If Amp 2-1-2-1-3- then Z represents a point on

Solution: argz 1/z+1=π/3 arg(z 1) arg(z+1)=π/3 tan^ 1(y/x 1) tan^ 1(y/x+1)=π/3 tan^ 1y{1/x 1 1/x+1}/1+y^2/x^2 1=π/3 tan^ 12y/x^2+y^2 1=π/3 x^2+y^2 1=2y/√(3) W ...

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Byju's Answer

Standard XII

Mathematics

Fractional Part Function

If Amp 2-1/2+...

Question

If Amp $(\frac{z - 1}{z + 1}) = \frac{π}{3}$ , then Z represents a point on

Straight Line

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Circle

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Ellipse

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Parabola

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Solution

The correct option is B Circle

Solution: $a r g \frac{z - 1}{z + 1} = \frac{π}{3}$
arg $(z - 1) - a r g (z + 1) = \frac{π}{3}$
$t a n^{- 1} (\frac{y}{x - 1}) - t a n^{- 1} (\frac{y}{x + 1}) = \frac{π}{3}$
$t a n^{- 1} \frac{y {\frac{1}{x - 1} - \frac{1}{x + 1}}}{1 + \frac{y^{2}}{x^{2} - 1}} = \frac{π}{3}$
$t a n^{- 1} \frac{2 y}{x^{2} + y^{2} - 1} = \frac{π}{3}$
$x^{2} + y^{2} - 1 = \frac{2 y}{\sqrt{3}}$
Which is equation of circle.Answer B.

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If Amp (z−1z+1)=π3, then Z represents a point on

The correct option is B Circle Solution: argz−1z+1=π3 arg(z−1)−arg(z+1)=π3 tan−1(yx−1)−tan−1(yx+1)=π3 tan−1y{1x−1−1x+1}1+y2x2−1=π3 tan−12yx2+y2−1=π3 x2+y2−1=2y√3 Which is equation of circle.Answer B.

If Amp $(\frac{z - 1}{z + 1}) = \frac{π}{3}$ , then Z represents a point on