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Can someone explain - Complex numbers and quadratic equations - JEE Main-10

All the points in the set 

S=\left \{ \frac{\alpha +i}{\alpha -i}:\alpha\; \epsilon\; \mathbf{R} \right \}\left ( i=\sqrt{-1} \right )  lie on a :

  • Option 1)

    straight line whose slope is 1.

  • Option 2)

    circle whose radius is 1.

  • Option 3)

    circle whose radius is \sqrt{2} .

  • Option 4)

    straight line whose slop is -1

 
Answers (1)
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S solutionqc

Given , S= \frac{\alpha +i}{\alpha -i}:\alpha\; \epsilon\; \mathbf{R}

  S=x+iy

x+iy=\frac{\left ( \alpha +i \right )^{2}}{\alpha ^{2}+1}=\frac{\alpha ^{2}-1+2\alpha i}{\alpha ^{2}+1}

x=\frac{\alpha ^{2}-1}{\alpha +1}=y=\frac{2\alpha}{\alpha ^{2}+1}

x^{2}+y^{2}=\frac{\left ( \alpha ^{2}-1^{2} \right )+4\alpha ^{2}}{\left ( \alpha ^{2}+1 \right )^{2}}=1

x^{2}+y^{2}=1


Option 1)

straight line whose slope is 1.

Option 2)

circle whose radius is 1.

Option 3)

circle whose radius is \sqrt{2} .

Option 4)

straight line whose slop is -1

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