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  • Give answer! - Complex numbers and quadratic equations - JEE Main-4

\left ( 1+i \right )^{8}+\left ( 1-i \right )^{9}  equals

  • Option 1)

    16i

  • Option 2)

    -16i

  • Option 3)

    32+16i

  • Option 4)

    32-16i

 

Answers (1)

best_answer

\left ( 1+i \right )^{8}=\left ( \left ( 1+i \right )^{2} \right )^{4}=\left ( 1+i^{2}+2i \right )^{4}=16

\left ( 1-i \right )^{9}=\left ( \left ( 1-i \right )^{2} \right )^{4}\cdot \left ( 1-i \right )^{1}=\left ( 1+i^{2}-2i \right )^{4}\cdot \left ( 1-i \right )

=16\left ( 1-i \right )

=16-16i

\therefore \left ( 1+i \right )^{8}+\left ( 1-i \right )^{9}=16+16-16i=32-16i

\therefore  Option (D)

 

Power of i in Complex Numbers -

i^{4n}=1,i^{4n+1}=i, i^{4n+2}=-1,i^{4n+3}=-i

- wherein

n\epsilon Integer

 

 


Option 1)

16i

This is incorrect

Option 2)

-16i

This is incorrect

Option 3)

32+16i

This is incorrect

Option 4)

32-16i

This is correct

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