Question Number 28190 by ajfour last updated on 21/Jan/18
Commented by ajfour last updated on 21/Jan/18
![Q. 28188 (solution) find least value of [∣z_1 +z_2 ∣]+1 if z_1 , z_2 , z_3 , z_4 are roots of z^4 +z^3 +z^2 +z+1=0 .](Q28191.png)
$${Q}.\:\mathrm{28188}\:\:\:\left({solution}\right) \\ $$$${find}\:{least}\:{value}\:{of}\:\left[\mid{z}_{\mathrm{1}} +{z}_{\mathrm{2}} \mid\right]+\mathrm{1} \\ $$$${if}\:\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} ,\:{z}_{\mathrm{3}} ,\:{z}_{\mathrm{4}} \:{are}\:{roots}\:{of} \\ $$$${z}^{\mathrm{4}} +{z}^{\mathrm{3}} +{z}^{\mathrm{2}} +{z}+\mathrm{1}=\mathrm{0}\:\:. \\ $$
Commented by abdo imad last updated on 21/Jan/18
![e)⇔ z^5 =1 and z≠1 the roots of the polynomial z^5 −1=0 are z_k = e^(i((2kπ)/5)) with k∈[[1 ,4]] if we take z_1 =e^(i((2π)/5)) and z_2 =^ e^(i((4π)/5)) we have z_2 = e^(i(π−(π/5))) =−z_1 ^− ⇒z_1 +z_2 =2i Im(z_1 ) = 2i sin( ((2π)/5)) and ∣z_1 +z_2 ∣= 2sin(2(π/5))and [∣z_1 +z_2 ∣]+1= [2sin(((2π)/(5))))]+1 but we know that cos((π/5))= ((1+(√5))/4) ⇒sin((π/5))=(√( 1−(((1+(√(5)^2 )))/(16)))) =((√(16−(6+2(√5))))/4)= ((√(10−2(√5)))/4) and sin(((2π)/5))=2sin((π/5))cos((π/5))=2 ((√(10−2(√5)))/4) ((1+(√5))/4) =(((1+(√(5)))(√(10−2(√5))))/8) ⇒2sin(((2π)/5))= (((1+(√5))(√(10−2(√5))))/4) = ((√((6+2(√5))(10−2(√5))))/4)=((√(60−12(√5)+20(√5)−20))/4) =((√(40+8(√5)))/4) =((√(10+2(√5)))/2)=(√((5/2) +((√5)/2)))=(√((5+(√5))/2)) ∼(√(3,5)) but we have 1<(√(3,5))<2 ⇒[(√(3,5)) ]=1 so [∣z_1 +z_2 ∣]+1=2 .](Q28204.png)
$$\left.{e}\right)\Leftrightarrow\:{z}^{\mathrm{5}} =\mathrm{1}\:{and}\:{z}\neq\mathrm{1}\:{the}\:{roots}\:{of}\:{the}\:{polynomial}\:{z}^{\mathrm{5}} −\mathrm{1}=\mathrm{0} \\ $$$${are}\:{z}_{{k}} =\:{e}^{{i}\frac{\mathrm{2}{k}\pi}{\mathrm{5}}} \:{with}\:{k}\in\left[\left[\mathrm{1}\:,\mathrm{4}\right]\right]\:{if}\:{we}\:{take}\:{z}_{\mathrm{1}} ={e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{5}}} \:{and} \\ $$$${z}_{\mathrm{2}} =^{} \:{e}^{{i}\frac{\mathrm{4}\pi}{\mathrm{5}}} \:{we}\:{have}\:\:{z}_{\mathrm{2}} =\:{e}^{{i}\left(\pi−\frac{\pi}{\mathrm{5}}\right)} =−{z}_{\mathrm{1}} ^{−} \:\:\Rightarrow{z}_{\mathrm{1}} \:+{z}_{\mathrm{2}} =\mathrm{2}{i}\:{Im}\left({z}_{\mathrm{1}} \right) \\ $$$$=\:\mathrm{2}{i}\:{sin}\left(\:\frac{\mathrm{2}\pi}{\mathrm{5}}\right)\:\:{and}\:\mid{z}_{\mathrm{1}} +{z}_{\mathrm{2}} \mid=\:\mathrm{2}{sin}\left(\mathrm{2}\frac{\pi}{\mathrm{5}}\right){and} \\ $$$$\left[\mid{z}_{\mathrm{1}} +{z}_{\mathrm{2}} \mid\right]+\mathrm{1}=\:\left[\mathrm{2}{sin}\left(\frac{\mathrm{2}\pi}{\left.