Menu Close

Question-81123




Question Number 81123 by M±th+et£s last updated on 09/Feb/20
Commented by mr W last updated on 09/Feb/20
let z=x+yi  ∣z∣=(√(x^2 +y^2 ))  let P=∣z∣^2 =x^2 +y^2   z+(1/z)=x+yi+(1/(x+yi))=x+yi+((x−yi)/(x^2 +y^2 ))  =x(1+(1/(x^2 +y^2 )))+y(1−(1/(x^2 +y^2 )))i  ∣z+(1/z)∣=(√(x^2 (1+(1/(x^2 +y^2 )))^2 +y^2 (1−(1/(x^2 +y^2 )))^2 ))=a  x^2 (1+(1/(x^2 +y^2 )))^2 +y^2 (1−(1/(x^2 +y^2 )))^2 =a^2   x^2 (x^2 +y^2 +1)^2 +y^2 (x^2 +y^2 −1)^2 =a^2 (x^2 +y^2 )^2   x^2 (P+1)^2 +(P−x^2 )(P−1)^2 =a^2 P^2   2x(P+1)^2 +2x^2 (P+1)(dP/dx)+((dP/dx)−2x)(P−1)^2 +2(P−x^2 )(P−1)(dP/dx)=2a^2 P(dP/dx)  min. or max. ∣z∣ ⇒min. or max. P ⇒(dP/dx)=0  2x(P+1)^2 +(−2x)(P−1)^2 =0  8xP=0  P≠0 ⇒x=0  y^2 (y^2 −1)^2 =a^2 (y^2 )^2   y≠0 ⇒(y^2 −1)^2 =a^2 y^2   y^4 −(2+a^2 )y^2 +1=0  y^2 =((2+a^2 ±a(√(4+a^2 )))/2)  P_(min) =0^2 +((2+a^2 −a(√(4+a^2 )))/2)=((2+a^2 −a(√(4+a^2 )))/2)  P_(max) =0^2 +((2+a^2 +a(√(4+a^2 )))/2)=((2+a^2 +a(√(4+a^2 )))/2)  ⇒∣z∣_(min) =(√((2+a^2 −a(√(4+a^2 )))/2))  ⇒∣z∣_(max) =(√((2+a^2 +a(√(4+a^2 )))/2))
$${let}\:{z}={x}+{yi} \\ $$$$\mid{z}\mid=\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} } \\ $$$${let}\:{P}=\mid{z}\mid^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} \\ $$$${z}+\frac{\mathrm{1}}{{z}}={x}+{yi}+\frac{\mathrm{1}}{{x}+{yi}}={x}+{yi}+\frac{{x}−{yi}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} } \\ $$$$={x}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\right)+{y}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\right){i} \\ $$$$\mid{z}+\frac{\mathrm{1}}{{z}}\mid=\sqrt{{x}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\right)^{\mathrm{2}} +{y}^{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\right)^{\mathrm{2}} }={a} \\ $$$${x}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\right)^{\mathrm{2}} +{y}^{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\right)^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} \left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} \left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} \left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} \left({P}+\mathrm{1}\right)^{\mathrm{2}} +\left({P}−{x}^{\mathrm{2}} \right)\left({P}−\mathrm{1}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} {P}^{\mathrm{2}} \\ $$$$\mathrm{2}{x}\left({P}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2}{x}^{\mathrm{2}} \left({P}+\mathrm{1}\right)\frac{{dP}}{{dx}}+\left(\frac{{dP}}{{dx}}−\mathrm{2}{x}\right)\left({P}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2}\left({P}−{x}^{\mathrm{2}} \right)\left({P}−\mathrm{1}\right)\frac{{dP}}{{dx}}=\mathrm{2}{a}^{\mathrm{2}} {P}\frac{{dP}}{{dx}} \\ $$$${min}.\:{or}\:{max}.\:\mid{z}\mid\:\Rightarrow{min}.\:{or}\:{max}.\:{P}\:\Rightarrow\frac{{dP}}{{dx}}=\mathrm{0} \\ $$$$\mathrm{2}{x}\left({P}+\mathrm{1}\right)^{\mathrm{2}} +\left(−\mathrm{2}{x}\right)\left({P}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{8}{xP}=\mathrm{0} \\ $$$${P}\neq\mathrm{0}\:\Rightarrow{x}=\mathrm{0} \\ $$$${y}^{\mathrm{2}} \left({y}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} \left({y}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$${y}\neq\mathrm{0}\:\Rightarrow\left({y}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} {y}^{\mathrm{2}} \\ $$$${y}^{\mathrm{4}} −\left(\mathrm{2}+{a}^{\mathrm{2}} \right){y}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$${y}^{\mathrm{2}} =\frac{\mathrm{2}+{a}^{\mathrm{2}} \pm{a}\sqrt{\mathrm{4}+{a}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$${P}_{{min}} =\mathrm{0}^{\mathrm{2}} +\frac{\mathrm{2}+{a}^{\mathrm{2}} −{a}\sqrt{\mathrm{4}+{a}^{\mathrm{2}} }}{\mathrm{2}}=\frac{\mathrm{2}+{a}^{\mathrm{2}} −{a}\sqrt{\mathrm{4}+{a}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$${P}_{{max}} =\mathrm{0}^{\mathrm{2}} +\frac{\mathrm{2}+{a}^{\mathrm{2}} +{a}\sqrt{\mathrm{4}+{a}^{\mathrm{2}} }}{\mathrm{2}}=\frac{\mathrm{2}+{a}^{\mathrm{2}} +{a}\sqrt{\mathrm{4}+{a}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\Rightarrow\mid{z}\mid_{{min}} =\sqrt{\frac{\mathrm{2}+{a}^{\mathrm{2}} −{a}\sqrt{\mathrm{4}+{a}^{\mathrm{2}} }}{\mathrm{2}}} \\ $$$$\Rightarrow\mid{z}\mid_{{max}} =\sqrt{\frac{\mathrm{2}+{a}^{\mathrm{2}} +{a}\sqrt{\mathrm{4}+{a}^{\mathrm{2}} }}{\mathrm{2}}} \\ $$
Commented by mr W last updated on 10/Feb/20
answer correct sir?
$${answer}\:{correct}\:{sir}? \\ $$
Commented by M±th+et£s last updated on 10/Feb/20
yes thank you nice work
$${yes}\:{thank}\:{you}\:{nice}\:{work} \\ $$

Leave a Reply

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