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Question Number 126553 by mathocean1 last updated on 21/Dec/20

s h o w t h a t f o r z \in C

∣ z + 1 ∣^{2} = 2 ∣ z ∣^{2} \Leftrightarrow∣ z - 1 ∣^{2} = 2 .

T h i s f o r m u l a c a n b e u s e d :

∣ z ∣^{2} = z \times \bar{z}

Answered by Olaf last updated on 21/Dec/20

∣ z + 1 ∣^{2} = ∣ z + 1 ∣ \times \overset{―}{∣ z + 1 ∣}

∣ z + 1 ∣^{2} = ∣ z + 1 ∣ \times ∣ \bar{z} + 1 ∣

∣ z + 1 ∣^{2} = ∣ (z + 1) (\bar{z} + 1) ∣

∣ z + 1 ∣^{2} = ∣ z \bar{z} + (z + \bar{z}) + 1 ∣

∣ z + 1 ∣^{2} = ∣ z \bar{z} + 2 Re (z) + 1 ∣

∣ z + 1 ∣^{2} = ∣ z ∣^{2} + 2 Re (z) + 1 (1)

And :

∣ z - 1 ∣^{2} = ∣ z ∣^{2} - 2 Re (z) + 1 (2)

(1) + (2) :

∣ z + 1 ∣^{2} + ∣ z - 1 ∣^{2} = 2 ∣ z ∣^{2} + 2 (3)

(3) : ∣ z + 1 ∣^{2} = 2 ∣ z ∣^{2} \Leftrightarrow ∣ z - 1 ∣^{2} = 2

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