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Question Number 140198 by EnterUsername last updated on 05/May/21
Let a>0 and ∣z+(1/z)∣=a (z≠0 is a complex number). Then the maximum and minimum values of ∣z∣ are (A) ((a+(√(a^2 +4)))/2) (B) ((2a+(√(a^2 +4)))/2) (C) (((√(a^2 +4))−a)/2) (D) (((√(a^2 +4))−2a)/2)
$$\mathrm{Let}\:{a}>\mathrm{0}\:\mathrm{and}\:\mid{z}+\left(\mathrm{1}/{z}\right)\mid={a}\:\left({z}\neq\mathrm{0}\:\mathrm{is}\:\mathrm{a}\:\mathrm{complex}\:\mathrm{number}\right). \\ $$$$\mathrm{Then}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{and}\:\mathrm{minimum}\:\mathrm{values}\:\mathrm{of}\:\mid{z}\mid\:\mathrm{are} \\ $$$$\left(\mathrm{A}\right)\:\frac{{a}+\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\frac{\mathrm{2}{a}+\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}} \\ $$$$\left(\mathrm{C}\right)\:\frac{\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}−{a}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\frac{\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}−\mathrm{2}{a}}{\mathrm{2}} \\ $$
Answered by Dwaipayan Shikari last updated on 05/May/21
∣z∣+∣(1/z)∣≥∣z+(1/z)∣ For Maximum ∣z+(1/z)∣=∣z∣+∣(1/z)∣ ∣z∣+(1/(∣z∣))=a⇒∣z∣=((a±(√(a^2 +4)))/2) Maximum Value ∣z∣_(max) =((a+(√(a^2 +4)))/2)
$$\mid{z}\mid+\mid\frac{\mathrm{1}}{{z}}\mid\geqslant\mid{z}+\frac{\mathrm{1}}{{z}}\mid \\ $$$${For}\:{Maximum}\:\mid{z}+\frac{\mathrm{1}}{{z}}\mid=\mid{z}\mid+\mid\frac{\mathrm{1}}{{z}}\mid \\ $$$$\mid{z}\mid+\frac{\mathrm{1}}{\mid{z}\mid}={a}\Rightarrow\mid{z}\mid=\frac{{a}\pm\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}} \\ $$$${Maximum}\:{Value}\:\mid{z}\mid_{{max}} =\frac{{a}+\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}} \\ $$
Commented by EnterUsername last updated on 05/May/21
Thanks !
$$\mathrm{Thanks}\:! \\ $$
Commented by mr W last updated on 05/May/21
∣(1/z)∣≠(1/(∣z∣))
$$\mid\frac{\mathrm{1}}{{z}}\mid\neq\frac{\mathrm{1}}{\mid{z}\mid} \\ $$
Answered by mr W last updated on 05/May/21
let z=re^(θi) ∣z∣=r (1/z)=(1/(re^(θi) ))=(1/r)e^(−θi) ∣z+(1/z)∣=∣re^(θi) +(1/r)e^(−θi) ∣=(√(r^2 +(1/r^2 )+2cos 2θ))=a r^2 +(1/r^2 )=a^2 −2cos 2θ=λ max. or min. from r is when λ is as large as possible, i.e. λ=a^2 +2, and θ=(π/2). r^4 −(a^2 +2)r^2 +1=0 ⇒r_(max/min) ^2 =((a^2 +2±(√((a^2 +2)^2 −4)))/2) =((a^2 +2±a(√(a^2 +4)))/2) =((a^2 +4±2a(√(a^2 +4))+a^2 )/4) =((((√(a^2 +4)))^2 ±2a(√(a^2 +4))+a^2 )/4) =((((√(a^2 +4))±a)^2 )/4) ⇒r_(max/min) =(((√(a^2 +4))±a)/2) i.e. r_(max) =(((√(a^2 +4))+a)/2) r_(min) =(((√(a^2 +4))−a)/2)
$${let}\:{z}={re}^{\theta{i}} \\ $$$$\mid{z}\mid={r} \\ $$$$\frac{\mathrm{1}}{{z}}=\frac{\mathrm{1}}{{re}^{\theta{i}} }=\frac{\mathrm{1}}{{r}}{e}^{−\theta{i}} \\ $$$$\mid{z}+\frac{\mathrm{1}}{{z}}\mid=\mid{re}^{\theta{i}} +\frac{\mathrm{1}}{{r}}{e}^{−\theta{i}} \mid=\sqrt{{r}^{\mathrm{2}} +\frac{\mathrm{1}}{{r}^{\mathrm{2}} }+\mathrm{2cos}\:\mathrm{2}\theta}={a} \\ $$$${r}^{\mathrm{2}} +\frac{\mathrm{1}}{{r}^{\mathrm{2}} }={a}^{\mathrm{2}} −\mathrm{2cos}\:\mathrm{2}\theta=\lambda \\ $$$${max}.\:{or}\:{min}.\:{from}\:{r}\:{is}\:{when}\:\lambda\:{is} \\ $$$${as}\:{large}\:{as}\:{possible},\:{i}.{e}.\:\lambda={a}^{\mathrm{2}} +\mathrm{2},\:{and} \\ $$$$\theta=\frac{\pi}{\mathrm{2}}. \\ $$$${r}^{\mathrm{4}} −\left({a}^{\mathrm{2}} +\mathrm{2}\right){r}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{r}_{{max}/{min}} ^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} +\mathrm{2}\pm\sqrt{\left({a}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}} \\ $$$$=\frac{{a}^{\mathrm{2}} +\mathrm{2}\pm{a}\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}} \\ $$$$=\frac{{a}^{\mathrm{2}} +\mathrm{4}\pm\mathrm{2}{a}\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}+{a}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$=\frac{\left(\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}\right)^{\mathrm{2}} \pm\mathrm{2}{a}\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}+{a}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$=\frac{\left(\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}\pm{a}\right)^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\Rightarrow{r}_{{max}/{min}} =\frac{\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}\pm{a}}{\mathrm{2}} \\ $$$${i}.{e}. \\ $$$${r}_{{max}} =\frac{\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}+{a}}{\mathrm{2}} \\ $$$${r}_{{min}} =\frac{\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}−{a}}{\mathrm{2}} \\ $$
Commented by EnterUsername last updated on 06/May/21
Thank you Sir

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