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Question-46383

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Question Number 46383 by peter frank last updated on 24/Oct/18
Commented by maxmathsup by imad last updated on 24/Oct/18
z=0 is not solution for z≠0 (e) ⇔ (((z+1)/z))^n =1 ⇔(1+z^(−1) )^n =1 the roots of Z^n =1 are the compolex Z_k =e^(i((2kπ)/n)) with k∈[[1,n−1]] so the solutions for the equation are z_k / 1+z_k ^(−1) =Z_k ⇒ ((z_k +1)/z_k )=Z_k ⇒ z_k +1 =Z_k z_k ⇒ (1−Z_k )z_k =−1 ⇒z_k =−(1/(1−Z_k )) =−(1/(1−cos(((2kπ)/n))−isin(((2kπ)/n)))) = ((−1)/(2sin^2 (((kπ)/n))−2i sin(((kπ)/n))cos(((kπ)/n)))) = ((−1)/(−2i sin(((kπ)/n))e^(i((kπ)/n)) )) = ((−i)/(2sin(((kπ)/n)))) e^(−((ikπ)/n)) =((−i)/(2sin(((kπ)/n)))){cos(((kπ)/n))−isin(((kπ)/n))} =−(1/2) −(i/2) cotan(((kπ)/n)) =−(1/2)(1+cotan(((kπ)/n))} with k∈[[1,n−1]] .
$${z}=\mathrm{0}\:{is}\:{not}\:{solution}\:{for}\:{z}\neq\mathrm{0}\:\:\left({e}\right)\:\Leftrightarrow\:\left(\frac{{z}+\mathrm{1}}{{z}}\right)^{{n}} =\mathrm{1}\:\Leftrightarrow\left(\mathrm{1}+{z}^{−\mathrm{1}} \right)^{{n}} =\mathrm{1}\:\:{the}\:{roots}\:{of} \\ $$$${Z}^{{n}} =\mathrm{1}\:{are}\:{the}\:{compolex}\:{Z}_{{k}} ={e}^{{i}\frac{\mathrm{2}{k}\pi}{{n}}} \:\:{with}\:{k}\in\left[\left[\mathrm{1},{n}−\mathrm{1}\right]\right]\:{so}\:{the}\:{solutions}\:{for} \\ $$$${the}\:{equation}\:{are}\:{z}_{{k}} \:/\:\:\:\:\:\mathrm{1}+{z}_{{k}} ^{−\mathrm{1}} \:\:={Z}_{{k}} \:\Rightarrow\:\frac{{z}_{{k}} +\mathrm{1}}{{z}_{{k}} }={Z}_{{k}} \:\Rightarrow \\ $$$${z}_{{k}} +\mathrm{1}\:={Z}_{{k}} {z}_{{k}} \:\Rightarrow\:\left(\mathrm{1}−{Z}_{{k}} \right){z}_{{k}} =−\mathrm{1}\:\Rightarrow{z}_{{k}} =−\frac{\mathrm{1}}{\mathrm{1}−{Z}_{{k}} }\:=−\frac{\mathrm{1}}{\mathrm{1}−{cos}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)−{isin}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)} \\ $$$$=\:\frac{−\mathrm{1}}{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{k}\pi}{{n}}\right)−\mathrm{2}{i}\:{sin}\left(\frac{{k}\pi}{{n}}\right){cos}\left(\frac{{k}\pi}{{n}}\right)}\:=\:\frac{−\mathrm{1}}{−\mathrm{2}{i}\:{sin}\left(\frac{{k}\pi}{{n}}\right){e}^{{i}\frac{{k}\pi}{{n}}} } \\ $$$$=\:\frac{−{i}}{\mathrm{2}{sin}\left(\frac{{k}\pi}{{n}}\right)}\:{e}^{−\frac{{ik}\pi}{{n}}} \:\:=\frac{−{i}}{\mathrm{2}{sin}\left(\frac{{k}\pi}{{n}}\right)}\left\{{cos}\left(\frac{{k}\pi}{{n}}\right)−{isin}\left(\frac{{k}\pi}{{n}}\right)\right\} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\:−\frac{{i}}{\mathrm{2}}\:{cotan}\left(\frac{{k}\pi}{{n}}\right)\:=−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+{cotan}\left(\frac{{k}\pi}{{n}}\right)\right\}\:{with}\:{k}\in\left[\left[\mathrm{1},{n}−\mathrm{1}\right]\right]\:. \\ $$
