### Instruction: Say that a complex number $z$ is three-presentable if there is a complex number $w$ of absolute value $3$ such that $z = w - \frac{1}{w}$.  Let $T$ be the set of all three-presentable complex numbers.  The set $T$ forms a closed curve in the complex plane.  What is the area inside $T$?### Response: To solve this problem, I will try to find a parametric equation for the curve $T$ and then use the formula for the area of a parametric curve.
First, I will write $z$ and $w$ in polar form, using $r$ and $\theta$ for the modulus and argument of $w$, respectively.
Then I have $z = w - \frac{1}{w} = r(\cos \theta + i \sin \theta) - \frac{1}{r}(\cos \theta - i \sin \theta)$.
Expanding and simplifying, I get $z = (r - \frac{1}{r})\cos \theta + i(r + \frac{1}{r})\sin \theta$.
I can rewrite this as $z = f(\theta) + i g(\theta)$, where $f(\theta) = (r - \frac{1}{r})\cos \theta$ and $g(\theta) = (r + \frac{1}{r})\sin \theta$.
Now, I notice that the curve $T$ depends only on the value of $r$, which is fixed at $3$.
So, I can plug in $r = 3$ and simplify the functions $f$ and $g$.
I get $f(\theta) = (3 - \frac{1}{3})\cos \theta = \frac{8}{3}\cos \theta$ and $g(\theta) = (3 + \frac{1}{3})\sin \theta = \frac{10}{3}\sin \theta$.
So, the curve $T$ has the parametric equation $z = \frac{8}{3}\cos \theta + i \frac{10}{3}\sin \theta$, where $\theta$ ranges from $0$ to $2\pi$.
Next, I will use the formula for the area of a parametric curve, which is $A = \frac{1}{2} \int_{a}^{b} (f(\theta)g'(\theta) - f'(\theta)g(\theta)) \, d\theta$, where $a$ and $b$ are the endpoints of the parameter interval.
In this case, $a = 0$ and $b = 2\pi$, so I have $A = \frac{1}{2} \int_{0}^{2\pi} (f(\theta)g'(\theta) - f'(\theta)g(\theta)) \, d\theta$.
I can compute the derivatives of $f$ and $g$ using the chain rule.
I get $f'(\theta) = -\frac{8}{3}\sin \theta$ and $g'(\theta) = \frac{10}{3}\cos \theta$.
Substituting these into the formula, I get $A = \frac{1}{2} \int_{0}^{2\pi} (\frac{8}{3}\cos \theta \cdot \frac{10}{3}\cos \theta - (-\frac{8}{3}\sin \theta) \cdot \frac{10}{3}\sin \theta) \, d\theta$.
Simplifying, I get $A = \frac{40}{9} \int_{0}^{2\pi} (\cos^2 \theta + \sin^2 \theta) \, d\theta$.
Using the trigonometric identity $\cos^2 \theta + \sin^2 \theta = 1$, I get $A = \frac{40}{9} \int_{0}^{2\pi} 1 \, d\theta$.