Let S be the set of all complex numbers z satisfying $|\mathrm{z}-2+\mathrm{i}| \geq \sqrt{5}$. If the complex number $\mathrm{z}_{0}$ is such that $\frac{1}{\left|z_{0}-1\right|}$ is the maximum of the set $\left\{\frac{1}{|z-1|}: z \in S\right\}$, then the principal argument of $\frac{4-z_{0}-\bar{z}_{0}}{z_{0}-\bar{z}_{0}+2 i}$ is

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Complex Numbers

Let S be the set of all complex numbers z satisfying

∣ z - 2 + i ∣ \geq 5

. If the complex number

z_{0}

is such that

\frac{1}{∣ z _{0} - 1 ∣}

is the maximum of the set

{\frac{1}{∣ z - 1∣} : z \in S}

, then the principal argument of

\frac{4 - z _{0} - z ˉ _{0}}{z _{0} - z ˉ _{0} + 2 i}

Answer: (B)

The set S is the exterior (including boundary) of a circle centered at

(2, - 1)

with radius

5

. To maximize

\frac{1}{∣ z _{0} - 1∣}

, we minimize

∣ z_{0} - 1∣

, which means

z_{0}

is the point on the boundary of S closest to

(1, 0)

.

From the geometry,

z_{0}

lies on the line joining

(2, - 1)

and

(1, 0)

, at distance

5

from

(2, - 1)

toward

(1, 0)

.

Now:

ar g (\frac{4 - ( z _{0} + z _{0} )}{( z _{0} - z _{0} ) + 2 i}) = ar g (\frac{4 - 2 Re z _{0}}{2 i Im z _{0} + 2 i})

= ar g (\frac{2 - Re z _{0}}{( 1 + Im z _{0} ) i}) = ar g (- (\frac{2 - Re z _{0}}{1 + Im z _{0}}) i)

Since

Re z_{0} < 2

and

Im z_{0} > - 1

, the coefficient is positive, so we have

ar g (- k i)

where

k > 0

, which equals

- \frac{π}{2}

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Single correct option