JEE MAIN - Mathematics (2021 - 31st August Evening Shift - No. 10)
If z is a complex number such that $${{z - i} \over {z - 1}}$$ is purely imaginary, then the minimum value of | z $$-$$ (3 + 3i) | is :
$$2\sqrt 2 - 1$$
$$3\sqrt 2 $$
$$6\sqrt 2 $$
$$2\sqrt 2 $$
Explanation
$${{z - i} \over {z - 1}}$$ is purely imaginary number
Let $$z = x + iy$$
$$\therefore$$ $${{x + i(y - 1)} \over {(x - 1) + i(y)}} \times {{(x - 1) - iy} \over {(x - 1) - iy}}$$
$$ \Rightarrow {{x(x - 1) + y(y - 1) + i( - y - x + 1)} \over {{{(x - 1)}^2} + {y^2}}}$$ is purely imaginary number
$$ \Rightarrow x(x - 1) + y(y - 1) = 0$$
$$ \Rightarrow {\left( {x - {1 \over 2}} \right)^2} + {\left( {y - {1 \over 2}} \right)^2} = {1 \over 2}$$
_31st_August_Evening_Shift_en_10_2.png)
$$\therefore$$ $${\left| {z - (3 + 3i)} \right|_{\min }} = \left| {PC} \right| - {1 \over {\sqrt 2 }}$$
$$ = {5 \over {\sqrt 2 }} - {1 \over {\sqrt 2 }} = 2\sqrt 2 $$
Let $$z = x + iy$$
$$\therefore$$ $${{x + i(y - 1)} \over {(x - 1) + i(y)}} \times {{(x - 1) - iy} \over {(x - 1) - iy}}$$
$$ \Rightarrow {{x(x - 1) + y(y - 1) + i( - y - x + 1)} \over {{{(x - 1)}^2} + {y^2}}}$$ is purely imaginary number
$$ \Rightarrow x(x - 1) + y(y - 1) = 0$$
$$ \Rightarrow {\left( {x - {1 \over 2}} \right)^2} + {\left( {y - {1 \over 2}} \right)^2} = {1 \over 2}$$
$$\therefore$$ $${\left| {z - (3 + 3i)} \right|_{\min }} = \left| {PC} \right| - {1 \over {\sqrt 2 }}$$
$$ = {5 \over {\sqrt 2 }} - {1 \over {\sqrt 2 }} = 2\sqrt 2 $$
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