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Question Number 208959 by hardmath last updated on 28/Jun/24

$$\mathrm{If}\mathbf{z}=-\frac{1}{2}+\frac{\sqrt{3}}{2}\mathbf{i}$$$$\mathrm{Find}({\mathrm{z}}^4+2\mathrm{z})\cdot ({\mathrm{z}}^3+\mathrm{z})=?$$

Answered by grigoriy last updated on 29/Jun/24

$$$$$$f\left(z\right)=(z^4+2z)(z^3+z);$$$$z^4+2z=f_1\left(z\right);z^3+z=f_2\left(z\right);$$$$f\left(z\right)=f_1\left(z\right)f_2\left(z\right);$$$$z=-\frac{1}{2}+\frac{\sqrt{3}}{2}i=R(cos\left(\phi \right)+isin\left(\phi \right));$$$$R^2={(-\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2=\frac{1}{4}+\frac{3}{4}=1;\Rightarrow R=1;(R=\mid f\left(z\right)\mid );$$$$cos\left(\phi \right)=-\frac{1}{2};sin\left(\phi \right)=\frac{\sqrt{3}}{2};\Rightarrow cos\left(\phi \right)<0;sin\left(\phi \right)>0\Rightarrow \pi <\phi <\frac{\pi }{2};$$$$tg\left(\phi \right)=\frac{sin\left(\phi \right)}{cos\left(\phi \right)}=-\sqrt{3};\Rightarrow \phi =arctg(-\sqrt{3})=\pi -arctg\left(\sqrt{3}\right)=\pi -\frac{\pi }{3}=\frac{2\pi }{3};$$$$\pi <\phi <\frac{\pi }{2};\Rightarrow \phi =\frac{2\pi }{3};$$$$z=cos\left(\frac{2\pi }{3}\right)+isin\left(\frac{2\pi }{3}\right);$$$$z^4={(cos\left(\frac{2\pi }{3}\right)+isin\left(\frac{2\pi }{3}\right))}^4=cos\left(\frac{8\pi }{3}\right)+isin\left(\frac{8\pi }{3}\right);$$$$z^4=cos\left(\frac{8\pi }{3}\right)+isin\left(\frac{8\pi }{3}\right);$$$$f_1\left(z\right)=z^4+2z;$$$$f_1\left(z\right)=cos\left(\frac{8\pi }{3}\right)+isin\left(\frac{8\pi }{3}\right)+2cos\left(\frac{2\pi }{3}\right)+2isin\left(\frac{2\pi }{3}\right);$$$$f_1\left(z\right)=(cos\left(\frac{8\pi }{3}\right)+2cos\left(\frac{2\pi }{3}\right))+i(sin\left(\frac{8\pi }{3}\right)+2sin\left(\frac{2\pi }{3}\right));$$$$z^4={(-\frac{1}{2}+i\frac{\sqrt{3}}{2})}^4;$$$${(a+b)}^4=\underset{n=0}{\sum ^4}{C}_4^n{a}^{4-n}{b}^n=C_4^0{a}^4+C_4^1{a}^{3}b+C_4^2{a}^2{b}^2+C_4^3{ab}^3+C_4^4{b}^4;$$$$C_4^0=C_4^4=1;$$$$C_4^1=C_3^1+C_3^0=C_2^1+C_2^0+1=C_1^1+C_1^0+2=4;$$$$C_4^2=C_3^2+C_3^1=C_2^2+C_2^1+C_2^1+C_2^0=1+2C_2^1+1=2+2(C_1^1+C_1^0)=6;$$$$C_4^3=C_4^{4-3}=C_4^1=4;$$$${(a+b)}^4=a^4+4a^{3}b+6a^2{b}^2+4{ab}^3+b^4;$$$$z^4={(-\frac{1}{2})}^4+4\cdot {(-\frac{1}{2})}^3\cdot \frac{\sqrt{3}}{2}\cdot i+6\cdot {(-\frac{1}{2})}^2\cdot {\left(\frac{\sqrt{3}}{2}\right)}^2\cdot i^2+4\cdot (-\frac{1}{2})\cdot {\left(\frac{\sqrt{3}}{2}\right)}^3\cdot i^3+{\left(\frac{\sqrt{3}}{2}\right)}^4\cdot i^4;$$$$i=\sqrt{-1}={(-1)}^{\frac{1}{2}};i^2=-1;i^3=i\cdot i^2=-i;i^4={(-1)}^2=1;$$$$z^4=\frac{1}{16}-\frac{\sqrt{3}}{4}i-\frac{9}{8}-\frac{\sqrt{3}}{4}i+\frac{9}{16}=-\frac{1}{2}-\frac{\sqrt{3}}{2}i;$$$$z^4=-\frac{1}{2}-\frac{\sqrt{3}}{2}i;$$$$2z=2(-\frac{1}{2}+\frac{\sqrt{3}}{2}i)=-1+\sqrt{3}i;$$$$f_1\left(z\right)=-\frac{1}{2}-\frac{\sqrt{3}}{2}i-1+\sqrt{3}i=-\frac{3}{2}+\frac{\sqrt{3}}{2}i;$$$$f_1\left(z\right)=-\frac{3}{2}+\frac{\sqrt{3}}{2}i;$$$$f_2\left(z\right)=z^3+z;$$$$z^3={(-\frac{1}{2}+\frac{\sqrt{3}}{2}i)}^3;$$$${(a+b)}^3=\underset{n=0}{\sum ^3}{C}_3^n{a}^{3-n}{b}^n=C_3^0{a}^3+C_3^1{a}^{2}b+C_3^2{ab}^2+C_3^3{b}^3;$$$$C_3^0=C_3^3=1;$$$$C_3^1=C_2^1+C_2^0=C_1^1+C_1^0+1=3;$$$$C_3^2=C_3^{3-2}=C_3^1=3;$$$${(a+b)}^3=a^3+3a^{2}b+3{ab}^2+b^3;$$$$z^3={(-\frac{1}{2})}^3+3{(-\frac{1}{2})}^2\frac{\sqrt{3}}{2}i+3(-\frac{1}{2}){\left(\frac{\sqrt{3}}{2}\right)}^2{i}^2+{\left(\frac{\sqrt{3}}{2}\right)}^3{i}^3;$$$$z^3=-\frac{1}{8}+\frac{3\sqrt{3}}{8}i+\frac{9}{8}-\frac{3\sqrt{3}}{8}i=1;$$$$z^3=1;$$$$f_2\left(z\right)=1-\frac{1}{2}+\frac{\sqrt{3}}{2}i=\frac{1}{2}+\frac{\sqrt{3}}{2}i;$$$$f\left(z\right)=f_1\left(z\right)f_2\left(z\right)=(-\frac{3}{2}+\frac{\sqrt{3}}{2}i)(\frac{1}{2}+\frac{\sqrt{3}}{2}i)=-\frac{3}{2}(\frac{1}{2}+\frac{\sqrt{3}}{2}i)+\frac{\sqrt{3}}{2}(\frac{1}{2}i-\frac{\sqrt{3}}{2});$$$$f\left(z\right)=-\frac{3}{4}-\frac{3\sqrt{3}}{4}i+\frac{\sqrt{3}}{4}i-\frac{\sqrt{3}}{4}=-\frac{\sqrt{3}}{2}+\frac{\sqrt{3}-3\sqrt{3}}{4}i;$$$$Ansver:(z^4+2z)(z^3+z)=-\frac{\sqrt{3}}{2}+\frac{\sqrt{3}-3\sqrt{3}}{4}i;$$

