Question

if z = x + i y such that
$|z + 1| = |z - 1|$
and amp
$\frac{z - 1}{z + 1} = \frac{\pi}{4}$
​

Answer 1

hey
friends you question is
if
$z = x + i y \:$
such that
$|z + 1| = |z - 1| \:$

and
$amp(\frac{z - 1}{z + 1}) = \frac{\pi}{4}$
now we have to find the value of x and y

solution

hence it is given that

hence x=0
$amp{ (\frac{z - 1}{z + 1} )} = \frac{\pi}{4} \\ arg( \frac{(x - 1) + iy}{(x + 1) + iy} ) = \frac{\pi}{4}$
after multiplying deviding by conjugates numerator and denominator
$arg( \frac{(x - 1) + iy}{(x + 1) + iy} \times \frac{(x + 1) - iy}{(x + 1) - iy} ) = \frac{\pi}{4}$
=》
$arg( \frac{(x - 1)(x + 1) - iy(x - 1) + iy(x + 1) + {y}^{2} }{ {(x + 1)}^{2} + {y}^{2} } = \frac{\pi}{4 } \\ arg( \: \frac{(( {x}^{2} - 1) +{y}^{2} + 2iy }{ {(x + 1)}^{2} + {y}^{2} } ) = \frac{\pi}{4}$
after taking tan in Invernes and use
$\frac{imginory \: part \: }{real \: part}$
=
${tan}^{ - 1} ( \frac{ \frac{2y}{ {(x + 1)}^{2} + {y}^{2} } }{ \frac{ ({x}^{2} - 1) + {y}^{2} }{ {(x + 1)}^{2} + {y}^{2} } } ) = \frac{\pi}{4}$
now
${(x + 1)}^{2} + {y}^{2}$
will be cancelling
=
${tan}^{ - 1} ( \frac{2y}{ ({x}^{2} - 1) + {y}^{2} }) = \frac{\pi}{4}$

now take tan on opposite side then
$( \frac{2y}{ ({x}^{2} - 1) + {y}^{2} }) = tan\frac{\pi}{4}$
we know that
$tan \frac{\pi}{4} = 1$
so
$( \frac{2y}{ ({x}^{2} - 1) + {y}^{2} }) = 1$
after cross multiplying
${2y} = { ({x}^{2} - 1) + {y}^{2} }$
after submitting the value of x we obtain that
${2y} = { ({0}^{2} - 1) + {y}^{2} } \: \\ = > {2y} = { (0 - 1) + {y}^{2} } \\ = > 0= {y}^{2} - 2y - 1$
after comparing with
${ax}^{2} + bx + c = 0$
we obtain that
a= 1
b=-2
c=-1
now use that quadratic formula
$D = {{b}^{2} - 4ac} \\ so D = {( - 2)}^{2} - 4(1)( - 1) \\ = 4 + 4 \\ = 8$
now
$y = \frac{ - b + _{ - } \sqrt{D} }{2a} \\ = \frac{ - ( - 2) + _{ - } \sqrt{8} }{2(1)} \\ = \frac{2 + \sqrt{8} }{2} or \frac{2 - \sqrt{8} }{2} \\ = \frac{ 2(1 + \sqrt{2} \: )}{2} or \: \frac{2(1 - \sqrt{2} )}{2} \\ = (1 + \sqrt{2} )or \: (1 - \sqrt{2)}$
hope it will help you thanks
at last I want to say that the question was so easy and funny