Question

Let $S$ be the set of all complex numbers $Z$ satisfying $\left| {Z - 2 + i} \right| \geqslant \sqrt 5$ .If the complex number ${Z_0}$ is such that $\dfrac{1}{{\left| {{{\rm Z}_0} - 1} \right|}}$ is the maximum of the set \left\\{ {\dfrac{1}{{\left| {Z - 1} \right|}}\,\,:\,Z \in S} \right\\} , then the principal argument of $\dfrac{{4 - {Z_0} - \overline {{Z_0}} }}{{{Z_0} - \overline {{Z_0}} + 2i}}$ is
(A) $\dfrac{\pi }{4}$
(B) $\dfrac{{3\pi }}{4}$
(C) $- \dfrac{\pi }{2}$
(D) $\dfrac{\pi }{2}$

Answer 1

Here, the given equation $\left| {Z - 2 + i} \right| \geqslant \sqrt 5$ is the equation of the circle so firstly we have to find the centre and radius of this circle. It is given that $\dfrac{1}{{\left| {{{\rm Z}_0} - 1} \right|}}$ is maximum so we have to find the minimum value of $\left| {{Z_0} - 1} \right|$. We can write ${Z_0} = x + iy$and to find the minimum value of $\left| {{Z_0} - 1} \right|$ we have to put conditions on $x$and $y$.then the argument of $\dfrac{{4 - {Z_0} - \overline {{Z_0}} }}{{{Z_0} - \overline {{Z_0}} + 2i}}$ can be found.

Complete step-by-step answer:
Here, $S$ be the set of all complex numbers $Z$ satisfying $\left| {Z - 2 + i} \right| \geqslant \sqrt 5$.
We can write $\left| {Z - 2 + i} \right| \geqslant \sqrt 5$as $\left| {Z - \left( {2 - i} \right)} \right| \geqslant \sqrt 5$
From above equation we can say that the centre of the circle is $\left( {2, - 1} \right)$and the radius is $\sqrt 5$ .
It is given that complex number ${Z_0}$ is such that $\dfrac{1}{{\left| {{{\rm Z}_0} - 1} \right|}}$ is maximum. And it clearly visible that for this to be maximum $\left| {{Z_0} - 1} \right|$ should be minimum.
Let ${Z_0} = x + iy$ where $x < 1$ and $y > 0$for $\left| {{Z_0} - 1} \right|$to be minimum.
Now, $\overline {{Z_0}} = x - iy$
We have to find the argument of $\dfrac{{4 - {Z_0} - \overline {{Z_0}} }}{{{Z_0} - \overline {{Z_0}} + 2i}}$ .
We can write $\dfrac{{4 - {Z_0} - \overline {{Z_0}} }}{{{Z_0} - \overline {{Z_0}} + 2i}}$ as $\dfrac{{4 - x - iy - x + iy}}{{x + iy - x + iy + 2i}}$
$= \dfrac{{4 - 2x}}{{\left( {y + 1} \right)2i}} \\\ = \dfrac{{ - 2\left( {x - 2} \right)}}{{\left( {y + 1} \right)2i}} \\\$
Now, by multiplying by $i$ in both numerator and denominator we get,
$= \dfrac{{ - i\left( {2 - x} \right)}}{{\left( {y + 1} \right)}}$
Since, $\dfrac{{\left( {2 - x} \right)}}{{\left( {y + 1} \right)}}$ is a positive real number as $x < 1$ and $y > 0$.
and argument of negative of the imaginary numbers is $- \dfrac{\pi }{2}$
The argument of $\dfrac{{4 - {Z_0} - \overline {{Z_0}} }}{{{Z_0} - \overline {{Z_0}} + 2i}}$ is $- \dfrac{\pi }{2}$.

Thus, the correct option is (C).

Note: $\overline Z$ is a conjugate of a complex number $Z$ whose real part remains same but there is change of the sign from positive to negative or vice versa in imaginary part of the complex numbers.
Argument of any number present on the positive $X$ axis is zero and that of the positive $Y$ axis is $\dfrac{\pi }{2}$.