If Im2 z -1- iz -1-3- then locus of z isA- a straight lineB- none of theseC- a parabolaD- a circle

The correct option is D a circleLet z=x+iy Then 2z+1/iz+1= (2x+1)+2iy/(1 y)+ix×(1 y) ix/(1 y) ix =(2x+1)(1 y)+2xy +i(2y(1 y) x(2x+1))/(1 y)^2+x^2 nbsp;Give ...

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Byju's Answer

Standard XII

Mathematics

Equation of Circle in Complex Form

If Im2 z +1/ ...

Question

If $Im (\frac{2 z + 1}{i z + 1}) = - 3$ , then locus of $z$ is

a circle

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a parabola

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a straight line

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none of these

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Solution

The correct option is A a circle
$Let z = x + i y$
$Then \frac{2 z + 1}{i z + 1} = \frac{(2 x + 1) + 2 i y}{(1 - y) + i x} \times \frac{(1 - y) - i x}{(1 - y) - i x}$

$= \frac{(2 x + 1) (1 - y) + 2 x y + i (2 y (1 - y) - x (2 x + 1))}{(1 - y)^{2} + x^{2}}$
$Given, Im (\frac{2 z + 1}{i z + 1}) = - 3$
$\Rightarrow \frac{2 y (1 - y) - x (2 x + 1)}{(1 - y)^{2} + x^{2}} = - 3$
$\Rightarrow x^{2} + y^{2} - x - 4 y + 3 = 0$
So, The locus of $z$ is a circle.

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If Im(2z+1iz+1)=−3, then locus of z is

The correct option is A a circleLet z=x+iy Then 2z+1iz+1=(2x+1)+2iy(1−y)+ix×(1−y)−ix(1−y)−ix =(2x+1)(1−y)+2xy+i(2y(1−y)−x(2x+1))(1−y)2+x2 Given, Im(2z+1iz+1)=−3 ⇒2y(1−y)−x(2x+1)(1−y)2+x2=−3 ⇒x2+y2−x−4y+3=0 So, The locus of z is a circle.

If $Im (\frac{2 z + 1}{i z + 1}) = - 3$ , then locus of $z$ is