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- z0z0- z0 - 2i is

|z 2+i| ≥√(5), nbsp; nbsp; nbsp; nbsp; nbsp;|z_0 1|_min ⇒ P(z_0) lies on the line joining AC. Let z_0 =x+i y (4 z_0 z_0z_0 z_0+2i)=(4 2x2iy +2i ...

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Standard XII

Mathematics

Argument of a Complex Number

Let S be the ...

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 $\frac{4 - z_{0} -_{0}}{z_{0} -_{0} + 2 i}$ is

- \frac{π}{2}

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\frac{π}{4}

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\frac{π}{2}

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\frac{3 π}{4}

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Solution

The correct option is A $- \frac{π}{2}$
$| z - 2 + i |$ $\geq \sqrt{5}$ , $| z_{0} - 1 |_{m i n}$

$\Rightarrow P (z_{0})$ lies on the line joining $A C .$
Let $z_{0} = x + i y$
$(\frac{4 - z_{0} -_{0}}{z_{0} -_{0} + 2 i}) = (\frac{4 - 2 x}{2 i y + 2 i})$
$= \frac{2 - x}{i (1 + y)} = \frac{i (x - 2)}{(y + 1)}$
From the image, $x < 1, y > 0$
$\frac{x - 2}{y + 1} < 0, \frac{4 - z_{0} -_{0}}{z_{0} -_{0} + 2 i} = - R e (i)$
$∴ arg (\frac{4 - z_{0} -_{0}}{z_{0} -_{0} + 2 i}) = arg (- R e (i)) = - \frac{π}{2}$

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Similar questions

Q. Let

S

be the set of all complex numbers

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satisfying

| z - 2 + i |

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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−¯¯¯¯¯z0z0−¯¯¯¯¯z0+2i is

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 $\frac{4 - z_{0} -_{0}}{z_{0} -_{0} + 2 i}$ is