If-z-1-prove-that-z-1-z-1-z-1-is-a-pure-imaginary-

Question Number 51643 by Tawa1 last updated on 29/Dec/18

If ∣ z ∣ = 1, prove that \frac{z - 1}{\bar{z} - 1} (z \neq 1) is a pure imaginary

Answered by tanmay.chaudhury50@gmail.com last updated on 29/Dec/18

z = r c o s θ + i r s i n θ = r (c o s θ + i s i n θ)

∣ z ∣= r = 1

\frac{z - 1}{\bar{z} - 1} = \frac{c o s θ + i s i n θ - 1}{c o s θ - i s i n θ - 1} = \frac{- 2 {s i n}^{2} \frac{θ}{2} + i 2 s i n \frac{θ}{2} c o s \frac{θ}{2}}{- 2 {s i n}^{2} \frac{θ}{2} - i 2 s i n \frac{θ}{2} c o s \frac{θ}{2}}

= \frac{- 2 s i n \frac{θ}{2} (s i n \frac{θ}{2} - i c o s \frac{θ}{2})}{- 2 s i n \frac{θ}{2} (s i n \frac{θ}{2} + i c o s \frac{θ}{2})}

= \frac{{(s i n \frac{θ}{2} - i c o s \frac{θ}{2})}^{2}}{{(s i n \frac{θ}{2})}^{2} - i^{2} {(c o s \frac{θ}{2})}^{2}}

= \frac{{s i n}^{2} \frac{θ}{2} - 2 i s i n \frac{θ}{2} c o s \frac{θ}{2} + i^{2} {c o s}^{2} \frac{θ}{2}}{{s i n}^{2} \frac{θ}{2} + {c o s}^{2} \frac{θ}{2}}

= - ({c o s}^{2} \frac{θ}{2} - {s i n}^{2} \frac{θ}{2} + i \times 2 s i n \frac{θ}{2} c o s \frac{θ}{2})

= - (c o s θ + i s i n θ)

= - z

p l s c h e c k t h e q u e s t i o n \dots

Commented by Tawa1 last updated on 29/Dec/18

God bless you sir . The question is not true sir ?

Commented by peter frank last updated on 29/Dec/18

\Rightarrow \frac{z + 1}{z - 1} i t h i n k

Answered by peter frank last updated on 29/Dec/18

\frac{z + 1}{z - 1}

z = x + i y

\bar{z} = x - i y

\frac{x + i y - 1}{x - i y - 1}

\frac{(x + 1) + i y}{(x - 1) - i y}

[\frac{(x + 1) + i y}{(x - 1) - i y}] \times [\frac{(x - 1) + i y}{(x - 1) + i y}]

\frac{x^{2} + y^{2} - 1}{x^{2} + y^{2} - 2 x} + \frac{2 x i}{x^{2} + y^{2} - 2 x}

p u r e i m a g i n a r y

\frac{x^{2} + y^{2} - 1}{x^{2} + y^{2} - 2 x} = 0

x^{2} + y^{2} - 1 = 0

x^{2} + y^{2} = 1

∣ x + i y ∣= 1

∣ z ∣= 1

Commented by Tawa1 last updated on 29/Dec/18

God bless you sir

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