arg-z-a-arg-z-z-1-arg-z-z-1-kpi-a-R-z-1-C-z-1-0-z-C-k-Z-the-locus-of-z-are-

Question Number 651 by 123456 last updated on 19/Feb/15

\arg (z - a) - \arg (z - z_{1}) - \arg (z - {\bar{z}}_{1}) = k π

a \in R

z_{1} \in C, ℑ ({\bar{z}}_{1}) \neq 0

z \in C

k \in Z

t h e l o c u s o f z a r e ?

Commented by 123456 last updated on 20/Feb/15

\frac{z - a}{(z - z_{1}) (z - {\bar{z}}_{1})}

= \frac{z - a}{[z - (x_{1} + y_{1} ı)] [z - (x_{1} - y_{1} ı)]}

= \frac{z - a}{[(z - x_{1}) - y_{1} ı] [(z - x_{1}) + y_{1} ı]}

= \frac{z - a}{{(z - x_{1})}^{2} - {(y_{1} ı)}^{2}}

= \frac{z - a}{z^{2} - 2 {z x}_{1} + x_{1}^{2} + y_{1}^{2}}

= \frac{x + y ı - a}{{(x + y ı)}^{2} - 2 (x + y ı) x_{1} + x_{1}^{2} + y_{1}^{2}}

= \frac{(x - a) + y ı}{x^{2} + 2 x y ı - y^{2} - 2 {x x}_{1} - 2 {y x}_{1} ı + x_{1}^{2} + y_{1}^{2}}

= \frac{(x - a) + y ı}{(x^{2} + x_{1}^{2} + y_{1}^{2} - y^{2} - 2 {x x}_{1}) + (2 x y - 2 {y x}_{1}) ı}

= \frac{u_{1} + v_{1} ı}{u_{2} + v_{2} ı}

u_{1} = x - a

v_{1} = y

u_{2} = x^{2} + x_{1}^{2} + y_{1}^{2} - y^{2} - 2 {x x}_{1}

v_{2} = 2 x y - 2 {y x}_{1}

= \frac{(u_{1} + v_{1} ı) (u_{2} - v_{2} ı)}{(u_{2} + v_{2} ı) (u_{2} - v_{2} ı)}

= \frac{u_{1} u_{1} + v_{1} v_{2}}{u_{2}^{2} + v_{2}^{2}} + \frac{- u_{1} v_{2} + v_{1} u_{2}}{u_{2}^{2} + v_{2}^{2}} ı

- u_{1} v_{2} + v_{1} u_{2} = 0

- (x - a) (2 x y - 2 {y x}_{1}) + y (x^{2} + x_{1}^{2} + y_{1}^{2} - y^{2} - 2 {x x}_{1}) = 0

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