5

Let $z$ be a complex no. satisfying $\displaystyle \frac{1}{2}\leq |z|\leq 4\;,$ then the Sum of greatest and

least value of $\displaystyle \left|z+\frac{1}{z}\right|$ is

$\bf{My\; Try::}$ Let $z=re^{i\theta} = r\left(\cos \theta+i\sin \theta\right)\;,$ Where $-\pi<\theta \leq \pi.$ and $\displaystyle \frac{1}{2}\leq r \leq 4$

Then $$\displaystyle \left|z+\frac{1}{z}\right| = \left|re^{i\alpha}+\frac{e^{-i\alpha}}{r}\right| = \left|\frac{r^2\cos \alpha+r^2\cdot i\sin \alpha+\cos \alpha-i\sin \alpha}{r}\right|$$

so we get $$\displaystyle \left|\frac{(r^2+1)\cos \alpha+i(r^2-1)\sin \alpha}{r}\right| = \frac{\sqrt{(r^2+1)^2\cos^2 \alpha+(r^2-1)^2\sin^2 \alpha}}{r}$$

Now Let $$\displaystyle f(r,\alpha) = \frac{\sqrt{r^4+1+2r^2\cos 2\alpha}}{r}$$

Now i did not understand how can i solve after that, Help me , Thanks

Community
  • 1
juantheron
  • 53,015

2 Answers2

4

Take $z=re^{i\theta}$ then $|z+z^{-1}|= | re^{2i\theta} - r^{-1}|$ with constraint $\frac{1}{2}\le r\le 4$ and arbitrary $\theta$. By triangle inequality and reverse triangle inequality we get $$|r-r^{-1}|\le | re^{2i\theta} - r^{-1}| \le r + r^{-1}.$$ You can check that $r + r^{-1}$ is maximized at $r=4$ and $|r-r^{-1}|$ is minimized at $r=1$ so $$0\le | re^{2i\theta} - r^{-1}|\le 4.$$ Take $(r,\theta) = (4, \pi/2)$ then you will get the upper bound 4. Similarly, if you take $(r,\theta) = (1,\pi)$, you will get the lower bound 0, so they are maximum and minimum of $|z+z^{-1}|$.

Hanul Jeon
  • 27,376
3

$g(r,\alpha)= r^2+\dfrac{1}{r^2}+2cos2\alpha$

for min:$r^2+\dfrac{1}{r^2} \ge 2,cos2\alpha \ge -1$

for max: $r^2+\dfrac{1}{r^2}$ will get max at two ends, $cos2\alpha \le 1$

chenbai
  • 7,581