QJEEAdv19Ma4978: JEE Advanced 2019 Mathematics - Class 11 JEEAdv19Ma4978 Solution: Let S be the set of all complex numbers z satisfying . If the…

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.

$\Rightarrow$ $\arg \left( {{{4 - {z_0} - {{\overline z }_0}} \over {{z_0} - {{\overline z }_0} + 2i}}} \right) = - {\pi \over 2}$