Subfield: Geometry


Question: On the complex plane, the parallelogram formed by the points $0, z, \frac{1}{z}$, and $z+\frac{1}{z}$ has area $\frac{35}{37}$, and the real part of $z$ is positive. If $d$ is the smallest possible value of $\left|z+\frac{1}{z}\right|$, compute $d^{2}$.

Solution:
As is usual, let $\arg z$ refer to measure of the directed angle whose vertex is the origin, whose initial ray passes through 1 (i.e., the point $(1,0)$ ), and whose terminal ray passes through $z$. Then $\arg 1 / z=-\arg z$. Using the formula $a b \sin \gamma$ for the area of the parallelogram with sides $a$ and $b$ and included angle $\gamma$ yields the equation

$$
\frac{35}{37}=|z| \cdot\left|\frac{1}{z}\right| \cdot \sin (2 \arg z)
$$

However, $|1 / z|=1 /|z|$, so the right side simplifies to $\sin (2 \arg z)$.

To compute the length $c$ of the diagonal from 0 to $z+1 / z$, use the Law of Cosines and the fact that consecutive angles of a parallelogram are supplementary:

$$
\begin{aligned}
c^{2} & =|z|^{2}+\left|\frac{1}{z}\right|^{2}-2|z| \cdot\left|\frac{1}{z}\right| \cos (\pi-2 \arg z) \\
& =|z|^{2}+\left|\frac{1}{z}\right|^{2}-2 \cos (\pi-2 \arg z) \\
& =|z|^{2}+\left|\frac{1}{z}\right|^{2}+2 \cos (2 \arg z) .
\end{aligned}
$$

This expression separates into two parts: the first, $|z|^{2}+|1 / z|^{2}$, is independent of the argument (angle) of $z$, while the second, $2 \cos (2 \arg z)$, is determined by the condition that $\sin (2 \arg z)=$ 35/37. The minimum value of $|z|^{2}+|1 / z|^{2}$ is 2 , as can be shown by the Arithmetic MeanGeometric Mean inequality applied to $|z|^{2}$ and $|1 / z|^{2}$ :

$$
|z|^{2}+|1 / z|^{2} \geq 2 \sqrt{|z|^{2} \cdot|1 / z|^{2}}=2
$$

The value of $\cos (2 \arg z)$ is given by the Pythagorean Identity:

$$
\cos (2 \arg z)= \pm \sqrt{1-\left(\frac{35}{37}\right)^{2}}= \pm \sqrt{1-\frac{1225}{1369}}= \pm \sqrt{\frac{144}{1369}}= \pm \frac{12}{37}
$$

Because the goal is to minimize the diagonal's length, choose the negative value to obtain

$$
d^{2}=2-2 \cdot \frac{12}{37}=\frac{50}{37}
$$
Using polar coordinates, write

$$
z=r(\cos \theta+i \sin \theta)
$$

so that

$$
\frac{1}{z}=r^{-1}(\cos \theta-i \sin \theta)
$$

Without loss of generality, assume that $z$ is in the first quadrant, so that $\theta>0$. Then the angle between the sides $\overline{0 z}$ and $\overline{0 z^{-1}}$ is $2 \theta$, and the side lengths are $r$ and $r^{-1}$, so the area of the parallelogram is

$$
\frac{35}{37}=r \cdot r^{-1} \cdot \sin (2 \theta)=\sin 2 \theta
$$

Note that $0<\theta<\pi / 2$, so $0<2 \theta<\pi$, and there are two values of $\theta$ that satisfy this equation. Adding the expressions for $z$ and $z^{-1}$ and calculating the absolute value yields

$$
\begin{aligned}
\left|z+\frac{1}{z}\right|^{2} & =\left(r+r^{-1}\right)^{2} \cos ^{2} \theta+\left(r-r^{-1}\right)^{2} \sin ^{2} \theta \\
& =\left(r^{2}+r^{-2}\right)\left(\cos ^{2} \theta+\sin ^{2} \theta\right)+2 r \cdot r^{-1}\left(\cos ^{2} \theta-\sin ^{2} \theta\right) \\
& =r^{2}+r^{-2}+2 \cos 2 \theta .
\end{aligned}
$$

