\bar Z=i(-1+i)=-i-1=-1-i,$$ thus $$\frac{1}{i\bar Z}=\frac{1}{-1-i}=\frac{-1+i}{2}.$$ Hence $$iZ+\frac{1}{i\bar Z}=(1-i)+\frac{-1+i}{2}=\frac{1-i}{2}.$$ --- 6. Sum over all $(\alpha,\beta)\in S$: $$\frac{1+i}{2}+\frac{1-i}{2}=1.$$ Therefore, $$\sum_{(\alpha,\beta)\in S}\left(iZ_{\alpha\beta}+\frac{1}{i\bar Z_{\alpha\beta}}\right)=1.$$ So the correct option is: $$\boxed{\text{C}}$$

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View written solutionFree

Correct answer: C

Find all admissible $\alpha$ from $\frac{1-i\sin\alpha}{1+2i\sin\alpha}$ being purely imaginary.

Let $x=\sin\alpha$ . Then $\frac{1-ix}{1+2ix}.$ Multiply numerator and denominator by the conjugate of the denominator:

=\frac{(1-ix)(1-2ix)}{1+4x^2}.$$ Now, $$ (1-ix)(1-2ix)=1-3ix-2x^2. $$ So $$\frac{1-ix}{1+2ix}=\frac{1-2x^2}{1+4x^2}-i\frac{3x}{1+4x^2}.$$ For this to be purely imaginary, its real part must be zero: $$1-2x^2=0 \implies x^2=\frac12.$$ Hence $$\sin\alpha=\pm \frac{1}{\sqrt2}.$$ Given $\pi<\alpha<2\pi$, we get $$\alpha=\frac{5\pi}{4},\ \frac{7\pi}{4}.$$ --- 2. **Find all admissible $\beta$** from $$\frac{1+i\cos\beta}{1-2i\cos\beta}$$ being purely real. Let $y=\cos\beta$. Then $$\frac{1+iy}{1-2iy}.$$ Multiply numerator and denominator by $1+2iy$: $$\frac{1+iy}{1-2iy}\cdot\frac{1+2iy}{1+2iy} =\frac{(1+iy)(1+2iy)}{1+4y^2}.$$ Now, $$ (1+iy)(1+2iy)=1+3iy-2y^2. $$ So $$\frac{1+i\cos\beta}{1-2i\cos\beta}=\frac{1-2y^2}{1+4y^2}+i\frac{3y}{1+4y^2}.$$ For this to be purely real, imaginary part must be zero: $$3y=0 \implies y=0.$$ Thus $$\cos\beta=0.$$ Given $\pi<\beta<2\pi$, we get $$\beta=\frac{3\pi}{2}.$$ --- 3. Therefore the set $S$ consists of two pairs: $$S=\left\{\left(\frac{5\pi}{4},\frac{3\pi}{2}\right),\left(\frac{7\pi}{4},\frac{3\pi}{2}\right)\right\}.$$ --- 4. Compute $Z_{\alpha\beta}=\sin2\alpha+i\cos2\beta$ for each pair. Since $$\beta=\frac{3\pi}{2}\implies 2\beta=3\pi \implies \cos2\beta=\cos3\pi=-1,$$ we have $$Z_{\alpha\beta}=\sin2\alpha-i.$$ - For $\alpha=\frac{5\pi}{4}$: $$2\alpha=\frac{5\pi}{2},\quad \sin\frac{5\pi}{2}=1,$$ so $$Z=1-i.$$ - For $\alpha=\frac{7\pi}{4}$: $$2\alpha=\frac{7\pi}{2},\quad \sin\frac{7\pi}{2}=-1,$$ so $$Z=-1-i.$$ --- 5. Evaluate $$iZ+\frac{1}{i\bar Z}$$ for each value. ## Case 1: $Z=1-i$ Then $$\bar Z=1+i,$$ so $$iZ=i(1-i)=1+i.$$ Also, $$i\bar Z=i(1+i)=i+i^2=-1+i,$$ thus $$\frac{1}{i\bar Z}=\frac{1}{-1+i}=\frac{-1-i}{(-1)^2+1^2}=\frac{-1-i}{2}.$$ Hence $$iZ+\frac{1}{i\bar Z}=(1+i)+\frac{-1-i}{2}=\frac{1+i}{2}.$$ ## Case 2: $Z=-1-i$ Then $$\bar Z=-1+i,$$ so $$iZ=i(-1-i)=1-i.$$ Also, $$i\bar Z=i(-1+i)=-i-1=-1-i,$$ thus $$\frac{1}{i\bar Z}=\frac{1}{-1-i}=\frac{-1+i}{2}.$$ Hence $$iZ+\frac{1}{i\bar Z}=(1-i)+\frac{-1+i}{2}=\frac{1-i}{2}.$$ --- 6. Sum over all $(\alpha,\beta)\in S$: $$\frac{1+i}{2}+\frac{1-i}{2}=1.$$ Therefore, $$\sum_{(\alpha,\beta)\in S}\left(iZ_{\alpha\beta}+\frac{1}{i\bar Z_{\alpha\beta}}\right)=1.$$ So the correct option is: $$\boxed{\text{C}}$$

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