If z - x - iy - x - y - R and Im2 z -1- i z -1-2- thenA- 4 x2-6 y-2-y2-0- 4 y 2-6 y -2- x -0- 4 y 2-6 y -2- x -0D- 4 y 2-6 y -2- x -0

Given : z = x+iy, so z = x iy Now, nbsp; 2z+1iz+1 nbsp;=2x+1+2iyi(x iy)+1 nbsp;=(2x+1)+2iy(1+y)+ix× nbsp;(1+y) ix(1+y) ix nbsp;= (2x+1)(1+y)+2xy+i ...

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Standard XII

Mathematics

Algebra of Complex Numbers

If z = x + iy...

Question

If $z = x + i y, x, y \in R$ and $I m (\frac{2 z + 1}{i ¯ ¯ ¯ z + 1}) = - 2$ , then

4 x^{2} + 6 y + 2 + y^{2} = 0

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4 y^{2} + 6 y - 2 + x = 0

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4 y^{2} + 6 y + 2 - x = 0

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4 y^{2} + 6 y + 2 + x = 0

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Solution

The correct option is C $4 y^{2} + 6 y + 2 - x = 0$
Given : $z = x + i y$ , so $¯ ¯ ¯ z = x - i y$
Now,
$\frac{2 z + 1}{i ¯ ¯ ¯ z + 1} = \frac{2 x + 1 + 2 i y}{i (x - i y) + 1} = \frac{(2 x + 1) + 2 i y}{(1 + y) + i x} \times \frac{(1 + y) - i x}{(1 + y) - i x} = \frac{(2 x + 1) (1 + y) + 2 x y + i [2 y (1 + y) - x (2 x + 1)]}{(1 + y)^{2} + x^{2}}$
So,
$I m (\frac{2 z + 1}{i ¯ ¯ ¯ z + 1}) = - 2 \Rightarrow \frac{2 y (1 + y) - x (2 x + 1)}{(1 + y)^{2} + x^{2}} = - 2 \Rightarrow 2 x^{2} + x - 2 y - 2 y^{2} = 2 y^{2} + 4 y + 2 + 2 x^{2} \Rightarrow 4 y^{2} + 6 y + 2 - x = 0$

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z = x + i y, x, y \in R

and

I m (\frac{2 z + 1}{i ¯ ¯ ¯ z + 1}) = - 2

, then

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II: $x^{2} + y^{2} = 1$ $x^{2} + y^{2} - 2 x - 6 y + 6 = 0$	b) $1$
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Algebra of Complex Numbers

Standard XII Mathematics

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If z=x+iy, x,y∈R and Im(2z+1i¯¯¯z+1)=−2, then

If $z = x + i y, x, y \in R$ and $I m (\frac{2 z + 1}{i ¯ ¯ ¯ z + 1}) = - 2$ , then