Let z - x - iy- where x and y are real- The points x - y in the X - Y plane for which Z - i - Z - i is purely imaginary lie onA- a straight lineB- an ellipseC- a hyperbolaD- a circle

The correct option is D a circlez+iz i =x+i(y+1)x+i(y 1) =[x+i(y+1)][x i(y 1)]x^2+(y 1)^2 =[x^2+(y^2 1)+2ix]x^2+(y 1)^2 Since nbsp;above expression is pu ...

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Byju's Answer

Standard XII

Mathematics

Complex Numbers

Let z = x + i...

Question

Let $z = x + i y$ , where $x$ and $y$ are real. The points $(x, y)$ in the $X - Y$ plane for which
$\frac{z + i}{z - i}$ is purely imaginary lie on

a straight line

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an ellipse

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a hyperbola

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a circle

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Solution

The correct option is D a circle
$\frac{z + i}{z - i} = \frac{x + i (y + 1)}{x + i (y - 1)} = \frac{[x + i (y + 1)] [x - i (y - 1)]}{x^{2} + (y - 1)^{2}} = \frac{[x^{2} + (y^{2} - 1) + 2 i x]}{x^{2} + (y - 1)^{2}}$

Since above expression is purely imaginary,
$∴ x^{2} + y^{2} - 1 = 0$
$\Rightarrow x^{2} + y^{2} = 1$ which is a circle

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Let z=x+iy, where x and y are real. The points (x,y) in the X−Y plane for which z+iz−i is purely imaginary lie on

The correct option is D a circlez+iz−i=x+i(y+1)x+i(y−1)=[x+i(y+1)][x−i(y−1)]x2+(y−1)2=[x2+(y2−1)+2ix]x2+(y−1)2 Since above expression is purely imaginary, ∴x2+y2−1=0 ⇒x2+y2=1 which is a circle

Let $z = x + i y$ , where $x$ and $y$ are real. The points $(x, y)$ in the $X - Y$ plane for which
$\frac{z + i}{z - i}$ is purely imaginary lie on