Question Number 49272 by rahul 19 last updated on 05/Dec/18
$${Z}\epsilon\mathbb{C}\:{satisfies}\:{the}\:{condition}\:\mid{Z}\mid\geqslant\mathrm{3}. \\ $$$${Then}\:{find}\:{the}\:{least}\:{value}\:{of}\:\mid{Z}+\frac{\mathrm{1}}{{Z}}\mid\:? \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 05/Dec/18
Commented by tanmay.chaudhury50@gmail.com last updated on 05/Dec/18
Commented by rahul 19 last updated on 05/Dec/18
thank you sir! ��
Answered by mr W last updated on 05/Dec/18
![Z=r(cos θ+i sin θ) ∣Z∣=r≥3 Z+(1/Z)=r(cos θ+i sin θ)+(1/(r(cos θ+i sin θ))) Z+(1/Z)=r(cos θ+i sin θ)+((cos θ−i sin θ)/r) Z+(1/Z)=(r+(1/r))cos θ+i (r−(1/r))sin θ Z+(1/Z)=(1/r)[(r^2 +1)cos θ+i (r^2 −1)sin θ] ∣Z+(1/Z)∣=(1/r)(√((r^2 +1)^2 cos^2 θ+(r^2 −1)^2 sin^2 θ)) ∣Z+(1/Z)∣=(1/r)(√((r^2 +1)^2 −[(r^2 +1)^2 −(r^2 −1)^2 ]sin^2 θ)) ∣Z+(1/Z)∣=(1/r)(√((r^2 +1)^2 −4r^2 sin^2 θ)) ∣Z+(1/Z)∣=(√((r+(1/r))^2 −4sin^2 θ)) ≥(√((r+(1/r))^2 −4)) =(√((r−(1/r))^2 )) =∣r−(1/r)∣ (increasing function upon r=1) ≥∣3−(1/3)∣=(8/3)=2.67](https://www.tinkutara.com/question/Q49290.png)
$${Z}={r}\left(\mathrm{cos}\:\theta+{i}\:\mathrm{sin}\:\theta\right) \\ $$$$\mid{Z}\mid={r}\geqslant\mathrm{3} \\ $$$$ \\ $$$${Z}+\frac{\mathrm{1}}{{Z}}={r}\left(\mathrm{cos}\:\theta+{i}\:\mathrm{sin}\:\theta\right)+\frac{\mathrm{1}}{{r}\left(\mathrm{cos}\:\theta+{i}\:\mathrm{sin}\:\theta\right)} \\ $$$${Z}+\frac{\mathrm{1}}{{Z}}={r}\left(\mathrm{cos}\:\theta+{i}\:\mathrm{sin}\:\theta\right)+\frac{\mathrm{cos}\:\theta−{i}\:\mathrm{sin}\:\theta}{{r}} \\ $$$${Z}+\frac{\mathrm{1}}{{Z}}=\left({r}+\frac{\mathrm{1}}{{r}}\right)\mathrm{cos}\:\theta+{i}\:\left({r}−\frac{\mathrm{1}}{{r}}\right)\mathrm{sin}\:\theta \\ $$$${Z}+\frac{\mathrm{1}}{{Z}}=\frac{\mathrm{1}}{{r}}\left[\left({r}^{\mathrm{2}} +\mathrm{1}\right)\mathrm{cos}\:\theta+{i}\:\left({r}^{\mathrm{2}} −\mathrm{1}\right)\mathrm{sin}\:\theta\right] \\ $$$$\mid{Z}+\frac{\mathrm{1}}{{Z}}\mid=\frac{\mathrm{1}}{{r}}\sqrt{\left({r}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta+\left({r}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta} \\ $$$$\mid{Z}+\frac{\mathrm{1}}{{Z}}\mid=\frac{\mathrm{1}}{{r}}\sqrt{\left({r}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −\left[\left({r}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −\left({r}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \right]\mathrm{sin}^{\mathrm{2}} \:\theta} \\ $$$$\mid{Z}+\frac{\mathrm{1}}{{Z}}\mid=\frac{\mathrm{1}}{{r}}\sqrt{\left({r}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}{r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta} \\ $$$$\mid{Z}+\frac{\mathrm{1}}{{Z}}\mid=\sqrt{\left({r}+\frac{\mathrm{1}}{{r}}\right)^{\mathrm{2}} −\mathrm{4sin}^{\mathrm{2}} \:\theta} \\ $$$$\geqslant\sqrt{\left({r}+\frac{\mathrm{1}}{{r}}\right)^{\mathrm{2}} −\mathrm{4}} \\ $$$$=\sqrt{\left({r}−\frac{\mathrm{1}}{{r}}\right)^{\mathrm{2}} } \\ $$$$=\mid{r}−\frac{\mathrm{1}}{{r}}\mid\:\left({increasing}\:{function}\:{upon}\:{r}=\mathrm{1}\right) \\ $$$$\geqslant\mid\mathrm{3}−\frac{\mathrm{1}}{\mathrm{3}}\mid=\frac{\mathrm{8}}{\mathrm{3}}=\mathrm{2}.\mathrm{67} \\ $$
Commented by rahul 19 last updated on 05/Dec/18
thank you sir! ��