\mathrm{5}\right)}\right)\right]+\mathrm{1}\:\:\:{but}\:{we}\:{know}\:{that} \\ $$$${cos}\left(\frac{\pi}{\mathrm{5}}\right)=\:\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}\:\:\Rightarrow{sin}\left(\frac{\pi}{\mathrm{5}}\right)=\sqrt{\:\mathrm{1}−\frac{\left(\mathrm{1}+\sqrt{\left.\mathrm{5}\right)^{\mathrm{2}} }\right.}{\mathrm{16}}} \\ $$$$=\frac{\sqrt{\mathrm{16}−\left(\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}\right)}}{\mathrm{4}}=\:\frac{\sqrt{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{4}}\:{and} \\ $$$${sin}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)=\mathrm{2}{sin}\left(\frac{\pi}{\mathrm{5}}\right){cos}\left(\frac{\pi}{\mathrm{5}}\right)=\mathrm{2}\:\frac{\sqrt{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{4}}\:\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$$=\frac{\left(\mathrm{1}+\sqrt{\left.\mathrm{5}\right)}\sqrt{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{5}}}\right.}{\mathrm{8}}\:\Rightarrow\mathrm{2}{sin}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)=\:\frac{\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)\sqrt{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{4}} \\ $$$$=\:\frac{\sqrt{\left(\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}\right)\left(\mathrm{10}−\mathrm{2}\sqrt{\mathrm{5}}\right)}}{\mathrm{4}}=\frac{\sqrt{\mathrm{60}−\mathrm{12}\sqrt{\mathrm{5}}+\mathrm{20}\sqrt{\mathrm{5}}−\mathrm{20}}}{\mathrm{4}} \\ $$$$=\frac{\sqrt{\mathrm{40}+\mathrm{8}\sqrt{\mathrm{5}}}}{\mathrm{4}}\:=\frac{\sqrt{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{2}}=\sqrt{\frac{\mathrm{5}}{\mathrm{2}}\:+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}}=\sqrt{\frac{\mathrm{5}+\sqrt{\mathrm{5}}}{\mathrm{2}}}\:\:\:\sim\sqrt{\mathrm{3},\mathrm{5}}\:\:{but} \\ $$$${we}\:{have}\:\:\:\:\mathrm{1}<\sqrt{\mathrm{3},\mathrm{5}}<\mathrm{2}\:\Rightarrow\left[\sqrt{\mathrm{3},\mathrm{5}}\:\right]=\mathrm{1}\:{so} \\ $$$$\left[\mid{z}_{\mathrm{1}} +{z}_{\mathrm{2}} \mid\right]+\mathrm{1}=\mathrm{2}\:\:. \\ $$
Answered by ajfour last updated on 21/Jan/18
![((1−z^5 )/(1−z))=0 so z_1 , z_2 , z_3 , z_4 , 1 are roots of z^5 =1 (fifth roots of unity) Least value of ∣z_1 +z_2 ∣=2cos 72° =2sin 18° < 2sin 30° (=1) So least value of [∣z_1 +z_2 ∣]+1=1 .](Q28192.png)
$$\frac{\mathrm{1}−{z}^{\mathrm{5}} }{\mathrm{1}−{z}}=\mathrm{0} \\ $$$${so}\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} ,\:{z}_{\mathrm{3}} ,\:{z}_{\mathrm{4}} ,\:\mathrm{1}\:{are}\:{roots}\:{of} \\ $$$${z}^{\mathrm{5}} =\mathrm{1}\:\:\:\left({fifth}\:{roots}\:{of}\:{unity}\right) \\ $$$${Least}\:{value}\:{of}\:\mid{z}_{\mathrm{1}} +{z}_{\mathrm{2}} \mid=\mathrm{2cos}\:\mathrm{72}° \\ $$$$\:\:\:=\mathrm{2sin}\:\mathrm{18}°\:<\:\mathrm{2sin}\:\mathrm{30}°\:\:\:\left(=\mathrm{1}\right) \\ $$$${So}\:{least}\:{value}\:{of}\:\left[\mid{z}_{\mathrm{1}} +{z}_{\mathrm{2}} \mid\right]+\mathrm{1}=\mathrm{1}\:. \\ $$$$ \\ $$
