Question
If $$z = x + iy$$ such that $$\left| {z + 1} \right| = \left| {z - 1} \right|$$ and $${\text{amp}}\frac{{z - 1}}{{z + 1}} = \frac{\pi }{4}$$ then
A.
$$x = \sqrt 2 + 1,y = 0$$
B.
$$x = 0,y = \sqrt 2 + 1$$
C.
$$x = 0,y = \sqrt 2 - 1$$
D.
$$x = \sqrt 2 - 1,y = 0$$
Answer :
$$x = 0,y = \sqrt 2 + 1$$
Solution :
$$\eqalign{
& \left| {\frac{{z - 1}}{{z + 1}}} \right| = 1.\,{\text{So, }}\frac{{z - 1}}{{z + 1}} = 1 \cdot \left( {\cos \frac{\pi }{4} + i\sin \frac{\pi }{4}} \right) = \frac{{1 + i}}{{\sqrt 2 }}. \cr
& \therefore \,\,z = \frac{{\sqrt 2 + 1 + i}}{{\sqrt 2 - 1 - i}} = \frac{{\left( {\sqrt 2 + 1 + i} \right)\left( {\sqrt 2 - 1 + i} \right)}}{{{{\left( {\sqrt 2 - 1} \right)}^2} + 1}},\,{\text{e}}{\text{.t}}{\text{.c}}{\text{.}} \cr} $$