Question

If z = x + iy such that |z + 1| = |z – 1| and amp $\frac{z - 1}{z + 1}$=$\frac{\pi}{4}$ then

• A. x = Ö2 + 1, y = 0
• B. x = 0, y = Ö2 + 1
• C. x = 0, y = Ö2 = 1
• D. x = Ö2 – 1, y = 0

Answer 1

Answer: (B)

x = 0, y = Ö2 + 1

Answer 2

Sol. |z + 1| = |z – 1|  (x + 1)² + y² = (x – 1)² + y²

 x = 0 … (1)

Arg $\left( \frac{z - 1}{z + 1} \right)$=$\frac{\pi}{4}$

 arg $\left\{ \frac{\left\lbrack (x - 1) + iy \right\rbrack\lbrack x + 1 - iy\rbrack}{(x + 1)^{2} + y^{2}} \right\}$= $\frac{\pi}{4}$

 arg $\left\{ \frac{(x^{2} + y^{2} - 1) + 2iy}{(x + 1)^{2} + y^{2}} \right\}$= $\frac{\pi}{4}$

 y > 0, x² + y² – 1 > 0, $\frac{2y}{x^{2} + y^{2} - 1}$ = 1

 x² + y² – 2y – 1 = 0, y > 0, x² + y² – 1 > 0.

\ arg $\left( \frac{z - 1}{z + 1} \right)$=$\frac{\pi}{4}$

is the arc ABC of the circle x² + y² – 2y – 1 = 0

Solving with x = 0, we get

y = $\frac{2 \pm \sqrt{8}}{2}$ = 1 ฑ $\sqrt{2}$, y > 0, \ y = 1 + ึ2.