The locus of point z satisfying Re 1z - k- where k is a non-zero real number is

Let z = x + iy, then nbsp; Re ( 1z ) = k ⇒ Re (1x+iy ) = k ⇒ Re ( x iyx^2+y^2) = k ⇒xx^2+ y^2 = k ⇒ x^2+ y^2 xk = 0 Hence, the locus is a circle. ...

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Standard XIII

Mathematics

Equation of Conics in Complex Form

The locus of ...

Question

The locus of point $z$ satisfying $R e (\frac{1}{z}) = k$ , where $k$ is a non-zero real number is

a straight line

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a circle

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an ellipse

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a hyperbola

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Solution

The correct option is B a circle
Let $z = x + i y$ , then
$R e (\frac{1}{z}) = k \Rightarrow R e (\frac{1}{x + i y}) = k \Rightarrow R e (\frac{x - i y}{x^{2} + y^{2}}) = k \Rightarrow \frac{x}{x^{2} + y^{2}} = k \Rightarrow x^{2} + y^{2} - \frac{x}{k} = 0$

Hence, the locus is a circle.

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Equation of Conics in Complex Form

Standard XIII Mathematics

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The locus of point z satisfying Re(1z)=k, where k is a non-zero real number is

The correct option is B a circleLet z=x+iy, then Re(1z)=k⇒Re(1x+iy)=k⇒Re(x−iyx2+y2)=k⇒xx2+y2=k⇒x2+y2−xk=0 Hence, the locus is a circle.

The locus of point $z$ satisfying $R e (\frac{1}{z}) = k$ , where $k$ is a non-zero real number is

The correct option is B a circle
Let $z = x + i y$ , then
$R e (\frac{1}{z}) = k \Rightarrow R e (\frac{1}{x + i y}) = k \Rightarrow R e (\frac{x - i y}{x^{2} + y^{2}}) = k \Rightarrow \frac{x}{x^{2} + y^{2}} = k \Rightarrow x^{2} + y^{2} - \frac{x}{k} = 0$

Hence, the locus is a circle.