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Question Number 70704 by 20190927 last updated on 07/Oct/19

z^4 −2z^2 +2=0

$$\mathrm{z}^{\mathrm{4}} −\mathrm{2z}^{\mathrm{2}} +\mathrm{2}=\mathrm{0} \\ $$

Commented by mathmax by abdo last updated on 08/Oct/19

z^4 −2z^2 +2 =0 ⇒t^2 −2t+2=0 with t=z^2 Δ^′ =1−2=−1 ⇒t_1 =1+i and t_2 =1−i t_1 =(√2)e^((iπ)/4) and t_2 =(√2)e^(−((iπ)/4)) z^2 =t_1 ⇒z=+^− (√t_1 )=+^− (^4 (√2))e^((iπ)/8) z^2 =t_2 ⇒z =+^− (√t_2 )=+^− (^4 (√2))e^(−((iπ)/8)) so the roots are +^− (^4 (√2))e^((iπ)/8) and +^− (^4 (√2))e^(−((iπ)/8)) also the factorization is z^4 −2z^2 +2 =(z−(^4 (√2))e^((iπ)/8) )(z+^4 (√2)e^((iπ)/8) )(z−^4 (√2)e^(−((iπ)/8)) )(z+^4 (√2)e^(−((iπ)/8)) )

$${z}^{\mathrm{4}} −\mathrm{2}{z}^{\mathrm{2}} +\mathrm{2}\:=\mathrm{0}\:\Rightarrow{t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{2}=\mathrm{0}\:\:{with}\:{t}={z}^{\mathrm{2}} \\ $$$$\Delta^{'} =\mathrm{1}−\mathrm{2}=−\mathrm{1}\:\Rightarrow{t}_{\mathrm{1}} =\mathrm{1}+{i}\:{and}\:{t}_{\mathrm{2}} =\mathrm{1}−{i} \\ $$$${t}_{\mathrm{1}} =\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \:\:{and}\:{t}_{\mathrm{2}} =\sqrt{\mathrm{2}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \: \\ $$$${z}^{\mathrm{2}} ={t}_{\mathrm{1}} \:\Rightarrow{z}=\overset{−} {+}\sqrt{{t}_{\mathrm{1}} }=\overset{−} {+}\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right){e}^{\frac{{i}\pi}{\mathrm{8}}} \\ $$$${z}^{\mathrm{2}} ={t}_{\mathrm{2}} \:\Rightarrow{z}\:=\overset{−} {+}\sqrt{{t}_{\mathrm{2}} }=\overset{−} {+}\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right){e}^{−\frac{{i}\pi}{\mathrm{8}}} \:{so}\:{the}\:{roots}\:{are}\:\overset{−} {+}\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right){e}^{\frac{{i}\pi}{\mathrm{8}}} \\ $$$${and}\:\overset{−} {+}\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right){e}^{−\frac{{i}\pi}{\mathrm{8}}} \:\:\:{also}\:{the}\:{factorization}\:{is} \\ $$$${z}^{\mathrm{4}} −\mathrm{2}{z}^{\mathrm{2}} \:+\mathrm{2}\:=\left({z}−\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right){e}^{\frac{{i}\pi}{\mathrm{8}}} \right)\left({z}+^{\mathrm{4}} \sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{8}}} \right)\left({z}−^{\mathrm{4}} \sqrt{\mathrm{2}}{e}^{−\frac{{i}\pi}{\mathrm{8}}} \right)\left({z}+^{\mathrm{4}} \sqrt{\mathrm{2}}{e}^{−\frac{{i}\pi}{\mathrm{8}}} \right) \\ $$

Commented by 20190927 last updated on 09/Oct/19

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

Answered by Rasheed.Sindhi last updated on 07/Oct/19

z^4 −2z^2 +2=0 (z^2 )^2 −2z^2 +2=0 z^2 =y⇒y^2 −2y+2=0 y=((−(−2)±(√((−2)^2 −4(1)(2))))/(2(1))) =((2±(√((4−8)))/2)=((2±(√(−4)))/2)=((2±2i)/2) y=1±i z^2 =1±i z=±(√(1±i))

$$\mathrm{z}^{\mathrm{4}} −\mathrm{2z}^{\mathrm{2}} +\mathrm{2}=\mathrm{0} \\ $$$$\left({z}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}{z}^{\mathrm{2}} +\mathrm{2}=\mathrm{0} \\ $$$${z}^{\mathrm{2}} ={y}\Rightarrow{y}^{\mathrm{2}} −\mathrm{2}{y}+\mathrm{2}=\mathrm{0} \\ $$$$\:\:\:{y}=\frac{−\left(−\mathrm{2}\right)\pm\sqrt{\left(−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}\right)\left(\mathrm{2}\right)}}{\mathrm{2}\left(\mathrm{1}\right)} \\ $$$$\:\:\:\:\:=\frac{\mathrm{2}\pm\sqrt{\left(\mathrm{4}−\mathrm{8}\right.}}{\mathrm{2}}=\frac{\mathrm{2}\pm\sqrt{−\mathrm{4}}}{\mathrm{2}}=\frac{\mathrm{2}\pm\mathrm{2}{i}}{\mathrm{2}} \\ $$$$\:\:{y}=\mathrm{1}\pm{i} \\ $$$$\:{z}^{\mathrm{2}} =\mathrm{1}\pm{i} \\ $$$${z}=\pm\sqrt{\mathrm{1}\pm{i}} \\ $$

Commented by 20190927 last updated on 09/Oct/19

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

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