Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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 byEnterUsername last updated on 05/May/21

Thanks !

$$\mathrm{Thanks}\:! \\ $$

Commented bymr 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 byEnterUsername last updated on 06/May/21

Thank you Sir

Terms of Service

Privacy Policy

Contact: info@tinkutara.com