if Z is a complex number such that Z-1 / Z+1 is purely imaginary prove that |Z|=1 - Maths - Complex Numbers and Quadratic Equations

Question

if Z is a complex number such that Z-1 / Z+1 is purely imaginary
prove that |Z|=1

Harleen Birdi · Accepted Answer

@Sreejababu: Here is the answer to your question.

$$ = x + iy

$\frac{ - 1}{ + 1}$ is purely imaginary number.

$\Rightarrow Re (\frac{ - 1}{ + 1}) = 0 \Rightarrow Re (\frac{x + i y - 1}{x + i y + 1}) = 0 (Put  = x + i y) \Rightarrow Re [\frac{(x - 1) + i y}{(x + 1) + i y}] = 0 \Rightarrow Re [\frac{(x - 1) + i y}{(x + 1) + i y} \times \frac{(x + 1) - i y}{(x + 1) - i y}] = 0 \Rightarrow Re [\frac{(x - 1) (x + 1) - i (x - 1) y + i y (x + 1) + y^{2}}{(x + 1)^{2} + y^{2}}] = 0 \Rightarrow Re [\frac{(x^{2} - 1) + y^{2} + i {y (x + 1) - (x - 1) y}}{(x + 1)^{2} + y^{2}}] = 0 \Rightarrow Re (\frac{x^{2} + y^{2} - 1 + i (x y + y - x y + y)}{(x + 1)^{2} + y^{2}}) = 0 \Rightarrow Re [\frac{x^{2} + y^{2} - 1}{(x + 1)^{2} + y^{2}} + i \frac{2 y}{(x + 1)^{2} + y^{2}}] = 0$

$\Rightarrow \frac{x^{2} + y^{2} - 1}{(x + 1)^{2} + y^{2}} = 0 \Rightarrow x^{2} + y^{2} - 1 = 0 \Rightarrow x^{2} + y^{2} = 1 \Rightarrow |  | = 1 (as  = x + i y \Rightarrow |  | = \sqrt{x^{2} + y^{2}}) \Rightarrow |  | = 1$

if Z is a complex number such that Z-1 / Z+1 is purely imaginary.prove that |Z|=1