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Question Number 56146 by gunawan last updated on 11/Mar/19
Given complex number z_1 , z_2 , and z_3 satiesfied z_1 +z_2 +z_3 =0 and ∣z_1 ∣=∣z_2 ∣=∣z_3 ∣=1. Prove that z_1 ^2 +z_2 ^2 +z_3 ^2 =0
$$\mathrm{Given}\:\mathrm{complex}\:\mathrm{number} \\ $$$${z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} ,\:\mathrm{and}\:{z}_{\mathrm{3}} \:\mathrm{satiesfied}\:{z}_{\mathrm{1}} +{z}_{\mathrm{2}} +{z}_{\mathrm{3}} =\mathrm{0} \\ $$$$\mathrm{and}\:\mid{z}_{\mathrm{1}} \mid=\mid{z}_{\mathrm{2}} \mid=\mid{z}_{\mathrm{3}} \mid=\mathrm{1}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$${z}_{\mathrm{1}} ^{\mathrm{2}} +{z}_{\mathrm{2}} ^{\mathrm{2}} +{z}_{\mathrm{3}} ^{\mathrm{2}} =\mathrm{0} \\ $$
Commented by MJS last updated on 11/Mar/19
z_1 , z_2 , z_3 are the solutions of z^3 =cos 3α +isin 3α z^3 =e^(3iα) ⇒ z_1 =e^(iα) ; z_2 =e^(i(α−((2π)/3))) ; z_3 =e^(i(α+((2π)/3)))
$${z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} ,\:{z}_{\mathrm{3}} \:\mathrm{are}\:\mathrm{the}\:\mathrm{solutions}\:\mathrm{of} \\ $$$${z}^{\mathrm{3}} =\mathrm{cos}\:\mathrm{3}\alpha\:+{i}\mathrm{sin}\:\mathrm{3}\alpha \\ $$$${z}^{\mathrm{3}} =\mathrm{e}^{\mathrm{3i}\alpha} \:\Rightarrow\:{z}_{\mathrm{1}} =\mathrm{e}^{\mathrm{i}\alpha} ;\:{z}_{\mathrm{2}} =\mathrm{e}^{\mathrm{i}\left(\alpha−\frac{\mathrm{2}\pi}{\mathrm{3}}\right)} ;\:{z}_{\mathrm{3}} =\mathrm{e}^{\mathrm{i}\left(\alpha+\frac{\mathrm{2}\pi}{\mathrm{3}}\right)} \\ $$
Answered by MJS last updated on 11/Mar/19
we can choose z_3 =1+0i z_1 =a+bi z_2 =c+di (1) a+c+1=0 (2) b+d=0 (3) a^2 +b^2 =1 (4) c^2 +d^2 =1 (1) ⇒ c=−(a+1) (2) ⇒ d=−b (3) a^2 +b^2 =1 (4) a^2 +2a+b^2 +1=1 (4)−(3) 2a+1=0 ⇒ a=−(1/2) ⇒ c=−(1/2) ⇒ b^2 =(3/4) ⇒ b=±((√3)/2) ⇒ d=∓((√3)/2) ⇒ z_1 =−(1/2)±((√3)/2); z_2 =−(1/2)∓((√3)/2) ⇒ z_1 ^2 =z_2 ; z_2 ^2 =z_1 ⇒ z_1 ^2 +z_2 ^2 +z_3 ^2 =0 we have got the roots of unity ⇒ for any given z_3 =cos α +isin α we get z_(1, 2) =cos (α±((2π)/3)) +isin (α±((2π)/3)) because it′s just a rotation by α the sum of the squares is zero which can be shown by adding them and using trigonometric formulas for additions etc.
$$\mathrm{we}\:\mathrm{can}\:\mathrm{choose}\:{z}_{\mathrm{3}} =\mathrm{1}+\mathrm{0i} \\ $$$${z}_{\mathrm{1}} ={a}+{b}\mathrm{i} \\ $$$${z}_{\mathrm{2}} ={c}+{d}\mathrm{i} \\ $$$$\left(\mathrm{1}\right)\:\:{a}+{c}+\mathrm{1}=\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:\:{b}+{d}=\mathrm{0} \\ $$$$\left(\mathrm{3}\right)\:\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{1} \\ $$$$\left(\mathrm{4}\right)\:\:{c}^{\mathrm{2}} +{d}^{\mathrm{2}} =\mathrm{1} \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\:\Rightarrow\:{c}=−\left({a}+\mathrm{1}\right) \\ $$$$\left(\mathrm{2}\right)\:\Rightarrow\:{d}=−{b} \\ $$$$\left(\mathrm{3}\right)\:\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{1} \\ $$$$\left(\mathrm{4}\right)\:\:{a}^{\mathrm{2}} +\mathrm{2}{a}+{b}^{\mathrm{2}} +\mathrm{1}=\mathrm{1} \\ $$$$\left(\mathrm{4}\right)−\left(\mathrm{3}\right)\:\:\mathrm{2}{a}+\mathrm{1}=\mathrm{0}\:\Rightarrow\:{a}=−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:{c}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\:{b}^{\mathrm{2}} =\frac{\mathrm{3}}{\mathrm{4}}\:\Rightarrow\:{b}=\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\Rightarrow\:{d}=\mp\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\Rightarrow\:{z}_{\mathrm{1}} =−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}};\:{z}_{\mathrm{2}} =−\frac{\mathrm{1}}{\mathrm{2}}\mp\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\Rightarrow\:{z}_{\mathrm{1}} ^{\mathrm{2}} ={z}_{\mathrm{2}} ;\:{z}_{\mathrm{2}} ^{\mathrm{2}} ={z}_{\mathrm{1}} \:\Rightarrow\:{z}_{\mathrm{1}} ^{\mathrm{2}} +{z}_{\mathrm{2}} ^{\mathrm{2}} +{z}_{\mathrm{3}} ^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{got}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{unity}\:\Rightarrow\:\mathrm{for}\:\mathrm{any} \\ $$$$\mathrm{given}\:{z}_{\mathrm{3}} =\mathrm{cos}\:\alpha\:+{i}\mathrm{sin}\:\alpha\:\mathrm{we}\:\mathrm{get} \\ $$$${z}_{\mathrm{1},\:\mathrm{2}} =\mathrm{cos}\:\left(\alpha\pm\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\:+{i}\mathrm{sin}\:\left(\alpha\pm\frac{\mathrm{2}\pi}{\mathrm{3}}\right) \\ $$$$\mathrm{because}\:\mathrm{it}'\mathrm{s}\:\mathrm{just}\:\mathrm{a}\:\mathrm{rotation}\:\mathrm{by}\:\alpha \\ $$$$\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{squares}\:\mathrm{is}\:\mathrm{zero}\:\mathrm{which}\:\mathrm{can}\:\mathrm{be} \\ $$$$\mathrm{shown}\:\mathrm{by}\:\mathrm{adding}\:\mathrm{them}\:\mathrm{and}\:\mathrm{using}\:\mathrm{trigonometric} \\ $$$$\mathrm{formulas}\:\mathrm{for}\:\mathrm{additions}\:\mathrm{etc}. \\ $$

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