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Question Number 182011 by mathocean1 last updated on 03/Dec/22

Is this true for complex number: 4Re(z_1 z_2 ^(−) )=∣z_1 +z_2 ^(−) ∣^2 −∣z_1 −z_2 ^(−) ∣^2

$${Is}\:{this}\:{true}\:{for}\:{complex}\:{number}: \\ $$$$\mathrm{4}\mathscr{R}{e}\left({z}_{\mathrm{1}} \overline {{z}_{\mathrm{2}} }\right)=\mid{z}_{\mathrm{1}} +\overline {{z}_{\mathrm{2}} }\mid^{\mathrm{2}} −\mid{z}_{\mathrm{1}} −\overline {{z}_{\mathrm{2}} }\mid^{\mathrm{2}} \\ $$

Commented by Frix last updated on 03/Dec/22

The question is strange. If you only use z_2 ^− you could instead use z_3 because it will make no difference. Anyway as posted it′s wrong.

$$\mathrm{The}\:\mathrm{question}\:\mathrm{is}\:\mathrm{strange}.\:\mathrm{If}\:\mathrm{you}\:\mathrm{only}\:\mathrm{use} \\ $$$$\overset{−} {{z}}_{\mathrm{2}} \:\mathrm{you}\:\mathrm{could}\:\mathrm{instead}\:\mathrm{use}\:{z}_{\mathrm{3}} \:\mathrm{because}\:\mathrm{it}\:\mathrm{will} \\ $$$$\mathrm{make}\:\mathrm{no}\:\mathrm{difference}. \\ $$$$\mathrm{Anyway}\:\mathrm{as}\:\mathrm{posted}\:\mathrm{it}'\mathrm{s}\:\mathrm{wrong}. \\ $$

Answered by alephzero last updated on 03/Dec/22

z_1 = a+bi z_2 = c+di z_2 ^� = c−di z_1 z_2 ^� = (a+bi)(c−di) = = a(c−di)+bi(c−di) = = ac−adi+bci+bd = = ac+bd−adi+bci = = ac+bd−i(ad+bc) 4Re(z_1 z_2 ^� ) = 4(ac+bd) ∣z_1 +z_2 ^� ∣^2 −∣z_1 −z_2 ^� ∣^2 = = ∣a+bi+c−di∣^2 −∣a+bi−c+di∣^2 = = ∣a+c+i(b−d)∣^2 −∣a−c+i(b+d)∣^2 = ((√((a+c)^2 +(b−d)^2 )))^2 −((√((a−c)^2 +(b+d)^2 )))^2 = (a+c)^2 +(b−d)^2 −(a−c)^2 −(b+d)^2 = = a^2 +2ac+c^2 +b^2 −2bd+d^2 −a^2 +2ac−c^2 −b^2 −2bd−d^2 = = 2ac+2ac = 4ac ≠ 4(ac+bd) ⇒this is false

$${z}_{\mathrm{1}} \:=\:{a}+{bi} \\ $$$${z}_{\mathrm{2}} \:=\:{c}+{di} \\ $$$$\bar {{z}}_{\mathrm{2}} \:=\:{c}−{di} \\ $$$${z}_{\mathrm{1}} \bar {{z}}_{\mathrm{2}} \:=\:\left({a}+{bi}\right)\left({c}−{di}\right)\:= \\ $$$$=\:{a}\left({c}−{di}\right)+{bi}\left({c}−{di}\right)\:= \\ $$$$=\:{ac}−{adi}+{bci}+{bd}\:= \\ $$$$=\:{ac}+{bd}−{adi}+{bci}\:= \\ $$$$=\:{ac}+{bd}−{i}\left({ad}+{bc}\right) \\ $$$$\mathrm{4Re}\left({z}_{\mathrm{1}} \bar {{z}}_{\mathrm{2}} \right)\:=\:\mathrm{4}\left({ac}+{bd}\right) \\ $$$$\: \\ $$$$\mid{z}_{\mathrm{1}} +\bar {{z}}_{\mathrm{2}} \mid^{\mathrm{2}} −\mid{z}_{\mathrm{1}} −\bar {{z}}_{\mathrm{2}} \mid^{\mathrm{2}} \:= \\ $$$$=\:\mid{a}+{bi}+{c}−{di}\mid^{\mathrm{2}} −\mid{a}+{bi}−{c}+{di}\mid^{\mathrm{2}} \:= \\ $$$$=\:\mid{a}+{c}+{i}\left({b}−{d}\right)\mid^{\mathrm{2}} −\mid{a}−{c}+{i}\left({b}+{d}\right)\mid^{\mathrm{2}} \:= \\ $$$$\left(\sqrt{\left({a}+{c}\right)^{\mathrm{2}} +\left({b}−{d}\right)^{\mathrm{2}} }\right)^{\mathrm{2}} −\left(\sqrt{\left({a}−{c}\right)^{\mathrm{2}} +\left({b}+{d}\right)^{\mathrm{2}} }\right)^{\mathrm{2}} \\ $$$$=\:\left({a}+{c}\right)^{\mathrm{2}} +\left({b}−{d}\right)^{\mathrm{2}} −\left({a}−{c}\right)^{\mathrm{2}} −\left({b}+{d}\right)^{\mathrm{2}} \:= \\ $$$$=\:\cancel{{a}^{\mathrm{2}} }+\mathrm{2}{ac}+\cancel{{c}^{\mathrm{2}} }+\cancel{{b}^{\mathrm{2}} }−\cancel{\mathrm{2}{bd}}+\cancel{{d}^{\mathrm{2}} }−\cancel{{a}^{\mathrm{2}} }+\mathrm{2}{ac}−\cancel{{c}^{\mathrm{2}} }−\cancel{{b}^{\mathrm{2}} }−\cancel{\mathrm{2}{bd}}−\cancel{{d}^{\mathrm{2}} }\:= \\ $$$$=\:\mathrm{2}{ac}+\mathrm{2}{ac}\:=\:\mathrm{4}{ac}\:\neq\:\mathrm{4}\left({ac}+{bd}\right) \\ $$$$\Rightarrow{this}\:{is}\:{false} \\ $$

Commented by alephzero last updated on 03/Dec/22

Sorry, I′ve corrected now, thank you!

$${Sorry},\:{I}'{ve}\:{corrected}\:{now},\:{thank}\:{you}! \\ $$

Commented by Frix last updated on 03/Dec/22

Error of sign: = (a+c)^2 +(b−d)^2 −(a−c)^2 −^! (b+d)^2 =

$$\mathrm{Error}\:\mathrm{of}\:\mathrm{sign}: \\ $$$$=\:\left({a}+{c}\right)^{\mathrm{2}} +\left({b}−{d}\right)^{\mathrm{2}} −\left({a}−{c}\right)^{\mathrm{2}} \overset{!} {−}\left({b}+{d}\right)^{\mathrm{2}} \:= \\ $$

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