<s>[INST] If complex no. ($z$) satisfying $\frac{1}{2}\leq |z|\leq 4\;,$ Then Max. and Min. of $\left|z+\frac{1}{z}\right|$ 
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
 [/INST] $g(r,\alpha)= r^2+\dfrac{1}{r^2}+2cos2\alpha$
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, $cos2\alpha \le 1$
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