Question Number 155572 by peter frank last updated on 02/Oct/21
$$\mathrm{Find}\:\mathrm{Z}^{\mathrm{4}} =\mathrm{1}. \\ $$$$\mathrm{Hence}\:\mathrm{show}\:\mathrm{that}\:\mathrm{1}+\mathrm{w}+\mathrm{w}^{\mathrm{2}} +\mathrm{w}^{\mathrm{3}} =\mathrm{0} \\ $$
Answered by puissant last updated on 02/Oct/21
![z^4 =1 ⇒ z=e^(i((2kπ)/4)) ; k∈[∣0;3∣].. → k=0 ⇒ z=1.. → k=1 ⇒ z=e^(i(π/2)) = i.. → k=2 ⇒ z=e^(i((4π)/4)) = e^(iπ) =−1.. → k=3 ⇒z=e^(i((3π)/2)) = −i.. ∵ z ∈ {−1 ; −i ; i ; 1 }.. ★... 1+w+w^2 +w^3 =((1−w^4 )/(1−w))= ((w^4 −1)/(w−1)) w is a solution, we have w^4 = 1 ⇒ w^4 −1=0 ⇒((w^4 −1)/(w−1))=0 ⇒ 1+w+w^2 +w^3 =0..](https://www.tinkutara.com/question/Q155581.png)
$${z}^{\mathrm{4}} =\mathrm{1}\:\:\Rightarrow\:\:{z}={e}^{{i}\frac{\mathrm{2}{k}\pi}{\mathrm{4}}} \:;\:\:{k}\in\left[\mid\mathrm{0};\mathrm{3}\mid\right].. \\ $$$$\:\rightarrow\:{k}=\mathrm{0}\:\Rightarrow\:{z}=\mathrm{1}.. \\ $$$$\rightarrow\:{k}=\mathrm{1}\:\Rightarrow\:{z}={e}^{{i}\frac{\pi}{\mathrm{2}}} =\:{i}.. \\ $$$$\rightarrow\:{k}=\mathrm{2}\:\Rightarrow\:{z}={e}^{{i}\frac{\mathrm{4}\pi}{\mathrm{4}}} =\:{e}^{{i}\pi} =−\mathrm{1}.. \\ $$$$\rightarrow\:{k}=\mathrm{3}\:\Rightarrow{z}={e}^{{i}\frac{\mathrm{3}\pi}{\mathrm{2}}} =\:−{i}.. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\because\:{z}\:\in\:\left\{−\mathrm{1}\:;\:−{i}\:;\:{i}\:;\:\mathrm{1}\:\right\}.. \\ $$$$ \\ $$$$\:\:\:\bigstar…\:\mathrm{1}+{w}+{w}^{\mathrm{2}} +{w}^{\mathrm{3}} =\frac{\mathrm{1}−{w}^{\mathrm{4}} }{\mathrm{1}−{w}}=\:\frac{{w}^{\mathrm{4}} −\mathrm{1}}{{w}−\mathrm{1}} \\ $$$${w}\:{is}\:{a}\:{solution},\:{we}\:{have}\:{w}^{\mathrm{4}} =\:\mathrm{1} \\ $$$$\Rightarrow\:{w}^{\mathrm{4}} −\mathrm{1}=\mathrm{0}\: \\ $$$$\Rightarrow\frac{{w}^{\mathrm{4}} −\mathrm{1}}{{w}−\mathrm{1}}=\mathrm{0}\:\Rightarrow\:\mathrm{1}+{w}+{w}^{\mathrm{2}} +{w}^{\mathrm{3}} =\mathrm{0}.. \\ $$
Commented by peter frank last updated on 02/Oct/21
$$\mathrm{great}\:\mathrm{sir}\:.\mathrm{thanks} \\ $$
Answered by Rasheed.Sindhi last updated on 02/Oct/21
$$\mathrm{z}^{\mathrm{4}} −\mathrm{1}=\mathrm{0} \\ $$$$\left(\mathrm{z}−\mathrm{1}\right)\left(\mathrm{z}+\mathrm{1}\right)\left(\mathrm{z}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathrm{z}=\mathrm{1},\mathrm{z}=−\mathrm{1},\mathrm{z}=\pm{i} \\ $$$$\omega={i}\:\left({say}\right) \\ $$$$\omega^{\mathrm{2}} ={i}^{\mathrm{2}} =−\mathrm{1} \\ $$$$\omega^{\mathrm{3}} ={i}^{\mathrm{3}} ={i}^{\mathrm{2}} .{i}=−{i} \\ $$$$\omega^{\mathrm{4}} ={i}^{\mathrm{4}} ={i}^{\mathrm{2}} .{i}^{\mathrm{2}} =−\mathrm{1}.−\mathrm{1}=\mathrm{1} \\ $$$$\mathrm{1}+\omega+\omega^{\mathrm{2}} +\omega^{\mathrm{3}} =\mathrm{1}+{i}+\left(−\mathrm{1}\right)+\left(−{i}\right)=\mathrm{0} \\ $$
Commented by peter frank last updated on 02/Oct/21
$$\mathrm{thank}\:\mathrm{you} \\ $$