Let S be the set of all complex numbers z satisfying | z - 2 + i | 5 . If the complex number z 0 is such that 1 | z 0 - 1 | is the maximum of the set { 1 | z - 1 | : z S } , then the principal argument of 4 - z 0 - z 0 z 0 - z 0 + 2 i is

Question

Let $S$ be the set of all complex numbers $z$ satisfying $| z - 2 + i |$ $\geq \sqrt{5}$ . 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 $\frac{4 - z_{0} - \bar{z_{0}}}{z_{0} - \bar{z_{0}} + 2 i}$ is [2019]

Smart Achievers · Accepted Answer

(4)

$S :$ $| z - 2 + i |$ $\geq \sqrt{5} represents boundary and outer region of circle with centre (2, - 1) and radius \sqrt{5} units.$

$z_{0} \in S, such that \frac{1}{| z_{0} - 1 |} is the maximum.$

$∴$ $| z_{0} - 1 |$ $is minimum.$

$z_{0} \in S with | z_{0} - 1 | as minimum will be a point on boundary of circle of region S which lies on radius of this circle, which passes through (1, 0) .$

$∴$ $z_{0}, 1, 2 - i are collinear, or (x_{0}, y_{0}), (1, 0), (2, - 1) are collinear.$

$∴$ $Using slopes of parallel lines,$ $x'$

$\frac{y_{0}}{x_{0} - 1} = \frac{- 1}{2 - 1} \Rightarrow y_{0} = 1 - x_{0}$

$Now, \frac{4 - z_{0} - \bar{z_{0}}}{z_{0} - \bar{z_{0}} + 2 i} = \frac{4 - (z_{0} + \bar{z_{0}})}{(z_{0} - \bar{z_{0}}) + 2 i}$

$= \frac{4 - 2 x_{0}}{2 i y_{0} + 2 i} = \frac{4 - 2 x_{0}}{2 i (1 - x_{0}) + 2 i}$

$= \frac{2 (2 - x_{0})}{2 (2 - x_{0}) i} = \frac{1}{i} = - i$

$∴$ $Arg (\frac{4 - z_{0} - \bar{z_{0}}}{z_{0} - \bar{z_{0}} - 2 i}) = Arg (- i) = - \frac{π}{2}$

Solution

1 $\frac{π}{4}$

2 $\frac{3 π}{4}$

3 $\frac{π}{2}$

4 $- \frac{π}{2}$

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Solution

Q. Let S be the set of all complex numbers z satisfying |z-2+i|≥5. If the complex number z0 is such that 1|z0-1| is the maximum of the set {1|z-1|:z∈S}, then the principal argument of 4-z0-z0¯z0-z0¯+2i is [2019]

1 π4

2 3π4

3 π2

4 -π2

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1 $\frac{π}{4}$

2 $\frac{3 π}{4}$

3 $\frac{π}{2}$

4 $- \frac{π}{2}$