Minimize the terms involving $r$ using the Arithmetic-Geometric Mean inequality:

$$
r^{2}+r^{-2} \geq 2 \sqrt{r^{2} \cdot r^{-2}}=2
$$

with equality when $r^{2}=r^{-2}$, that is, when $r=1$. For the term involving $\theta$, recall that there are two possible values:

$$
\cos 2 \theta= \pm \sqrt{1-\sin ^{2} 2 \theta}= \pm \sqrt{\frac{37^{2}-35^{2}}{37^{2}}}= \pm \frac{\sqrt{(37+35)(37-35)}}{37}= \pm \frac{12}{37}
$$

To minimize this term, take the negative value, yielding

$$
d^{2}=2-2 \cdot \frac{12}{37}=\frac{\mathbf{5 0}}{\mathbf{3 7}}
$$
If $z=x+y i$, then compute $1 / z$ by rationalizing the denominator:

$$
\frac{1}{z}=\frac{x-y i}{x^{2}+y^{2}}=\frac{x}{x^{2}+y^{2}}+\frac{-y}{x^{2}+y^{2}} i
$$

The area of the parallelogram is given by the absolute value of the $2 \times 2$ determinant

$$
\left|\begin{array}{cc}
x & y \\
x /\left(x^{2}+y^{2}\right) & -y /\left(x^{2}+y^{2}\right)
\end{array}\right|=\frac{1}{x^{2}+y^{2}}\left|\begin{array}{cc}
x & y \\
x & -y
\end{array}\right|=\frac{-2 x y}{x^{2}+y^{2}}
$$



That is,

$$
\frac{2 x y}{x^{2}+y^{2}}=\frac{35}{37}
$$

Calculation shows that

$$
\left|z+\frac{1}{z}\right|^{2}=\left(x+\frac{x}{x^{2}+y^{2}}\right)^{2}+\left(y-\frac{y}{x^{2}+y^{2}}\right)^{2}=\left(x^{2}+y^{2}\right)+\frac{1}{x^{2}+y^{2}}+2\left(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\right) .
$$

As in the previous solution, the sum of the first two terms is at least 2 , when $x^{2}+y^{2}=1$. The trick for relating the third term to the area is to express both the third term and the area in terms of the ratio

$$
t=\frac{y}{x} .
$$

Indeed,

$$
\frac{2 x y}{x^{2}+y^{2}}=\frac{2 t}{1+t^{2}} \quad \text { and } \quad \frac{x^{2}-y^{2}}{x^{2}+y^{2}}=\frac{1-t^{2}}{1+t^{2}}=\frac{(1+t)(1-t)}{1+t^{2}}
$$

As in the previous solution, assume without loss of generality that $z$ is in the first quadrant, so that $t>0$. As found above,

$$
\frac{2 t}{1+t^{2}}=\frac{35}{37}
$$

It is not difficult to solve for $t$ using the quadratic formula, but the value of $t$ is not needed to solve the problem. Observe that

$$
\frac{(1 \pm t)^{2}}{1+t^{2}}=1 \pm \frac{2 t}{1+t^{2}}=1 \pm \frac{35}{37},
$$

so that

$$
\left(\frac{1-t^{2}}{1+t^{2}}\right)^{2}=\frac{(1+t)^{2}}{1+t^{2}} \cdot \frac{(1-t)^{2}}{1+t^{2}}=\frac{72}{37} \cdot \frac{2}{37}=\left(\frac{12}{37}\right)^{2}
$$

In order to minimize $d$, take the negative square root, leading to

$$
d^{2}=2+2 \cdot \frac{1-t^{2}}{1+t^{2}}=2-\frac{24}{37}=\frac{\mathbf{5 0}}{\mathbf{3 7}}
$$

Final answer: $\frac{50}{37}$