JEE Advance - Mathematics (2019 - Paper 1 Offline - No. 1)
Let S be the set of all complex numbers z satisfying $$\left| {z - 2 + i} \right| \ge \sqrt 5 $$. If the complex number z0 is such that $${1 \over {\left| {{z_0} - 1} \right|}}$$ is the maximum of the set $$\left\{ {{1 \over {\left| {{z_0} - 1} \right|}}:z \in S} \right\}$$, then the principal argument of $${{4 - {z_0} - {{\overline z }_0}} \over {{z_0} - {{\overline z }_0} + 2i}}$$ is
$${\pi \over 4}$$
$${3\pi \over 4}$$
$$ - $$$${\pi \over 2}$$
$${\pi \over 2}$$
Explanation
The complex number z satisfying $$\left| {z - 2 + i} \right| \ge \sqrt 5 $$, which represents the region outside the circle (including the circumference) having centre (2, $${ - 1}$$) and radius $$\sqrt 5 $$ units.
Now, for $${{z_0} \in S{1 \over {\left| {{z_0} - 1} \right|}}}$$ is maximum.
When $${\left| {{z_0} - 1} \right|}$$ is minimum. And for this it is required that $${{z_0} \in S}$$, such that z0 is collinear with the points (2, $$ - $$1) and (1, 0) and lies on the circumference of the circle $$\left| {z - 2 + i} \right|$$ = $$\sqrt 5 $$.
So let z0 = x + iy, and from the figure 0 < x < 1 and y >0.
So, $${{4 - {z_0} - {{\overline z }_0}} \over {{z_0} - {{\overline z }_0} + 2i}} = {{4 - x - iy - x + iy} \over {x + iy - x + iy + 2i}} = {{2(2 - x)} \over {2i(y + 1)}} = - i\left( {{{2 - x} \over {y + 1}}} \right)$$
$$ \because $$ $${{{2 - x} \over {y + 1}}}$$ is a positive real number, so $${{4 - {z_0} - {{\overline z }_0}} \over {{z_0} - {{\overline z }_0} + 2i}}$$ is purely negative imaginary number.
$$ \Rightarrow $$ $$\arg \left( {{{4 - {z_0} - {{\overline z }_0}} \over {{z_0} - {{\overline z }_0} + 2i}}} \right) = - {\pi \over 2}$$
Now, for $${{z_0} \in S{1 \over {\left| {{z_0} - 1} \right|}}}$$ is maximum.
When $${\left| {{z_0} - 1} \right|}$$ is minimum. And for this it is required that $${{z_0} \in S}$$, such that z0 is collinear with the points (2, $$ - $$1) and (1, 0) and lies on the circumference of the circle $$\left| {z - 2 + i} \right|$$ = $$\sqrt 5 $$.
So let z0 = x + iy, and from the figure 0 < x < 1 and y >0.
So, $${{4 - {z_0} - {{\overline z }_0}} \over {{z_0} - {{\overline z }_0} + 2i}} = {{4 - x - iy - x + iy} \over {x + iy - x + iy + 2i}} = {{2(2 - x)} \over {2i(y + 1)}} = - i\left( {{{2 - x} \over {y + 1}}} \right)$$
$$ \because $$ $${{{2 - x} \over {y + 1}}}$$ is a positive real number, so $${{4 - {z_0} - {{\overline z }_0}} \over {{z_0} - {{\overline z }_0} + 2i}}$$ is purely negative imaginary number.
$$ \Rightarrow $$ $$\arg \left( {{{4 - {z_0} - {{\overline z }_0}} \over {{z_0} - {{\overline z }_0} + 2i}}} \right) = - {\pi \over 2}$$
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