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Question Number 19313 by Tinkutara last updated on 09/Aug/17

Prove that ∣z_1 ± z_2 ∣^2 = ∣z_2 ∣^2 + ∣z_1 ∣^2 ± 2Re(z_1 z_2 ^� ) = ∣z_1 ∣^2 + ∣z_2 ∣^2 ± 2Re(z_1 ^� .z_2 )

$$\mathrm{Prove}\:\mathrm{that}\:\mid{z}_{\mathrm{1}} \:\pm\:{z}_{\mathrm{2}} \mid^{\mathrm{2}} \:=\:\mid{z}_{\mathrm{2}} \mid^{\mathrm{2}} \:+\:\mid{z}_{\mathrm{1}} \mid^{\mathrm{2}} \:\pm \\ $$$$\mathrm{2Re}\left({z}_{\mathrm{1}} \bar {{z}}_{\mathrm{2}} \right)\:=\:\mid{z}_{\mathrm{1}} \mid^{\mathrm{2}} \:+\:\mid{z}_{\mathrm{2}} \mid^{\mathrm{2}} \:\pm\:\mathrm{2Re}\left(\bar {{z}}_{\mathrm{1}} .{z}_{\mathrm{2}} \right) \\ $$

Answered by ajfour last updated on 09/Aug/17

let z_1 =x_1 +iy_1 z_2 =x_2 +iy_2 ∣z_1 ±z_2 ∣^2 =(x_1 ±x_2 )^2 +(y_1 ±y_2 )^2 =x_1 ^2 +y_1 ^2 +x_2 ^2 +y_2 ^2 ±2(x_1 x_2 +y_1 y_2 ) and z_1 z_2 ^� =(x_1 +iy_1 )(x_2 −iy_2 ) Re(z_1 z_2 ^� )=x_1 x_2 +y_1 y_2 So, ∣z_1 ±z_2 ∣^2 =∣z_1 ∣^2 +∣z_2 ∣^2 ±2Re(z_1 z_2 ^� ) .

$$\mathrm{let}\:\mathrm{z}_{\mathrm{1}} =\mathrm{x}_{\mathrm{1}} +\mathrm{iy}_{\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\mathrm{z}_{\mathrm{2}} =\mathrm{x}_{\mathrm{2}} +\mathrm{iy}_{\mathrm{2}} \\ $$$$\mid\mathrm{z}_{\mathrm{1}} \pm\mathrm{z}_{\mathrm{2}} \mid^{\mathrm{2}} =\left(\mathrm{x}_{\mathrm{1}} \pm\mathrm{x}_{\mathrm{2}} \right)^{\mathrm{2}} +\left(\mathrm{y}_{\mathrm{1}} \pm\mathrm{y}_{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\:\:=\mathrm{x}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{y}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{x}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{y}_{\mathrm{2}} ^{\mathrm{2}} \pm\mathrm{2}\left(\mathrm{x}_{\mathrm{1}} \mathrm{x}_{\mathrm{2}} +\mathrm{y}_{\mathrm{1}} \mathrm{y}_{\mathrm{2}} \right) \\ $$$$\mathrm{and}\:\mathrm{z}_{\mathrm{1}} \bar {\mathrm{z}}_{\mathrm{2}} =\left(\mathrm{x}_{\mathrm{1}} +\mathrm{iy}_{\mathrm{1}} \right)\left(\mathrm{x}_{\mathrm{2}} −\mathrm{iy}_{\mathrm{2}} \right) \\ $$$$\:\:\:\mathrm{Re}\left(\mathrm{z}_{\mathrm{1}} \bar {\mathrm{z}}_{\mathrm{2}} \right)=\mathrm{x}_{\mathrm{1}} \mathrm{x}_{\mathrm{2}} +\mathrm{y}_{\mathrm{1}} \mathrm{y}_{\mathrm{2}} \\ $$$$\mathrm{So},\:\:\:\:\mid\mathrm{z}_{\mathrm{1}} \pm\mathrm{z}_{\mathrm{2}} \mid^{\mathrm{2}} =\mid\mathrm{z}_{\mathrm{1}} \mid^{\mathrm{2}} +\mid\mathrm{z}_{\mathrm{2}} \mid^{\mathrm{2}} \pm\mathrm{2Re}\left(\mathrm{z}_{\mathrm{1}} \bar {\mathrm{z}}_{\mathrm{2}} \right)\:. \\ $$

Commented by Tinkutara last updated on 09/Aug/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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