Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • I need help with - Complex numbers and quadratic equations - JEE Main-5

\left (\frac{-1}{2}+\frac{i\sqrt{3}}{2} \right )^{100}+\left (\frac{-1}{2}-\frac{i\sqrt{3}}{2} \right )^{200}  equals

  • Option 1)

    -1+i\sqrt{3}

  • Option 2)

    -1-i\sqrt{3}

  • Option 3)

    1+i\sqrt{3}

  • Option 4)

    1-i\sqrt{3}

 

Answers (1)

best_answer

given is \rightarrow w^{100}+\left ( w^{2} \right )^{200}=w^{100}+w^{400}

=\left ( w^{3} \right )^{33}\cdot \left ( w^{3} \right )^{133}\cdot w=2w=-1+i\sqrt{3}

 

Cube roots of unity -

z=\left ( 1 \right )^{\frac{1}{3}}\Rightarrow z=\cos \frac{2k\pi }{3}+i\sin \frac{2k\pi }{3}

k=0,1,2 so z gives three roots 

\Rightarrow 1,\frac{-1}{2}+i\frac{\sqrt{3}}{2}\left ( \omega \right ),\frac{-1}{2}-i\frac{\sqrt{3}}{2}\left ( \omega^{2} \right )

- wherein

\omega=\frac{-1}{2} +\frac{i\sqrt{3}}{2},\omega^{2}=\frac{-1}{2} -\frac{i\sqrt{3}}{2},\omega^{3}=1, 1+\omega+\omega^{2}=0

1,\omega,\omega^{2} are cube roots of unity.

 

 


Option 1)

-1+i\sqrt{3}

This is correct

Option 2)

-1-i\sqrt{3}

This is incorrect

Option 3)

1+i\sqrt{3}

This is incorrect

Option 4)

1-i\sqrt{3}

This is incorrect

Posted by

Aadil

View full answer

JEE Main high-scoring chapters and topics

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

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025: Previous year NIT Calicut cut offs; opening, closing ranks
anam-khan

JEE Main 2025 Admit Card Live: Session 1 hall ticket soon at jeemain.nta.nic.in; photo correction, cut-offs
anam-khan

NTA finds errors in JEE Main 2025 photos; asks to upload correct images by January 17

Latest Question