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Question Number 208959 by hardmath last updated on 28/Jun/24
If z = − (1/2) + ((√3)/2) i Find (z^4 + 2z)∙(z^3 + z) = ?
$$\mathrm{If}\:\:\:\boldsymbol{\mathrm{z}}\:=\:−\:\frac{\mathrm{1}}{\mathrm{2}}\:\:+\:\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\boldsymbol{\mathrm{i}} \\ $$$$\mathrm{Find}\:\:\:\left(\mathrm{z}^{\mathrm{4}} \:+\:\mathrm{2z}\right)\centerdot\left(\mathrm{z}^{\mathrm{3}} \:+\:\mathrm{z}\right)\:=\:? \\ $$
Answered by grigoriy last updated on 29/Jun/24
f(z) = (z^4 + 2z)(z^3 + z); z^4 + 2z = f_1 (z); z^3 + z = f_2 (z); f(z) = f_1 (z)f_2 (z); z = −(1/2) + ((√3)/2)i = R(cos(ϕ) + isin(ϕ)); R^2 = (−(1/2))^2 + (((√3)/2))^2 = (1/4) + (3/4) = 1; ⇒ R = 1; (R = ∣f(z)∣); cos(ϕ) = −(1/2); sin(ϕ) = ((√3)/2); ⇒ cos(ϕ) < 0; sin(ϕ) > 0 ⇒ π < ϕ < (π/2); tg(ϕ) = ((sin(ϕ))/(cos(ϕ))) = −(√3); ⇒ ϕ = arctg(−(√3)) = π − arctg((√3)) = π − (π/3) = ((2π)/3); π < ϕ < (π/2); ⇒ ϕ = ((2π)/3); z = cos(((2π)/3)) + isin(((2π)/3)); z^4 = (cos(((2π)/3)) + isin(((2π)/3)))^4 = cos(((8π)/3)) + isin(((8π)/3)); z^4 = cos(((8π)/3)) + isin(((8π)/3)); f_1 (z) = z^4 + 2z; f_1 (z) = cos(((8π)/3)) + isin(((8π)/3)) + 2cos(((2π)/3)) + 2isin(((2π)/3)); f_1 (z) = (cos(((8π)/3)) + 2cos(((2π)/3))) + i(sin(((8π)/3)) + 2sin(((2π)/3))); z^4 = (−(1/2) + i((√3)/2))^4 ; (a + b)^4 = Σ_(n = 0) ^4 C_4 ^n a^(4− n) b^n = C_4 ^0 a^4 + C_4 ^1 a^3 b + C_4 ^2 a^2 b^2 + C_4 ^3 ab^3 + C_4 ^4 b^4 ; C_4 ^0 = C_4 ^4 = 1; C_4 ^1 = C_3 ^1 + C_3 ^0 = C_2 ^1 + C_2 ^0 + 1 = C_1 ^1 + C_1 ^0 + 2 = 4; C_4 ^2 = C_3 ^2 + C_3 ^1 = C_2 ^2 + C_2 ^1 + C_2 ^1 + C_2 ^0 = 1 + 2C_2 ^1 + 1 = 2 + 2(C_1 ^1 + C_1 ^0 ) = 6; C_4 ^3 = C_4 ^(4 − 3) = C_4 ^1 = 4; (a + b)^4 = a^4 + 4a^3 b + 6a^2 b^2 + 4ab^3 + b^4 ; z^4 = (−(1/2))^4 + 4∙(−(1/2))^3 ∙((√3)/2)∙i + 6∙(−(1/2))^2 ∙(((√3)/2))^2 ∙i^2 + 4∙(−(1/2))∙(((√3)/2))^3 ∙i^3 + (((√3)/2))^4 ∙i^4 ; i = (√(−1)) = (−1)^(1/2) ; i^2 = −1; i^3 = i∙i^2 = −i; i^4 = (−1)^2 = 1; z^4 = (1/(16)) − ((√3)/4)i − (9/8) − ((√3)/4)i + (9/(16)) = −(1/2) − ((√3)/2)i; z^4 = −(1/2) − ((√3)/2)i; 2z = 2(−(1/2) + ((√3)/2)i) =−1 + (√3)i; f_1 (z) = −(1/2) − ((√3)/2)i − 1 + (√3)i = −(3/2) + ((√3)/2)i; f_1 (z) = −(3/2) + ((√3)/2)i; f_2 (z) = z^3 + z; z^3 = (−(1/2) + ((√3)/2)i)^3 ; (a + b)^3 = Σ_(n = 0) ^3 C_3 ^n a^(3 − n) b^n = C_3 ^0 a^3 + C_3 ^1 a^2 b + C_3 ^2 ab^2 + C_3 ^3 b^3 ; C_3 ^0 = C_3 ^3 = 1; C_3 ^1 = C_2 ^1 + C_2 ^0 = C_1 ^1 + C_1 ^0 + 1 = 3; C_3 ^2 = C_3 ^(3 − 2) = C_3 ^1 = 3; (a + b)^3 = a^3 + 3a^2 b + 3ab^2 + b^3 ; z^3 = (−(1/2))^3 + 3(−(1/2))^2 ((√3)/2)i + 3(−(1/2))(((√3)/2))^2 i^2 + (((√3)/2))^3 i^3 ; z^3 = −(1/8) + ((3(√3))/8)i + (9/8) − ((3(√3))/8)i = 1; z^3 = 1; f_2 (z) = 1 − (1/2) + ((√3)/2)i = (1/2) + ((√3)/2)i; f(z) = f_1 (z)f_2 (z) = (−(3/2) + ((√3)/2)i)((1/2) + ((√3)/2)i) = −(3/2)((1/2) + ((√3)/2)i) + ((√3)/2)((1/2)i − ((√3)/2)); f(z) = −(3/4) − ((3(√3))/4)i + ((√3)/4)i− ((√3)/4) = −((√3)/2) + (((√3) − 3(√3))/4)i; Ansver: (z^4 + 2z)(z^3 + z) = −((√3)/2) + (((√3) − 3(√3))/4)i;
$$ \\ $$$${f}\left({z}\right)\:=\:\left({z}^{\mathrm{4}} \:+\:\mathrm{2}{z}\right)\left({z}^{\mathrm{3}} \:+\:{z}\right); \\ $$$${z}^{\mathrm{4}} \:+\:\mathrm{2}{z}\:=\:{f}_{\mathrm{1}} \left({z}\right);\:{z}^{\mathrm{3}} \:+\:{z}\:=\:{f}_{\mathrm{2}} \left({z}\right); \\ $$$${f}\left({z}\right)\:=\:{f}_{\mathrm{1}} \left({z}\right){f}_{\mathrm{2}} \left({z}\right); \\ $$$${z}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\:+\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i}\:=\:{R}\left({cos}\left(\varphi\right)\:+\:{isin}\left(\varphi\right)\right); \\ $$$${R}^{\mathrm{2}} \:=\:\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\:\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} =\:\frac{\mathrm{1}}{\mathrm{4}}\:+\:\frac{\mathrm{3}}{\mathrm{4}}\:=\:\mathrm{1};\:\Rightarrow\:{R}\:=\:\mathrm{1};\:\left({R}\:=\:\mid{f}\left({z}\right)\mid\right); \\ $$$${cos}\left(\varphi\right)\:=\:−\frac{\mathrm{1}}{\mathrm{2}};\:{sin}\left(\varphi\right)\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}};\:\Rightarrow\:{cos}\left(\varphi\right)\:<\:\mathrm{0};\:{sin}\left(\varphi\right)\:>\:\mathrm{0}\:\Rightarrow\:\pi\:<\:\varphi\:<\:\frac{\pi}{\mathrm{2}};\: \\ $$$${tg}\left(\varphi\right)\:=\:\frac{{sin}\left(\varphi\right)}{{cos}\left(\varphi\right)}\:=\:−\sqrt{\mathrm{3}};\:\Rightarrow\:\varphi\:=\:{arctg}\left(−\sqrt{\mathrm{3}}\right)\:=\:\pi\:−\:{arctg}\left(\sqrt{\mathrm{3}}\right)\:=\:\pi\:−\:\frac{\pi}{\mathrm{3}}\:=\:\frac{\mathrm{2}\pi}{\mathrm{3}}; \\ $$$$\pi\:<\:\varphi\:<\:\frac{\pi}{\mathrm{2}};\:\Rightarrow\:\varphi\:=\:\frac{\mathrm{2}\pi}{\mathrm{3}}; \\ $$$${z}\:=\:{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\:+\:{isin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right); \\ $$$${z}^{\mathrm{4}} \:=\:\left({cos}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\:+\:{isin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right)^{\mathrm{4}} \:=\:{cos}\left(\frac{\mathrm{8}\pi}{\mathrm{3}}\right)\:+\:{isin}\left(\frac{\mathrm{8}\pi}{\mathrm{3}}\right); \\ $$$${z}^{\mathrm{4}} \:=\:{cos}\left(\frac{\mathrm{8}\pi}{\mathrm{3}}\right)\:+\:{isin}\left(\frac{\mathrm{8}\pi}{\mathrm{3}}\right); \\ $$$${f}_{\mathrm{1}} \left({z}\right)\:=\:{z}^{\mathrm{4}} \:+\:\mathrm{2}{z}; \\ $$$${f}_{\mathrm{1}} \left({z}\right)\:=\:{cos}\left(\frac{\mathrm{8}\pi}{\mathrm{3}}\right)\:+\:{isin}\left(\frac{\mathrm{8}\pi}{\mathrm{3}}\right)\:+\:\mathrm{2}{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\:+\:\mathrm{2}{isin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right); \\ $$$${f}_{\mathrm{1}} \left({z}\right)\:=\:\left({cos}\left(\frac{\mathrm{8}\pi}{\mathrm{3}}\right)\:+\:\mathrm{2}{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right)\:+\:{i}\left({sin}\left(\frac{\mathrm{8}\pi}{\mathrm{3}}\right)\:+\:\mathrm{2}{sin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right); \\ $$$${z}^{\mathrm{4}} \:=\:\left(−\frac{\mathrm{1}}{\mathrm{2}}\:+\:{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{4}} ; \\ $$$$\left({a}\:+\:{b}\right)^{\mathrm{4}} \:=\:\underset{{n}\:=\:\mathrm{0}} {\overset{\mathrm{4}} {\sum}}{C}_{\mathrm{4}} ^{{n}} {a}^{\mathrm{4}−\:{n}} {b}^{{n}} \:=\:{C}_{\mathrm{4}} ^{\mathrm{0}} {a}^{\mathrm{4}} \:+\:{C}_{\mathrm{4}} ^{\mathrm{1}} {a}^{\mathrm{3}} {b}\:+\:{C}_{\mathrm{4}} ^{\mathrm{2}} {a}^{\mathrm{2}} {b}^{\mathrm{2}} \:+\:{C}_{\mathrm{4}} ^{\mathrm{3}} {ab}^{\mathrm{3}} \:+\:{C}_{\mathrm{4}} ^{\mathrm{4}} {b}^{\mathrm{4}} ; \\ $$$${C}_{\mathrm{4}} ^{\mathrm{0}} \:=\:{C}_{\mathrm{4}} ^{\mathrm{4}} \:=\:\mathrm{1}; \\ $$$${C}_{\mathrm{4}} ^{\mathrm{1}} \:=\:{C}_{\mathrm{3}} ^{\mathrm{1}} \:+\:{C}_{\mathrm{3}} ^{\mathrm{0}} \:=\:{C}_{\mathrm{2}} ^{\mathrm{1}} \:+\:{C}_{\mathrm{2}} ^{\mathrm{0}} \:+\:\mathrm{1}\:=\:{C}_{\mathrm{1}} ^{\mathrm{1}} +\:{C}_{\mathrm{1}} ^{\mathrm{0}} \:+\:\mathrm{2}\:=\:\mathrm{4}; \\ $$$${C}_{\mathrm{4}} ^{\mathrm{2}} \:=\:{C}_{\mathrm{3}} ^{\mathrm{2}} \:+\:{C}_{\mathrm{3}} ^{\mathrm{1}} \:=\:{C}_{\mathrm{2}} ^{\mathrm{2}} \:+\:{C}_{\mathrm{2}} ^{\mathrm{1}} \:+\:{C}_{\mathrm{2}} ^{\mathrm{1}} \:+\:{C}_{\mathrm{2}} ^{\mathrm{0}} \:=\:\mathrm{1}\:+\:\mathrm{2}{C}_{\mathrm{2}} ^{\mathrm{1}} \:+\:\mathrm{1}\:=\:\mathrm{2}\:+\:\mathrm{2}\left({C}_{\mathrm{1}} ^{\mathrm{1}} \:+\:{C}_{\mathrm{1}} ^{\mathrm{0}} \right)\:=\:\mathrm{6}; \\ $$$${C}_{\mathrm{4}} ^{\mathrm{3}} \:=\:{C}_{\mathrm{4}} ^{\mathrm{4}\:−\:\mathrm{3}} \:=\:{C}_{\mathrm{4}} ^{\mathrm{1}} \:=\:\mathrm{4}; \\ $$$$\left({a}\:+\:{b}\right)^{\mathrm{4}} \:=\:{a}^{\mathrm{4}} \:+\:\mathrm{4}{a}^{\mathrm{3}} {b}\:+\:\mathrm{6}{a}^{\mathrm{2}} {b}^{\mathrm{2}} \:+\:\mathrm{4}{ab}^{\mathrm{3}} \:+\:{b}^{\mathrm{4}} ; \\ $$$${z}^{\mathrm{4}} \:=\:\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{4}} +\:\mathrm{4}\centerdot\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} \centerdot\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\centerdot{i}\:+\:\mathrm{6}\centerdot\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \centerdot\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} \centerdot{i}^{\mathrm{2}} +\:\mathrm{4}\centerdot\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\centerdot\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{3}} \centerdot{i}^{\mathrm{3}} +\:\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{4}} \centerdot{i}^{\mathrm{4}} ; \\ $$$${i}\:=\:\sqrt{−\mathrm{1}}\:=\:\left(−\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} ;\:{i}^{\mathrm{2}} \:=\:−\mathrm{1};\:{i}^{\mathrm{3}} \:=\:{i}\centerdot{i}^{\mathrm{2}} \:=\:−{i};\:{i}^{\mathrm{4}} \:=\:\left(−\mathrm{1}\right)^{\mathrm{2}} \:=\:\mathrm{1}; \\ $$$${z}^{\mathrm{4}} \:=\:\frac{\mathrm{1}}{\mathrm{16}}\:−\:\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{i}\:−\:\frac{\mathrm{9}}{\mathrm{8}}\:−\:\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{i}\:+\:\frac{\mathrm{9}}{\mathrm{16}}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\:−\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i}; \\ $$$${z}^{\mathrm{4}} \:=\:−\frac{\mathrm{1}}{\mathrm{2}}\:−\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i}; \\ $$$$\mathrm{2}{z}\:=\:\mathrm{2}\left(−\frac{\mathrm{1}}{\mathrm{2}}\:+\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i}\right)\:=−\mathrm{1}\:+\:\sqrt{\mathrm{3}}{i}; \\ $$$${f}_{\mathrm{1}} \left({z}\right)\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\:−\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i}\:−\:\mathrm{1}\:+\:\sqrt{\mathrm{3}}{i}\:=\:−\frac{\mathrm{3}}{\mathrm{2}}\:+\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i}; \\ $$$${f}_{\mathrm{1}} \left({z}\right)\:=\:−\frac{\mathrm{3}}{\mathrm{2}}\:+\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i}; \\ $$$${f}_{\mathrm{2}} \left({z}\right)\:=\:{z}^{\mathrm{3}} \:+\:{z}; \\ $$$${z}^{\mathrm{3}} \:=\:\left(−\frac{\mathrm{1}}{\mathrm{2}}\:+\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i}\right)^{\mathrm{3}} ; \\ $$$$\left({a}\:+\:{b}\right)^{\mathrm{3}} \:=\:\underset{{n}\:=\:\mathrm{0}} {\overset{\mathrm{3}} {\sum}}{C}_{\mathrm{3}} ^{{n}} {a}^{\mathrm{3}\:−\:{n}} {b}^{{n}} \:=\:{C}_{\mathrm{3}} ^{\mathrm{0}} {a}^{\mathrm{3}} \:+\:{C}_{\mathrm{3}} ^{\mathrm{1}} {a}^{\mathrm{2}} {b}\:+\:{C}_{\mathrm{3}} ^{\mathrm{2}} {ab}^{\mathrm{2}} \:+\:{C}_{\mathrm{3}} ^{\mathrm{3}} {b}^{\mathrm{3}} ; \\ $$$${C}_{\mathrm{3}} ^{\mathrm{0}} \:=\:{C}_{\mathrm{3}} ^{\mathrm{3}} \:=\:\mathrm{1}; \\ $$$${C}_{\mathrm{3}} ^{\mathrm{1}} \:=\:{C}_{\mathrm{2}} ^{\mathrm{1}} \:+\:{C}_{\mathrm{2}} ^{\mathrm{0}} \:=\:{C}_{\mathrm{1}} ^{\mathrm{1}} \:+\:{C}_{\mathrm{1}} ^{\mathrm{0}} \:+\:\mathrm{1}\:=\:\mathrm{3}; \\ $$$${C}_{\mathrm{3}} ^{\mathrm{2}} \:=\:{C}_{\mathrm{3}} ^{\mathrm{3}\:−\:\mathrm{2}} \:=\:{C}_{\mathrm{3}} ^{\mathrm{1}} \:=\:\mathrm{3}; \\ $$$$\left({a}\:+\:{b}\right)^{\mathrm{3}} \:=\:{a}^{\mathrm{3}} \:+\:\mathrm{3}{a}^{\mathrm{2}} {b}\:+\:\mathrm{3}{ab}^{\mathrm{2}} \:+\:{b}^{\mathrm{3}} ; \\ $$$${z}^{\mathrm{3}} \:=\:\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} +\:\mathrm{3}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i}\:+\:\mathrm{3}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} {i}^{\mathrm{2}} +\:\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{3}} {i}^{\mathrm{3}} ; \\ $$$${z}^{\mathrm{3}} \:=\:−\frac{\mathrm{1}}{\mathrm{8}}\:+\:\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{8}}{i}\:+\:\frac{\mathrm{9}}{\mathrm{8}}\:−\:\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{8}}{i}\:=\:\mathrm{1}; \\ $$$${z}^{\mathrm{3}} \:=\:\mathrm{1}; \\ $$$${f}_{\mathrm{2}} \left({z}\right)\:=\:\mathrm{1}\:−\:\frac{\mathrm{1}}{\mathrm{2}}\:+\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:+\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i}; \\ $$$${f}\left({z}\right)\:=\:{f}_{\mathrm{1}} \left({z}\right){f}_{\mathrm{2}} \left({z}\right)\:=\:\left(−\frac{\mathrm{3}}{\mathrm{2}}\:+\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}\:+\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i}\right)\:=\:−\frac{\mathrm{3}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}\:+\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i}\right)\:+\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}{i}\:−\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right); \\ $$$${f}\left({z}\right)\:=\:−\frac{\mathrm{3}}{\mathrm{4}}\:−\:\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{4}}{i}\:+\:\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{i}−\:\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\:=\:−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:+\:\frac{\sqrt{\mathrm{3}}\:−\:\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{4}}{i}; \\ $$$${Ansver}:\:\left({z}^{\mathrm{4}} \:+\:\mathrm{2}{z}\right)\left({z}^{\mathrm{3}} \:+\:{z}\right)\:=\:−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:+\:\frac{\sqrt{\mathrm{3}}\:−\:\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{4}}{i}; \\ $$
Answered by grigoriy last updated on 29/Jun/24
feedback: mail: grigoriybrewer@gmail.com; telegram(!): @wraith _� machine;
$$ \\ $$$${feedback}:\:{mail}:\:{grigoriybrewer}@{gmail}.{com}; \\ $$$${telegram}\left(!\right):\:@{wraith}\underset{} {\:}{machine}; \\ $$$$ \\ $$
Answered by mr W last updated on 29/Jun/24
z=−(1/2)+((√3)/2)i=cos ((2π)/3)+i sin ((2π)/3)=e^(((2π)/3)i) (z^4 +2z)(z^3 +z) =z^7 +z^5 +2z^4 +2z^2 =e^((14πi)/3) +e^((10πi)/3) +2e^((8πi)/3) +2e^((4πi)/3) =e^((2πi)/3) +e^(−((2πi)/3)) +2e^((2πi)/3) +2e^(−((2πi)/3)) =3(e^((2πi)/3) +e^(−((2πi)/3)) ) =3(−(1/2)+(((√3)i)/2)−(1/2)−(((√3)i)/2)) =−3 ✓
$${z}=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i}=\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{3}}+{i}\:\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{3}}={e}^{\frac{\mathrm{2}\pi}{\mathrm{3}}{i}} \\ $$$$\left({z}^{\mathrm{4}} +\mathrm{2}{z}\right)\left({z}^{\mathrm{3}} +{z}\right) \\ $$$$={z}^{\mathrm{7}} +{z}^{\mathrm{5}} +\mathrm{2}{z}^{\mathrm{4}} +\mathrm{2}{z}^{\mathrm{2}} \\ $$$$={e}^{\frac{\mathrm{14}\pi{i}}{\mathrm{3}}} +{e}^{\frac{\mathrm{10}\pi{i}}{\mathrm{3}}} +\mathrm{2}{e}^{\frac{\mathrm{8}\pi{i}}{\mathrm{3}}} +\mathrm{2}{e}^{\frac{\mathrm{4}\pi{i}}{\mathrm{3}}} \\ $$$$={e}^{\frac{\mathrm{2}\pi{i}}{\mathrm{3}}} +{e}^{−\frac{\mathrm{2}\pi{i}}{\mathrm{3}}} +\mathrm{2}{e}^{\frac{\mathrm{2}\pi{i}}{\mathrm{3}}} +\mathrm{2}{e}^{−\frac{\mathrm{2}\pi{i}}{\mathrm{3}}} \\ $$$$=\mathrm{3}\left({e}^{\frac{\mathrm{2}\pi{i}}{\mathrm{3}}} +{e}^{−\frac{\mathrm{2}\pi{i}}{\mathrm{3}}} \right) \\ $$$$=\mathrm{3}\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}{i}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}{i}}{\mathrm{2}}\right) \\ $$$$=−\mathrm{3}\:\checkmark \\ $$
Answered by Rasheed.Sindhi last updated on 29/Jun/24
z = − (1/2) + ((√3)/2) i=ω (cuberoot of unity) z^3 =1 ⇒z^3 −1=0 ⇒(z−1)(z^2 +z+1)=0 ⇒z≠1⇒z^2 +z+1=0⇒z^2 +z=−1 ▶(z^4 + 2z)∙(z^3 + z) = (z + 2z)∙(1 + z) = 3z+3z^2 =3(z+z^2 )=3(−1)=−3 ✓