Commented by maxmathsup by imad last updated on 24/Oct/18
2 ) (z−1)^n +z^n =0 ⇔ (z−1)^n =−z^n ⇔ (((z−1)/z))^n =−1 =e^(i(2k+1)π) ⇒ ((z_k −1)/z_k ) = e^(i(((2k+1)π)/n)) ⇒ 1−z_k ^(−1) = e^(i((2k+1)/n)π) ⇒ 1−e^(i(((2k+1)π)/n)) =z_k ^(−1) ⇒ z_k = (1/(1−cos((((2k+1)π)/n))−i sin((((2k+1)π)/n))cos((((2k+1)π)/n)))) = (1/(2sin^2 ((((2k+1)π)/(2n)))−2i sin((((2k+1)π)/(2n)))cos((((2k+1)π)/(2n))))) =(1/(−2i sin((((2k+1)π)/(2n))) e^(i(((2k+1)/(2n))π)) )) = (i/(2sin((((2k+1)π)/(2n))))){ cos((((2k+1)π)/(2n)))−isin((((2k+1)π)/(2n)))} =(1/2) +(i/2) cotan((((2k+1)π)/(2n))) with k∈[[0,n−1]]
$$\left.\mathrm{2}\:\right)\:\left({z}−\mathrm{1}\right)^{{n}} \:+{z}^{{n}} \:=\mathrm{0}\:\Leftrightarrow\:\left({z}−\mathrm{1}\right)^{{n}} \:=−{z}^{{n}} \:\Leftrightarrow\:\left(\frac{{z}−\mathrm{1}}{{z}}\right)^{{n}} \:=−\mathrm{1}\:\:={e}^{{i}\left(\mathrm{2}{k}+\mathrm{1}\right)\pi} \:\Rightarrow\: \\ $$$$\frac{{z}_{{k}} −\mathrm{1}}{{z}_{{k}} }\:=\:{e}^{{i}\left(\frac{\left.\mathrm{2}{k}+\mathrm{1}\right)\pi}{{n}}\right.} \:\Rightarrow\:\mathrm{1}−{z}_{{k}} ^{−\mathrm{1}} \:=\:{e}^{{i}\frac{\mathrm{2}{k}+\mathrm{1}}{{n}}\pi} \:\Rightarrow\:\mathrm{1}−{e}^{{i}\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{{n}}} \:={z}_{{k}} ^{−\mathrm{1}} \:\Rightarrow \\ $$$${z}_{{k}} =\:\frac{\mathrm{1}}{\mathrm{1}−{cos}\left(\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{{n}}\right)−{i}\:{sin}\left(\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{{n}}\right){cos}\left(\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{{n}}\right)} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{2}{n}}\right)−\mathrm{2}{i}\:{sin}\left(\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{2}{n}}\right){cos}\left(\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{2}{n}}\right)} \\ $$$$=\frac{\mathrm{1}}{−\mathrm{2}{i}\:{sin}\left(\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{2}{n}}\right)\:{e}^{{i}\left(\frac{\mathrm{2}{k}+\mathrm{1}}{\mathrm{2}{n}}\pi\right)} } \\ $$$$=\:\frac{{i}}{\mathrm{2}{sin}\left(\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{2}{n}}\right)}\left\{\:{cos}\left(\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{2}{n}}\right)−{isin}\left(\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{2}{n}}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{{i}}{\mathrm{2}}\:{cotan}\left(\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{2}{n}}\right)\:\:{with}\:\:{k}\in\left[\left[\mathrm{0},{n}−\mathrm{1}\right]\right] \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 24/Oct/18
