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

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

$${show}\:{thatfor}\:{z}\in\mathbb{C}\: \\ $$$$\mid{z}+\mathrm{1}\mid^{\mathrm{2}} =\mathrm{2}\mid{z}\mid^{\mathrm{2}\:} \Leftrightarrow\mid{z}−\mathrm{1}\mid^{\mathrm{2}} =\mathrm{2}. \\ $$$${This}\:{formula}\:{can}\:{be}\:{used}: \\ $$$$\mid{z}\mid^{\mathrm{2}} ={z}×\overset{−} {{z}} \\ $$

Answered by Olaf last updated on 21/Dec/20

∣z+1∣^2  = ∣z+1∣×∣z+1∣^(−)   ∣z+1∣^2  = ∣z+1∣×∣z^− +1∣  ∣z+1∣^2  = ∣(z+1)(z^− +1)∣  ∣z+1∣^2  = ∣zz^− +(z+z^− )+1∣  ∣z+1∣^2  = ∣zz^− +2Re(z)+1∣  ∣z+1∣^2  = ∣z∣^2 +2Re(z)+1 (1)    And :  ∣z−1∣^2  = ∣z∣^2 −2Re(z)+1 (2)    (1)+(2) :  ∣z+1∣^2 +∣z−1∣^2  = 2∣z∣^2 +2 (3)    (3) : ∣z+1∣^2  = 2∣z∣^2  ⇔ ∣z−1∣^2  = 2

$$\mid{z}+\mathrm{1}\mid^{\mathrm{2}} \:=\:\mid{z}+\mathrm{1}\mid×\overline {\mid{z}+\mathrm{1}\mid} \\ $$$$\mid{z}+\mathrm{1}\mid^{\mathrm{2}} \:=\:\mid{z}+\mathrm{1}\mid×\mid\overset{−} {{z}}+\mathrm{1}\mid \\ $$$$\mid{z}+\mathrm{1}\mid^{\mathrm{2}} \:=\:\mid\left({z}+\mathrm{1}\right)\left(\overset{−} {{z}}+\mathrm{1}\right)\mid \\ $$$$\mid{z}+\mathrm{1}\mid^{\mathrm{2}} \:=\:\mid{z}\overset{−} {{z}}+\left({z}+\overset{−} {{z}}\right)+\mathrm{1}\mid \\ $$$$\mid{z}+\mathrm{1}\mid^{\mathrm{2}} \:=\:\mid{z}\overset{−} {{z}}+\mathrm{2Re}\left({z}\right)+\mathrm{1}\mid \\ $$$$\mid{z}+\mathrm{1}\mid^{\mathrm{2}} \:=\:\mid{z}\mid^{\mathrm{2}} +\mathrm{2Re}\left({z}\right)+\mathrm{1}\:\left(\mathrm{1}\right) \\ $$$$ \\ $$$$\mathrm{And}\:: \\ $$$$\mid{z}−\mathrm{1}\mid^{\mathrm{2}} \:=\:\mid{z}\mid^{\mathrm{2}} −\mathrm{2Re}\left({z}\right)+\mathrm{1}\:\left(\mathrm{2}\right) \\ $$$$ \\ $$$$\left(\mathrm{1}\right)+\left(\mathrm{2}\right)\:: \\ $$$$\mid{z}+\mathrm{1}\mid^{\mathrm{2}} +\mid{z}−\mathrm{1}\mid^{\mathrm{2}} \:=\:\mathrm{2}\mid{z}\mid^{\mathrm{2}} +\mathrm{2}\:\left(\mathrm{3}\right) \\ $$$$ \\ $$$$\left(\mathrm{3}\right)\::\:\mid{z}+\mathrm{1}\mid^{\mathrm{2}} \:=\:\mathrm{2}\mid{z}\mid^{\mathrm{2}} \:\Leftrightarrow\:\mid{z}−\mathrm{1}\mid^{\mathrm{2}} \:=\:\mathrm{2} \\ $$

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