337. It is known that $\left|z+\frac{1}{z}\right|=a$; what are the greatest and the least values that the modulus $|z|$ of the complex number $z$ can have?
337. Since $|z|=|\bar{z}|=|-z|=|-\bar{z}|$ and $\left|z+\frac{1}{z}\right|=\left|\bar{z}+\frac{1}{\bar{z}}\right|=$ $\left|-z-\frac{1}{z}\right|=\left|-\bar{z}-\frac{1}{\bar{z}}\right|$, it is sufficient to consider only one of the numbers $\bar{z}, \bar{z}, -z$ and $-\bar{z}$ that lies in the first quadrant. If $|z|$ reaches its maximum possible value, then $\left|\frac{1}{z}\right|=\frac{1}{|z|}$ is the smallest. Therefore, it is sufficient to find those $z$ whose modulus is the largest possible; in this case, $|z| \geqslant\left|\frac{1}{2}\right|$. Let the argument $\varphi$ of the number $z$ be given; $0 \leqslant \varphi \leqslant \frac{\pi}{2}$. Then from Fig. 44 we have (recall that $\left|z+\frac{1}{z}\right|=a$; below, $|z|=r$ is denoted):

$a^{2}=r^{2}+\frac{1}{r^{2}}-2 \cos 2 \varphi=$

$$
\begin{aligned}
& =r^{2}+\frac{1}{r^{2}}-2+4 \cos ^{2} \varphi= \\
& =\left(r-\frac{1}{r}\right)^{2}+4 \cos ^{2} \varphi
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_b937612abb2773f7abffg-348.jpg?height=361&width=466&top_left_y=974&top_left_x=599)

Fig. 44.

Since, according to our assumption, $r \geqslant \frac{1}{r}$, the difference $r-\frac{1}{r}$ increases with $r$, and vice versa. But $\left(r-\frac{1}{r}\right)^{2}=a^{2}-4 \cos ^{2} \varphi \leqslant$ $\leqslant a^{2} ;$ when $\varphi=\frac{\pi}{2}$, we have $\left(r-\frac{1}{r}\right)^{2}=a^{2}$. In this case, $r-\frac{1}{r}=a, r=$ $=\frac{1}{2}\left(a+\sqrt{a^{2}+4}\right)$.

Accordingly, the maximum value of $|z|=\frac{1}{2}\left(a+\sqrt{a^{2}+4}\right)$ is achieved when $z=\frac{i}{2}\left(a+\sqrt{a^{2}+4}\right)$, and the minimum value of $|z|=$ $=\frac{1}{2}\left(\sqrt{a^{2}+4}-a\right)$ when $z=-\frac{i}{2}\left(\sqrt{a^{2}+4}-a\right)$.