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WB JEE 2008

Subjective

Prove that if the ratio $${{z - i} \over {z - 1}}$$ is purely imaginary, the point z lies on the circle in the Argand plane whose centre is at the point $${1 \over 2}(1 + i)$$ and radius is $${1 \over {\sqrt 2 }}$$.

Answer

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Explanation

Let z = x + iy, where x and y are both real

$$\therefore$$ $${{z - i} \over {z - 1}} = {{x + iy - i} \over {x + iy - 1}} = {{x + i(y - 1)} \over {(x - 1) + iy}}$$

$$ = {{[x + i(y - 1)][(x - 1) - iy]} \over {[(x - 1) + iy][(x - 1) - iy]}}$$ [by rationalisation]

$$ = {{({x^2} + {y^2} - x - y) + i(1 - x - y)} \over {{{(x - 1)}^2} + {y^2}}}$$

$$ = {{{x^2} + {y^2} - x - y} \over {{{(x - 1)}^2} + {y^2}}} + i{{1 - x - y} \over {{{(x - 1)}^2} + {y^2}}}$$

$$\therefore$$ If $${{z - i} \over {z - 1}}$$ is purely imaginary, its real part is zero.

$$\therefore$$ $${x^2} + {y^2} - x - y = 0$$

or

$$\left[ {{x^2} - 2\,.\,x\,.\,{1 \over 2} + {{\left( {{1 \over 2}} \right)}^2}} \right] + \left[ {{y^2} - 2\,.\,y\,.\,{1 \over 2} + {{\left( {{1 \over 2}} \right)}^2}} \right] = {\left( {{1 \over 2}} \right)^2} + {\left( {{1 \over 2}} \right)^2}$$

[Adding (1/2)2 + (1/2)2 to both sides and rearranging]

$${\left( {x - {1 \over 2}} \right)^2} + {\left( {y - {1 \over 2}} \right)^2} = {1 \over 2} = {\left( {{1 \over {\sqrt 2 }}} \right)^2}$$

This is the equation of a circle in the cartesian form with its centre $$\left( {{1 \over 2},{1 \over 2}} \right)$$ and radius $${1 \over {\sqrt 2 }}$$.

Hence, centre is at $${1 \over 2} + i{1 \over 2}$$, i.e., at $${{1 + i} \over 2}$$ in the Argand plane and its radius is $${1 \over {\sqrt 2 }}$$.

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