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  • Help me solve this - Complex numbers and quadratic equations - JEE Main-12

If \alpha \; and\; \beta be the roots of the equation x^{2}-2x+2=0, then the least value of n for which \left ( \frac{\alpha }{\beta } \right )^{n}=1 is :

  • Option 1)

    2

  • Option 2)

    5

  • Option 3)

    4

  • Option 4)

    3

 

Answers (1)

best_answer

x^{2}-2x+2=0

x=\frac{2\pm \sqrt{4-8}}{2}=\frac{2\pm \sqrt{-4}}{2}=\frac{2\pm 2i}{2}

                                                                =1\pm i

\alpha =1+i

\beta =1-i          \left ( \frac{\alpha }{\beta } \right )=\left ( \frac{1+i}{1-i} \right )=\left ( \frac{(1+i) (1+i)}{(1+i)(1-i)} \right )

                                          =\left ( \frac{(1+i)^{2}}{1+1} \right )=\frac{(1+i)^{2}}{2}

=\frac{1-1+2i}{2}=i

So (i)^{n}=1\Rightarrow i=4

 


Option 1)

2

Option 2)

5

Option 3)

4

Option 4)

3

Posted by

solutionqc

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