if-Z-1-Then-1-Z-1-Z-is-equal-to-a-Z-b-Z-c-Z-Z-d-N-O-T-

Question Number 16540 by gourav~ last updated on 23/Jun/17

i f ∣ Z ∣= 1 T h e n \frac{1 + Z}{1 + \bar{Z}} i s e q u a l t o \dots

a) Z

b) \bar{Z}

c) Z + \bar{Z}

d) N . O . T

Answered by sma3l2996 last updated on 23/Jun/17

Z = a + i b \Leftrightarrow∣ Z ∣= \sqrt{a^{2} + b^{2}} = 1 \Leftrightarrow b = \sqrt{1 - a^{2}}

\frac{1 + a + i \sqrt{1 - a^{2}}}{1 + a - i \sqrt{1 - a^{2}}} = \frac{{(1 + a + i \sqrt{1 - a^{2}})}^{2}}{{(1 + a)}^{2} - {(i \sqrt{1 - a^{2}})}^{2}} = \frac{{(1 + a)}^{2} + 2 (1 + a) i \sqrt{1 - a^{2}} - 1 + a^{2}}{1 + a^{2} + 2 a + 1 - a^{2}}

= \frac{2 a^{2} + 2 a + 2 (1 + a) i \sqrt{1 - a^{2}}}{2 + 2 a} = \frac{2 (1 + a) (a + i \sqrt{1 - a^{2}})}{2 (1 + a)}

= a + i \sqrt{1 - a^{2}} = Z

s o t h e a n s w e r i s (a)

Answered by ajfour last updated on 23/Jun/17

l e t z = e^{i θ} t h e n \bar{z} = e^{- i θ}

\frac{1 + z}{1 + \bar{z}} = \frac{1 + e^{i θ}}{1 + e^{- i θ}} = (\frac{e^{i θ / 2}}{e^{- i θ / 2}}) (\frac{e^{- i θ / 2} + e^{i θ / 2}}{e^{i θ / 2} + e^{- i θ / 2}})

= e^{i θ} = z . o p t i o n (a) .

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