$$\boldsymbol{\mathrm{z}}\:=\:−\:\frac{\mathrm{1}}{\mathrm{2}}\:\:+\:\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\boldsymbol{\mathrm{i}}=\omega\:\left({cuberoot}\:{of}\:{unity}\right) \\ $$$$\mathrm{z}^{\mathrm{3}} =\mathrm{1} \\ $$$$\Rightarrow\mathrm{z}^{\mathrm{3}} −\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{z}−\mathrm{1}\right)\left(\mathrm{z}^{\mathrm{2}} +\mathrm{z}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{z}\neq\mathrm{1}\Rightarrow\mathrm{z}^{\mathrm{2}} +\mathrm{z}+\mathrm{1}=\mathrm{0}\Rightarrow\mathrm{z}^{\mathrm{2}} +\mathrm{z}=−\mathrm{1} \\ $$$$\blacktriangleright\left(\mathrm{z}^{\mathrm{4}} \:+\:\mathrm{2z}\right)\centerdot\left(\mathrm{z}^{\mathrm{3}} \:+\:\mathrm{z}\right)\:= \\ $$$$\left(\mathrm{z}\:+\:\mathrm{2z}\right)\centerdot\left(\mathrm{1}\:+\:\mathrm{z}\right)\:= \\ $$$$\mathrm{3z}+\mathrm{3z}^{\mathrm{2}} =\mathrm{3}\left(\mathrm{z}+\mathrm{z}^{\mathrm{2}} \right)=\mathrm{3}\left(−\mathrm{1}\right)=−\mathrm{3}\:\checkmark \\ $$
Answered by Rasheed.Sindhi last updated on 29/Jun/24
z = − (1/2) + ((√3)/2) i 2z+1=(√3) i 4z^2 +4z+1=−3 z^2 +z+1=0⇒z^2 +z=−1 (z−1)(z^2 +z+1)=0⇒z^3 −1=0 ⇒z^3 =1 ▶ (z^4 + 2z)∙(z^3 + z) =(z^3 .z+2z)(1+z) =(1.z+2z)(1+z) =3z(1+z)=3z.−z^2 =−3z^3 =−3
$$\boldsymbol{\mathrm{z}}\:=\:−\:\frac{\mathrm{1}}{\mathrm{2}}\:\:+\:\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\boldsymbol{\mathrm{i}} \\ $$$$\mathrm{2z}+\mathrm{1}=\sqrt{\mathrm{3}}\:\mathrm{i} \\ $$$$\mathrm{4z}^{\mathrm{2}} +\mathrm{4z}+\mathrm{1}=−\mathrm{3} \\ $$$$\mathrm{z}^{\mathrm{2}} +\mathrm{z}+\mathrm{1}=\mathrm{0}\Rightarrow\mathrm{z}^{\mathrm{2}} +\mathrm{z}=−\mathrm{1} \\ $$$$\left(\mathrm{z}−\mathrm{1}\right)\left(\mathrm{z}^{\mathrm{2}} +\mathrm{z}+\mathrm{1}\right)=\mathrm{0}\Rightarrow\mathrm{z}^{\mathrm{3}} −\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{z}^{\mathrm{3}} =\mathrm{1} \\ $$$$\blacktriangleright\:\left(\mathrm{z}^{\mathrm{4}} \:+\:\mathrm{2z}\right)\centerdot\left(\mathrm{z}^{\mathrm{3}} \:+\:\mathrm{z}\right) \\ $$$$\:\:\:\:\:=\left(\mathrm{z}^{\mathrm{3}} .\mathrm{z}+\mathrm{2z}\right)\left(\mathrm{1}+\mathrm{z}\right) \\ $$$$\:\:\:\:\:=\left(\mathrm{1}.\mathrm{z}+\mathrm{2z}\right)\left(\mathrm{1}+\mathrm{z}\right) \\ $$$$\:\:\:\:\:=\mathrm{3z}\left(\mathrm{1}+\mathrm{z}\right)=\mathrm{3z}.−\mathrm{z}^{\mathrm{2}} =−\mathrm{3z}^{\mathrm{3}} =−\mathrm{3} \\ $$

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