
PQ and PR are two infinite rays. QAR is an arc. Point lying in the shaded region excluding the boundary satisfies [IIT Screening \[2005\]]

A) \[|z - 1| > 2;|arg(z - 1)| < \dfrac{\pi }{4}\]
B) \[|z - 1| > 2;|arg(z - 1)| < \dfrac{\pi }{2}\]
C) \[|z + 1| > 2;|arg(z + 1)| < \dfrac{\pi }{4}\]
D) \[|z + 1| > 2;|arg(z + 1)| < \dfrac{\pi }{2}\]
Answer
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Hint: In this question we have to find points lying in a shaded region excluding the boundaries. Find the length of two infinite rays using distance formula and on the basis of length try to find out the properties of arc QAR.
Formula used: If \[({x_{1,}}{y_1})\]and \[({x_{2,}}{y_2})\]are two end point of line then length of line is given by
\[D = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} \]
D is the distance between two points
Complete step by step solution: Given: coordinates of each points on the curve
Now
Equation of infinite ray PQ\[\arg (z + 1) = \dfrac{\pi }{4}\]
Equation of infinite ray PR \[\arg (z + 1) = - \dfrac{\pi }{4}\]
Boundary of shaded region is given as
\[ - \dfrac{\pi }{4} < \arg (z + 1) < \dfrac{\pi }{4}\]
Coordinate of point P is \[( - 1,0)\]
Coordinate of point Q is \[( - 1 + \sqrt 2 ,\sqrt 2 i)\]
Length of PQ is given by
\[D = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} \]
\[\arg (z + 1) < \dfrac{\pi }{4}\]\[\left| {PQ} \right| = \sqrt {{{(\sqrt 2 )}^2} + {{(\sqrt 2 )}^2}} = 2\]
Similarly PA and PR is also equal to \[2\]
Now length of each ray is equal there given arc is a part of circle having center at P and radius is \[2\]
So all points in the shaded region is outside the circle \[\left| {z + 1} \right| < 2\]
Thus, Option (C) is correct.
Note: Don’t try to put the value of z in a combination of real and imaginary form. We must remember that if the length of all lines from common point to any point on the arc is equal then arc is a part of a circle having that common point as center and radius equal to length of line.
Formula used: If \[({x_{1,}}{y_1})\]and \[({x_{2,}}{y_2})\]are two end point of line then length of line is given by
\[D = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} \]
D is the distance between two points
Complete step by step solution: Given: coordinates of each points on the curve
Now
Equation of infinite ray PQ\[\arg (z + 1) = \dfrac{\pi }{4}\]
Equation of infinite ray PR \[\arg (z + 1) = - \dfrac{\pi }{4}\]
Boundary of shaded region is given as
\[ - \dfrac{\pi }{4} < \arg (z + 1) < \dfrac{\pi }{4}\]
Coordinate of point P is \[( - 1,0)\]
Coordinate of point Q is \[( - 1 + \sqrt 2 ,\sqrt 2 i)\]
Length of PQ is given by
\[D = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} \]
\[\arg (z + 1) < \dfrac{\pi }{4}\]\[\left| {PQ} \right| = \sqrt {{{(\sqrt 2 )}^2} + {{(\sqrt 2 )}^2}} = 2\]
Similarly PA and PR is also equal to \[2\]
Now length of each ray is equal there given arc is a part of circle having center at P and radius is \[2\]
So all points in the shaded region is outside the circle \[\left| {z + 1} \right| < 2\]
Thus, Option (C) is correct.
Note: Don’t try to put the value of z in a combination of real and imaginary form. We must remember that if the length of all lines from common point to any point on the arc is equal then arc is a part of a circle having that common point as center and radius equal to length of line.
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