Answered by grigoriy last updated on 29/Jun/24

$$$$$$feedback:mail:grigoriybrewer@gmail.com;$$$$telegram(!):@wraith\underset{}{}machine;$$$$$$

Answered by mr W last updated on 29/Jun/24

$$z=-\frac{1}{2}+\frac{\sqrt{3}}{2}i=\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}=e^{\frac{2\pi }{3}i}$$$$(z^4+2z)(z^3+z)$$$$=z^7+z^5+2z^4+2z^2$$$$=e^{\frac{14\pi i}{3}}+e^{\frac{10\pi i}{3}}+2e^{\frac{8\pi i}{3}}+2e^{\frac{4\pi i}{3}}$$$$=e^{\frac{2\pi i}{3}}+e^{-\frac{2\pi i}{3}}+2e^{\frac{2\pi i}{3}}+2e^{-\frac{2\pi i}{3}}$$$$=3(e^{\frac{2\pi i}{3}}+e^{-\frac{2\pi i}{3}})$$$$=3(-\frac{1}{2}+\frac{\sqrt{3}i}{2}-\frac{1}{2}-\frac{\sqrt{3}i}{2})$$$$=-3✓$$

Answered by Rasheed.Sindhi last updated on 29/Jun/24

$$\mathbf{z}=-\frac{1}{2}+\frac{\sqrt{3}}{2}\mathbf{i}=\omega \left(cuberootofunity\right)$$$${\mathrm{z}}^3=1$$$$\Rightarrow {\mathrm{z}}^3-1=0$$$$\Rightarrow (\mathrm{z}-1)({\mathrm{z}}^2+\mathrm{z}+1)=0$$$$\Rightarrow \mathrm{z}\ne 1\Rightarrow {\mathrm{z}}^2+\mathrm{z}+1=0\Rightarrow {\mathrm{z}}^2+\mathrm{z}=-1$$$$▸({\mathrm{z}}^4+2\mathrm{z})\cdot ({\mathrm{z}}^3+\mathrm{z})=$$$$(\mathrm{z}+2\mathrm{z})\cdot (1+\mathrm{z})=$$$$3\mathrm{z}+{3\mathrm{z}}^2=3(\mathrm{z}+{\mathrm{z}}^2)=3(-1)=-3✓$$

Answered by Rasheed.Sindhi last updated on 29/Jun/24

$$\mathbf{z}=-\frac{1}{2}+\frac{\sqrt{3}}{2}\mathbf{i}$$$$2\mathrm{z}+1=\sqrt{3}\mathrm{i}$$$${4\mathrm{z}}^2+4\mathrm{z}+1=-3$$$${\mathrm{z}}^2+\mathrm{z}+1=0\Rightarrow {\mathrm{z}}^2+\mathrm{z}=-1$$$$(\mathrm{z}-1)({\mathrm{z}}^2+\mathrm{z}+1)=0\Rightarrow {\mathrm{z}}^3-1=0$$$$\Rightarrow {\mathrm{z}}^3=1$$$$▸({\mathrm{z}}^4+2\mathrm{z})\cdot ({\mathrm{z}}^3+\mathrm{z})$$$$=({\mathrm{z}}^3.\mathrm{z}+2\mathrm{z})(1+\mathrm{z})$$$$=(1.\mathrm{z}+2\mathrm{z})(1+\mathrm{z})$$$$=3\mathrm{z}(1+\mathrm{z})=3\mathrm{z}.-{\mathrm{z}}^2=-{3\mathrm{z}}^3=-3$$

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