Commented by ajfour last updated on 24/Jan/18
$${German}\:{or}\:{Indian}\:{Sir}\:? \\ $$
Commented by math solver last updated on 24/Jan/18
$${thank}\:{you}\:{so}\:{much}\:{sir}\:! \\ $$
Commented by math solver last updated on 22/Jan/18
$$\mathrm{2}{cos}\:\mathrm{72}\:°{kaise}\:{aayi}\:{least}\:{value}\:? \\ $$
Commented by ajfour last updated on 22/Jan/18
$${to}\:{get}\:{least}\:{value}\:{of}\:\mid{z}_{\mathrm{1}} +{z}_{\mathrm{2}} \mid \\ $$$${we}\:{have}\:{to}\:{choose}\:\left({see}\:{diagram}\right) \\ $$$${such}\:{z}_{{i}} \:{and}\:{z}_{{j}} \:{that}\:{the}\:{manitude} \\ $$$${of}\:{their}\:{resultant}\:{is}\:{least} \\ $$$${as}\:{closer}\:{to}\:\mathrm{180}°\:{as}\:{can}\:{be}. \\ $$$${here}\:{we}\:{can}\:{only}\:{have}\:\mathrm{144}°\:. \\ $$$${then}\:\:\:\mid{z}_{\mathrm{1}} +{z}_{\mathrm{3}} \mid=\mid{z}_{\mathrm{1}} +{z}_{\mathrm{4}} \mid=\mid{z}_{\mathrm{2}} +{z}_{\mathrm{4}} \mid \\ $$$${all}\:{equal}\:\:\mathrm{2cos}\:\mathrm{72}°\:. \\ $$$${other}\:{pairs}\:{of}\:{z}_{{i}} ,\:{z}_{{j}} \:{yield} \\ $$$${greater}\:{values}\:{of}\:\mid{z}_{{i}} +{z}_{{j}} \mid\:. \\ $$
Commented by math solver last updated on 22/Jan/18
$${sorry}\:{sir}\:,\:{i}\:{m}\:{still}\:{not}\:{getting} \\ $$$$\mathrm{2}{cos}\mathrm{72}°\:. \\ $$
Commented by mrW2 last updated on 23/Jan/18
Commented by math solver last updated on 24/Jan/18
$${hey}\:,\:{r}\:{u}\:{mrw}\mathrm{1}\:{sir}\:? \\ $$$${the}\:{one}\:{from}\:{germany}. \\ $$
Commented by mrW2 last updated on 24/Jan/18
$${yes}\:{sir}. \\ $$$${mrW}\mathrm{2}={mrW}\mathrm{1}={mrW} \\ $$
Commented by mrW2 last updated on 24/Jan/18
$${I}'{m}\:{from}\:{Germany}\:{Sir}. \\ $$
Commented by ajfour last updated on 24/Jan/18
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Commented by mrW2 last updated on 24/Jan/18
$${Nice}\:{to}\:{know}\:{that}\:{Sir}. \\ $$
Commented by abdo imad last updated on 25/Jan/18
$${are}\:{you}\:{indian}\:{sir}\:{ajfour}\:?\:\:{me}\:\:{i}\:{am}\:{from}\:{morocco}... \\ $$
Commented by math solver last updated on 25/Jan/18
$${yes}\:,\:{he}\:{is}\:{from}\:{india}. \\ $$$$\left.{P}.{S}:\:{i}\:{am}\:{also}\:{an}\:{indian}\::\right) \\ $$
Commented by ajfour last updated on 25/Jan/18
$${never}\:{outside}\:{India}\:\left({physically}\right). \\ $$
Commented by abdo imad last updated on 25/Jan/18
$${nice}\:{to}\:{know}\:{you}\:{ajfour}\:,{p}.{s}.\:{prakash}\:{jain}\:.{mrw}....{wish}\:{you}\: \\ $$$${you}\:{good}\:{luck}.... \\ $$
Commented by mrW2 last updated on 25/Jan/18
$${Thanks}\:{Sir}! \\ $$