If $\alpha$ and $\beta$ are the roots of the equation $2 z^{2}-3 z-2 i=0$, where $i=\sqrt{-1}$, then $16\,\operatorname{Re}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right) \cdot \operatorname{Im}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right)$ is equal to

Complex Numbers — JEE Main · 2025 · 24 Jan 2025 · Shift 1 Question with Solution

α

and

β

are the roots of the equation

2 z^{2} - 3 z - 2 i = 0

, where

i = - 1

, then

16 Re (\frac{α ^{19} + β ^{19} + α ^{11} + β ^{11}}{α ^{15} + β ^{15}}) \cdot Im (\frac{α ^{19} + β ^{19} + α ^{11} + β ^{11}}{α ^{15} + β ^{15}})

is equal to

Answer: (D)

From

2 z^{2} - 3 z - 2 i = 0

, dividing by

2 z

(valid since

z \neq = 0

z - \frac{i}{z} = \frac{3}{2} .

Both roots

α, β

satisfy

z - \frac{i}{z} = \frac{3}{2}

. Squaring,

z^{2} - 2 i - \frac{1}{z ^{2}} = \frac{9}{4} \Rightarrow z^{2} - \frac{1}{z ^{2}} = \frac{9}{4} + 2 i .

Squaring again, using

(z^{2} - \frac{1}{z ^{2}})^{2} = z^{4} - 2 + \frac{1}{z ^{4}}

z^{4} + \frac{1}{z ^{4}} - 2 = (\frac{9}{4} + 2 i)^{2} = \frac{81}{16} + 9 i - 4 = \frac{17}{16} + 9 i \Rightarrow z^{4} + \frac{1}{z ^{4}} = \frac{49}{16} + 9 i,

and this holds equally for

α

and

β

.

Writing

α^{19} + α^{11} = α^{15} (α^{4} + \frac{1}{α ^{4}})

and similarly for

β

α^{19} + β^{19} + α^{11} + β^{11} = α^{15} (α^{4} + \frac{1}{α ^{4}}) + β^{15} (β^{4} + \frac{1}{β ^{4}}) = (α^{15} + β^{15}) (\frac{49}{16} + 9 i),

since both bracketed factors equal

\frac{49}{16} + 9 i

. Dividing by

α^{15} + β^{15}

\frac{α ^{19} + β ^{19} + α ^{11} + β ^{11}}{α ^{15} + β ^{15}} = \frac{49}{16} + 9 i .

Re = \frac{49}{16}

and

Im = 9

, and

16 \cdot Re (\cdot) \cdot Im (\cdot) = 16 \times \frac{49}{16} \times 9 = 441.

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Single correct option