Q

# 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 $-1$

Views

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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