the roots of (z−1)^5 +z^5 =0 are z_k =(1/2){1+i cotan((((2k+1)π)/(10))) with k∈[[0,4]] z_0 =(1/2){1+icotan((π/(10)))} , z_1 =(1/2){1+i cotan(((3π)/(10)))} z_2 =(1/2){1+i cotan((π/2))}dont exist and not solution z_3 =(1/2){1+i cotan(((7π)/(10)))} z_4 =(1/2){1+i cotan(((9π)/(10)))} also z =(1/2) is solution when cotan((((2k+1)π)/(10)))=0
$${the}\:{roots}\:{of}\:\left({z}−\mathrm{1}\right)^{\mathrm{5}} \:+{z}^{\mathrm{5}} =\mathrm{0}\:{are}\:{z}_{{k}} =\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{1}+{i}\:{cotan}\left(\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{10}}\right)\:\:\:{with}\:{k}\in\left[\left[\mathrm{0},\mathrm{4}\right]\right]\right. \\ $$$${z}_{\mathrm{0}} =\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{1}+{icotan}\left(\frac{\pi}{\mathrm{10}}\right)\right\}\:\:\:,\:{z}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{1}+{i}\:{cotan}\left(\frac{\mathrm{3}\pi}{\mathrm{10}}\right)\right\} \\ $$$${z}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{1}+{i}\:{cotan}\left(\frac{\pi}{\mathrm{2}}\right)\right\}{dont}\:{exist}\:{and}\:{not}\:{solution} \\ $$$${z}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{1}+{i}\:{cotan}\left(\frac{\mathrm{7}\pi}{\mathrm{10}}\right)\right\}\:\:\:{z}_{\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{1}+{i}\:{cotan}\left(\frac{\mathrm{9}\pi}{\mathrm{10}}\right)\right\} \\ $$$${also}\:{z}\:=\frac{\mathrm{1}}{\mathrm{2}}\:{is}\:{solution}\:{when}\:{cotan}\left(\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{10}}\right)=\mathrm{0} \\ $$
Answered by Smail last updated on 25/Oct/18
a) (z+1)^n =z^n z+1=ze^((2ikπ)/n) z(e^((2ikπ)/n) −1)=1 z=(1/(e^((2ikπ)/n) −1))=(1/(cos(((2kπ)/n))−1+isin(((2ikπ)/n)))) =(1/(2cos^2 (((kπ)/n))−1−1+i(2sin(((kπ)/n))cos(((kπ)/n))))) =(1/(2(cos^2 (((kπ)/n))−1+isin(((kπ)/n))cos(((kπ)/n))))) =−(1/(2(sin^2 (((kπ)/n))−isin(((kπ)/n))cos(((kπ)/n))))) =−(1/(2sin(((kπ)/n))(sin(((kπ)/n))−icos(((kπ)/n))))) =−((sin(((kπ)/n))+icos(((kπ)/n)))/(2sin(((kπ)/n))))=((−1)/2)(1+icot(((kπ)/n))) z=−(1/2)(1+icot(((kπ)/n))) b) (z−1)^n =−z^n =z^n e^(i(π+2kπ)) z−1=ze^(iπ((2k+1)/n)) ⇔z(e^(iπ((2k+1)/n)) −1)=−1 z=((−1)/(cos(((2k+1)/n)π)−1+isin(((2k+1)/n)π))) =−((cos(((2k+1)/n)π)−1−isin(((2k+1)/n)π))/((cos(((2k+1)/n)π)−1)^2 +sin^2 (((2k+1)/n)π))) =−((cos(((2k+1)/n)π)−1−isin(((2k+1)/n)π))/(cos^2 (((2k+1)/n)π)+1−2cos(((2k+1)/n)π)+sin^2 (((2k+1)/n)π))) =−((cos(((2k+1)/n)π)−1−isin(((2k+1)/n)π))/(2−2cos(((2k+1)/n)π))) =(1/2)(1+i((sin(((2k+1)/n)π))/(1−cos(((2k+1)/n)π)))) =(1/2)(1+i((2sin(((2k+1)/(2n))π)cos(((2k+1)/(2n))π))/(1−2cos^2 (((2k+1)/(2n))π)+1))) =(1/2)(1+i((sin(((2k+1)/(2n))π)cos(((2k+1)/(2n))π))/(1−cos^2 (((2k+1)/(2n))π)))) =(1/2)(1−i((sin(((2k+1)/(2n))π)cos(((2k+1)/(2n))π))/(sin^2 (((2k+1)/(2n))π)))) z=(1/2)(1−icot(((2k+1)/(2n))π)) When n=5 z_k =(1/2)(1−icot(((2k+1)/(10))π) with k=(0,1,2,3,4)
$$\left.{a}\right) \\ $$$$\left({z}+\mathrm{1}\right)^{{n}} ={z}^{{n}} \\ $$$${z}+\mathrm{1}={ze}^{\frac{\mathrm{2}{ik}\pi}{{n}}} \\ $$$${z}\left({e}^{\frac{\mathrm{2}{ik}\pi}{{n}}} −\mathrm{1}\right)=\mathrm{1} \\ $$$${z}=\frac{\mathrm{1}}{{e}^{\frac{\mathrm{2}{ik}\pi}{{n}}} −\mathrm{1}}=\frac{\mathrm{1}}{{cos}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)−\mathrm{1}+{isin}\left(\frac{\mathrm{2}{ik}\pi}{{n}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{{k}\pi}{{n}}\right)−\mathrm{1}−\mathrm{1}+{i}\left(\mathrm{2}{sin}\left(\frac{{k}\pi}{{n}}\right){cos}\left(\frac{{k}\pi}{{n}}\right)\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\left({cos}^{\mathrm{2}} \left(\frac{{k}\pi}{{n}}\right)−\mathrm{1}+{isin}\left(\frac{{k}\pi}{{n}}\right){cos}\left(\frac{{k}\pi}{{n}}\right)\right)} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}\left({sin}^{\mathrm{2}} \left(\frac{{k}\pi}{{n}}\right)−{isin}\left(\frac{{k}\pi}{{n}}\right){cos}\left(\frac{{k}\pi}{{n}}\right)\right)} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}{sin}\left(\frac{{k}\pi}{{n}}\right)\left({sin}\left(\frac{{k}\pi}{{n}}\right)−{icos}\left(\frac{{k}\pi}{{n}}\right)\right)} \\ $$$$=−\frac{{sin}\left(\frac{{k}\pi}{{n}}\right)+{icos}\left(\frac{{k}\pi}{{n}}\right)}{\mathrm{2}{sin}\left(\frac{{k}\pi}{{n}}\right)}=\frac{−\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+{icot}\left(\frac{{k}\pi}{{n}}\right)\right) \\ $$$${z}=−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+{icot}\left(\frac{{k}\pi}{{n}}\right)\right) \\ $$$$\left.{b}\right)\:\:\:\left({z}−\mathrm{1}\right)^{{n}} =−{z}^{{n}} ={z}^{{n}} {e}^{{i}\left(\pi+\mathrm{2}{k}\pi\right)} \\ $$$${z}−\mathrm{1}={ze}^{{i}\pi\frac{\mathrm{2}{k}+\mathrm{1}}{{n}}} \Leftrightarrow{z}\left({e}^{{i}\pi\frac{\mathrm{2}{k}+\mathrm{1}}{{n}}} −\mathrm{1}\right)=−\mathrm{1} \\ $$$${z}=\frac{−\mathrm{1}}{{cos}\left(\frac{\mathrm{2}{k}+\mathrm{1}}{{n}}\pi\right)−\mathrm{1}+{isin}\left(\frac{\mathrm{2}{k}+\mathrm{1}}{{n}}\pi\right)} \\ $$$$=−\frac{{cos}\left(\frac{\mathrm{2}{k}+\mathrm{1}}{{n}}\pi\right)−\mathrm{1}−{isin}\left(\frac{\mathrm{2}{k}+\mathrm{1}}{{n}}\pi\right)}{\left({cos}\left(\frac{\mathrm{2}{k}+\mathrm{1}}{{n}}\pi\right)−\mathrm{1}\right)^{\mathrm{2}} +{sin}^{\mathrm{2}} \left(\frac{\mathrm{2}{k}+\mathrm{1}}{{n}}\pi\right)} \\ $$$$=−\frac{{cos}\left(\frac{\mathrm{2}{k}+\mathrm{1}}{{n}}\pi\right)−\mathrm{1}−{isin}\left(\frac{\mathrm{2}{k}+\mathrm{1}}{{n}}\pi\right)}{{cos}^{\mathrm{2}} \left(\frac{\mathrm{2}{k}+\mathrm{1}}{{n}}\pi\right)+\mathrm{1}−\mathrm{2}{cos}\left(\frac{\mathrm{2}{k}+\mathrm{1}}{{n}}\pi\right)+{sin}^{\mathrm{2}} \left(\frac{\mathrm{2}{k}+\mathrm{1}}{{n}}\pi\right)} \\ $$$$=−\frac{{cos}\left(\frac{\mathrm{2}{k}+\mathrm{1}}{{n}}\pi\right)−\mathrm{1}−{isin}\left(\frac{\mathrm{2}{k}+\mathrm{1}}{{n}}\pi\right)}{\mathrm{2}−\mathrm{2}{cos}\left(\frac{\mathrm{2}{k}+\mathrm{1}}{{n}}\pi\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+{i}\frac{{sin}\left(\frac{\mathrm{2}{k}+\mathrm{1}}{{n}}\pi\right)}{\mathrm{1}−{cos}\left(\frac{\mathrm{2}{k}+\mathrm{1}}{{n}}\pi\right)}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+{i}\frac{\mathrm{2}{sin}\left(\frac{\mathrm{2}{k}+\mathrm{1}}{\mathrm{2}{n}}\pi\right){cos}\left(\frac{\mathrm{2}{k}+\mathrm{1}}{\mathrm{2}{n}}\pi\right)}{\mathrm{1}−\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{\mathrm{2}{k}+\mathrm{1}}{\mathrm{2}{n}}\pi\right)+\mathrm{1}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+{i}\frac{{sin}\left(\frac{\mathrm{2}{k}+\mathrm{1}}{\mathrm{2}{n}}\pi\right){cos}\left(\frac{\mathrm{2}{k}+\mathrm{1}}{\mathrm{2}{n}}\pi\right)}{\mathrm{1}−{cos}^{\mathrm{2}} \left(\frac{\mathrm{2}{k}+\mathrm{1}}{\mathrm{2}{n}}\pi\right)}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−{i}\frac{{sin}\left(\frac{\mathrm{2}{k}+\mathrm{1}}{\mathrm{2}{n}}\pi\right){cos}\left(\frac{\mathrm{2}{k}+\mathrm{1}}{\mathrm{2}{n}}\pi\right)}{{sin}^{\mathrm{2}} \left(\frac{\mathrm{2}{k}+\mathrm{1}}{\mathrm{2}{n}}\pi\right)}\right) \\ $$$${z}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−{icot}\left(\frac{\mathrm{2}{k}+\mathrm{1}}{\mathrm{2}{n}}\pi\right)\right) \\ $$$${When}\:\:{n}=\mathrm{5} \\ $$$${z}_{{k}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−{icot}\left(\frac{\mathrm{2}{k}+\mathrm{1}}{\mathrm{10}}\pi\right)\:\:{with}\:\:{k}=\left(\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4}\right)\right. \\ $$
Commented by peter frank last updated on 26/Oct/18
thanks sir
$${thanks}\:{sir} \\ $$

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