Question

It is known that $|z+\frac{1}{z}|=a$, where z is a complex number. What are the greatest and least positive values of the modulus |z| of the complex number z?

• A. $\left(\frac{\sqrt{(a^2+4)}+a}{2}\right)$,$\left(\frac{\sqrt{(a^2+4)}-a}{2}\right)$
• B. $\left(\frac{\sqrt{(a^2-4)}+a}{2}\right)$,$\left(\frac{\sqrt{(a^2-4)}-a}{2}\right)$
• C. $\left(\frac{-\sqrt{(a^2+4)}+a}{2}\right)$,$\left(\frac{-\sqrt{(a^2+4)}-a}{2}\right)$
• D. $\left(\frac{-\sqrt{(a^2-4)}+a}{2}\right)$,$(-\frac{\sqrt{(a^2-4)}-a}{2})$

Answer 1

The correct option is A $\left(\frac{\sqrt{(a^2+4)}+a}{2}\right)$,$\left(\frac{\sqrt{(a^2+4)}-a}{2}\right)$

$|z|=|\overline{z}|=|\overline{z}|=|-z|=|\overline{-z}|$
$\therefore |z+\frac{1}{z}|=|\overline{z}+\frac{1}{z}|$
$=|-z-\frac{1}{z}|=|-\overline{z}-\frac{1}{z}|$
it is sufficient to consider only one of the numbers z, $\overline{z}$, - z and -$\overline{z}$, namely the one lying in the first quadrant. When |z| assumes its maximum possible value of expression $\left|\frac{1}{z}\right|=\frac{1}{|z|}$ assume the minimum value, therefor it suffices to find those z whose modulus assume the greatest value under the assumption.
$|z|>\frac{1}{z}$
Let $z=r(cos\theta +isin\theta )$
$(0\le \theta \le \pi /2)$
Since $|z+\frac{1}{z}|=a$
We can write this relation as
$\left|r\right(cos\theta +isin\theta )+\frac{1}{r}(cos\theta -isin\theta \left)\right|=a$
or ${(rcos\theta +\frac{1}{r}cos\theta )}^2+{(rsin\theta -\frac{1}{r}sin\theta )}^2=a^2$
$\Rightarrow r^2+\frac{1}{r^2}+2cos2\theta =a^2$
$\Rightarrow {(r-\frac{1}{r})}^2+4co{s}^2\theta =a^2$
$\Rightarrow {(r-\frac{1}{r})}^2=a^2-4co{s}^2\theta \le a^2$
$\Rightarrow {(r-\frac{1}{r})}^2\le a^2$
For $\theta =\pi /2$, we have ${(r-\frac{1}{r})}^2=a^2$ and so $r-\frac{1}{r}=a$ which givens $r=\frac{a+\sqrt{a^2+4}}{2}$. It follows that the greatest value of $|z|=\left(\frac{a+\sqrt{a^2+4}}{2}\right)$ is attained for
$z=\left(\frac{a+\sqrt{a^2+4}}{2}\right)(cos\frac{\pi }{2}+isin\frac{\pi }{2})$
$=i\left(\frac{a+\sqrt{a^2+4}}{2}\right)$
Similarly smallest value of $|z|=\frac{\sqrt{(a^2+4)}-a}{2}$ is attained for $z=-i\left(\frac{\sqrt{(a^2+4)}-a}{2}\right)$
Ans: A