If α and β are the roots of the equation 2 z^2-3 z-2 i=0 , where i= -1 , then 16 · Re ( α^ 19 +β^ 19 +α^ 11 +β^ 11 α^ 15 +β^ 15 ) · lm ( α^ 19 +β^ 19 +α^ 11 +β^ 11 α^ 15 +β^ 15 ) is equal to

Correct answer: A. 441 2 z^2-3 z-2 i=0 ...(i) 2 (z-(i)/(z))=3 As α, β are roots of (i) aligned & α-(i)/(α)=(3)/(2) & α^2-(1)/(α^2)-2 i=(9)/(4) & α^2-(1)/(α^2)=(9)/(4)+2 i aligned Squaring both sides aligned & α^4+(1)/(α^4)-2=(81)/(16)-4+9 i & α^4+(1)/(α^4)=(49)/(16)+9 i aligned Similarly, β^4+(1)/(β^4)=(49)/(16)+9 i aligned & α^ 19 +β^ 19 +α^ 11 +β^ 11 α^ 15 +β^ 15 & = α^ 15 (α^4+(1)/(α^4))+β^ 15 (β^4+(1)/(β^4)) α^ 15 +β^ 15 & =(49)/(16)+9 i & Re( α^ 19 +β^ 19 +α^ 11 +β^ 11 α^ 15 +β^ 15 )=(49)/(16) & Im( α^ 19 +β^ 19 +α^ 11 +β^ 11 α^ 15 +β^ 15 )=9 & 16 Re( α^ 19 +β^ 19 +α^ 11 +β^ 11 α^ 15 +β^ 15 ) · lm( α^ 19 +β^ 19 +α^ 11 +β^ 11 α^ 15 +β^ 15 ) & =16 × (49)/(16) × 9 & =441 aligned

JEE Main 202524 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 24 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Complex Number

JEE Main 2025 · 24 January · Morning Shift · Mathematics · Complex Number

If α and β are the roots of the equation 2 z^2-3 z-2 i=0 , where i=...

\alpha

and

\beta

are the roots of the equation

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

, where

\mathrm{i}=\sqrt{-1}

, then

16 \cdot \operatorname{Re}

\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right) \cdot \operatorname{lm}

\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right)

is equal to

Options

A $441$
B $398$
C $312$
D $409$

Correct answer

A. $441$

Step-by-step solution

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

...(i)

\left(z-\frac{i}{z}\right)=3

\alpha, \beta

are roots of (i)

\begin{aligned} & \alpha-\frac{i}{\alpha}=\frac{3}{2} \\ & \Rightarrow \alpha^2-\frac{1}{\alpha^2}-2 i=\frac{9}{4} \\ & \Rightarrow \alpha^2-\frac{1}{\alpha^2}=\frac{9}{4}+2 i \end{aligned}

Squaring both sides

\begin{aligned} & \Rightarrow \alpha^4+\frac{1}{\alpha^4}-2=\frac{81}{16}-4+9 i \\ & \Rightarrow \alpha^4+\frac{1}{\alpha^4}=\frac{49}{16}+9 i \end{aligned}

Similarly,

\beta^4+\frac{1}{\beta^4}=\frac{49}{16}+9 i

\begin{aligned} & \frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}} \\ & =\frac{\alpha^{15}\left(\alpha^4+\frac{1}{\alpha^4}\right)+\beta^{15}\left(\beta^4+\frac{1}{\beta^4}\right)}{\alpha^{15}+\beta^{15}} \\ & =\frac{49}{16}+9 i \\ & \operatorname{Re}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right)=\frac{49}{16} \\ & \operatorname{Im}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right)=9 \\ & \Rightarrow 16 \operatorname{Re}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right) \cdot \operatorname{lm}\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right) \\ & =16 \times \frac{49}{16} \times 9 \\ & =441 \end{aligned}

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