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  • How to solve this problem- - Complex numbers and quadratic equations - JEE Main-2

1+i+i^{2}+i^{3}+i^{4}+i^{5}  equals

  • Option 1)

    1+i

  • Option 2)

    1-i

  • Option 3)

    i

  • Option 4)

    -i

 

Answers (2)

best_answer

i^{2}=-1,\, i^{3}=i^{2}\cdot i=-i

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

i^{5}=\left ( i^{4} \right )i=i

So,1+i+i^{2}+i^{3}+i^{4}+i^{5}=1+i-1-i+1+i= 1+i

\therefore Option (A)

 

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)

1+i

This is correct

Option 2)

1-i

This is incorrect

Option 3)

i

This is incorrect

Option 4)

-i

This is incorrect

Posted by

Himanshu

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JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

i^2 =-1

i^3 =-i

i^4 =1

i+i^2+i^3+i^4 = i - 1 - i + 1

                          = 0

1+0+i = 1+i

 

 

 

 

 

Posted by

Akanksha Kawate

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