If-z-2-z-2-2-where-z-and-are-complex-numbers-then-1-z-is-purely-real-2-z-is-purely-imaginary-3-z-z-0-4-amp-z-pi-2-

Question Number 20935 by Tinkutara last updated on 08/Sep/17

If ∣ z + ω ∣^{2} = ∣ z ∣^{2} + ∣ ω ∣^{2}, where z and ω

are complex numbers, then

(1) \frac{z}{ω} is purely real

(2) \frac{z}{ω} is purely imaginary

(3) z \bar{ω} + \bar{z} ω = 0

(4) amp (\frac{z}{ω}) = \frac{π}{2}

Answered by ajfour last updated on 08/Sep/17

∣ z + ω ∣^{2} =∣ z ∣^{2} + ∣ ω ∣^{2}

\Rightarrow ∣ \frac{z}{ω} + 1 ∣^{2} =∣ \frac{z}{ω} ∣^{2} + 1

b u t ∣ \frac{z}{ω} + 1 ∣^{2} =∣ \frac{z}{ω} ∣^{2} + 1 + 2 R e (\frac{z}{ω})

s o w e c a n i n f e r R e (\frac{z}{ω}) = 0

\Rightarrow \frac{z}{ω} i s p u r e l y i m a g i n a r y .

a m p (\frac{z}{ω}) = \pm \frac{π}{2}

A l s o (z + ω) (\bar{z} + \bar{ω}) = z \bar{z} + ω \bar{ω}

\Rightarrow z \bar{ω} + ω \bar{z} = 0 .

(2) a n d (3), (4) a r e c o r r e c t .

Commented by Tinkutara last updated on 09/Sep/17

amp (\frac{z}{ω}) = \pm \frac{π}{2} so why (4) is correct ?

Commented by Tinkutara last updated on 09/Sep/17

Thank you very much Sir!

So this is one of the two solutions .

Commented by ajfour last updated on 09/Sep/17

n o, y o u d o n t g e t w h a t i m e a n

a m p (\frac{z}{ω}) = \pm \frac{π}{2}, s o a m p (\frac{z}{ω}) = \frac{π}{2}

i s a p o s s i b l e s o l u t i o n .

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