QJEEM22Mafd44: JEE Main 2022 Mathematics - Class 11 JEEM22Mafd44 Solution: Let arg(z) represent the principal argument of the complex…

Let $z = x + iy$

$\therefore$ $|z| = \sqrt {{x^2} + {y^2}}$

Given, $|z| = 3$

$\therefore$ $\sqrt {{x^2} + {y^2}} = 3$

$\Rightarrow {x^2} + {y^2} = 9 = {3^2}$

This represent a circle with center at (0, 0) and radius = 3

Now, given

$\arg (z - 1) - \arg (z + 1) = {\pi \over 4}$

$\Rightarrow \arg (x + iy - 1) - \arg (x + iy + 1) = {\pi \over 4}$

$\Rightarrow \arg (x - 1 + iy) - \arg (x + 1 + iy) = {\pi \over 4}$

$\Rightarrow {\tan ^{ - 1}}\left( {{y \over {x - 1}}} \right) - {\tan ^{ - 1}}\left( {{y \over {x + 1}}} \right) = {\pi \over 4}$

$\Rightarrow {\tan ^{ - 1}}\left( {{{{y \over {x - 1}} - {y \over {x + 1}}} \over {1 + {y \over {x - 1}} \times {y \over {x + 1}}}}} \right) = {\pi \over 4}$

$\Rightarrow {\tan ^{ - 1}}\left( {{{{{xy + y - xy + y} \over {{x^2} - 1}}} \over {{{{x^2} - 1 + {y^2}} \over {{x^2} - 1}}}}} \right) = {\pi \over 4}$

$\Rightarrow {\tan ^{ - 1}}\left( {{{xy + y - xy + y} \over {{x^2} - 1 + {y^2}}}} \right) = {\pi \over 4}$

$\Rightarrow {{2y} \over {{x^2} - 1 + {y^2}}} = \tan \left( {{\pi \over 4}} \right)$

$\Rightarrow 2y = {x^2} + {y^2} - 1$

$\Rightarrow {x^2} + {y^2} - 2y - 1 = 0$

$\Rightarrow {x^2} + {(y - 1)^2} = {(\sqrt 2 )^2}$

This represent a circle with center at (0, 1) and radius $\sqrt 2$ .

From diagram you can see both the circles do not cut anywhere.