show thatfor z∈C ∣z+1∣^2 =2∣z∣^(2 ) ⇔∣z−1∣^2 =2. This formula can be used: ∣z∣^2 =z×z^−


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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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