Menu Close

Question-120389

Tinku Tara 0 Comments
Pin




Question Number 120389 by help last updated on 31/Oct/20
Commented by help last updated on 31/Oct/20
question 2a
$${question}\:\mathrm{2}{a} \\ $$
Commented by help last updated on 31/Oct/20
question 3a pls
$${question}\:\mathrm{3}{a}\:{pls} \\ $$
Answered by Dwaipayan Shikari last updated on 31/Oct/20
x^2 −((√2)+i+(√2)−i)x+((√2)+i)((√2)−i)=0 x^2 −2(√2)x+3=0
$${x}^{\mathrm{2}} −\left(\sqrt{\mathrm{2}}+{i}+\sqrt{\mathrm{2}}−{i}\right){x}+\left(\sqrt{\mathrm{2}}+{i}\right)\left(\sqrt{\mathrm{2}}−{i}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{2}}{x}+\mathrm{3}=\mathrm{0} \\ $$
Answered by bramlexs22 last updated on 31/Oct/20
(z−((√2)+i))(z−((√2)−i))=0 ⇒z^2 −((√2)−i)z−((√2)+i)z+((√2)+i)((√2)−i)=0 z^2 −2(√2) z+2+1=0 z^2 −2(√2)z+3 = 0
$$\left({z}−\left(\sqrt{\mathrm{2}}+{i}\right)\right)\left({z}−\left(\sqrt{\mathrm{2}}−{i}\right)\right)=\mathrm{0} \\ $$$$\Rightarrow{z}^{\mathrm{2}} −\left(\sqrt{\mathrm{2}}−{i}\right){z}−\left(\sqrt{\mathrm{2}}+{i}\right){z}+\left(\sqrt{\mathrm{2}}+{i}\right)\left(\sqrt{\mathrm{2}}−{i}\right)=\mathrm{0} \\ $$$$\:{z}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{2}}\:{z}+\mathrm{2}+\mathrm{1}=\mathrm{0} \\ $$$$\:{z}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{2}}{z}+\mathrm{3}\:=\:\mathrm{0} \\ $$
Answered by ebi last updated on 31/Oct/20
3a let f(z)=z^3 −4z^2 +7z−6 finding the factor of f(z) f(1)=1^3 −4(1)^2 +7(1)−6=−2≠0 ∴ (z−1) is not a factor f(2)=2^3 −4(2)^2 +7(2)−6=0 ∴ (z−2) is a factor ((z^3 −4z^2 +7z−6)/(z−2)) /long division z^2 −2z+3 z−2 (√(z^3 −4z^2 +7z−6)) ((−(z^3 −2z^2 ))/( −2z^2 +7z)) (( −(−2z^2 +4z))/( 3z−6)) (( −(3z−6))/( 0)) f(z)=(z−2)(z^2 −2z+3)=0 z−2=0 → z=2 (real root) z^2 −2z+3=0 z^2 −2z+(−1)^2 =(−1)^2 −3 (z−1)^2 =−2 z−1=±i(√2), i=(√(−1)) z=1±i(√2) z=1+i(√2) or z=1−i(√2) (complex root) ∴ the roots of f(z) are z=2, z=1+i(√2), z=1−i(√2)
$$\mathrm{3}{a} \\ $$$${let}\:{f}\left({z}\right)={z}^{\mathrm{3}} −\mathrm{4}{z}^{\mathrm{2}} +\mathrm{7}{z}−\mathrm{6} \\ $$$$ \\ $$$${finding}\:{the}\:{factor}\:{of}\:{f}\left({z}\right) \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{1}^{\mathrm{3}} −\mathrm{4}\left(\mathrm{1}\right)^{\mathrm{2}} +\mathrm{7}\left(\mathrm{1}\right)−\mathrm{6}=−\mathrm{2}\neq\mathrm{0} \\ $$$$\therefore\:\left({z}−\mathrm{1}\right)\:{is}\:{not}\:{a}\:{factor} \\ $$$${f}\left(\mathrm{2}\right)=\mathrm{2}^{\mathrm{3}} −\mathrm{4}\left(\mathrm{2}\right)^{\mathrm{2}} +\mathrm{7}\left(\mathrm{2}\right)−\mathrm{6}=\mathrm{0} \\ $$$$\therefore\:\left({z}−\mathrm{2}\right)\:{is}\:{a}\:{factor} \\ $$$$ \\ $$$$\frac{{z}^{\mathrm{3}} −\mathrm{4}{z}^{\mathrm{2}} +\mathrm{7}{z}−\mathrm{6}}{{z}−\mathrm{2}}\:/{long}\:{division} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{z}^{\mathrm{2}} −\mathrm{2}{z}+\mathrm{3} \\ $$$${z}−\mathrm{2}\:\sqrt{{z}^{\mathrm{3}} −\mathrm{4}{z}^{\mathrm{2}} +\mathrm{7}{z}−\mathrm{6}} \\ $$$$\frac{−\left({z}^{\mathrm{3}} −\mathrm{2}{z}^{\mathrm{2}} \right)}{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{2}{z}^{\mathrm{2}} +\mathrm{7}{z}} \\ $$$$\frac{\:\:\:\:\:\:−\left(−\mathrm{2}{z}^{\mathrm{2}} +\mathrm{4}{z}\right)}{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}{z}−\mathrm{6}} \\ $$$$\frac{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\left(\mathrm{3}{z}−\mathrm{6}\right)}{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}} \\ $$$${f}\left({z}\right)=\left({z}−\mathrm{2}\right)\left({z}^{\mathrm{2}} −\mathrm{2}{z}+\mathrm{3}\right)=\mathrm{0} \\ $$$${z}−\mathrm{2}=\mathrm{0}\:\:\rightarrow\:\:{z}=\mathrm{2}\:\left({real}\:{root}\right) \\ $$$${z}^{\mathrm{2}} −\mathrm{2}{z}+\mathrm{3}=\mathrm{0} \\ $$$${z}^{\mathrm{2}} −\mathrm{2}{z}+\left(−\mathrm{1}\right)^{\mathrm{2}} =\left(−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{3} \\ $$$$\left({z}−\mathrm{1}\right)^{\mathrm{2}} =−\mathrm{2} \\ $$$${z}−\mathrm{1}=\pm{i}\sqrt{\mathrm{2}},\:\:{i}=\sqrt{−\mathrm{1}} \\ $$$${z}=\mathrm{1}\pm{i}\sqrt{\mathrm{2}} \\ $$$${z}=\mathrm{1}+{i}\sqrt{\mathrm{2}}\:{or}\:{z}=\mathrm{1}−{i}\sqrt{\mathrm{2}}\:\left({complex}\:{root}\right) \\ $$$$ \\ $$$$\therefore\:{the}\:{roots}\:{of}\:{f}\left({z}\right)\:{are}\: \\ $$$${z}=\mathrm{2},\:{z}=\mathrm{1}+{i}\sqrt{\mathrm{2}},\:{z}=\mathrm{1}−{i}\sqrt{\mathrm{2}} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *