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 $T$?
Let $z$ be a member of the set $T$. Then $z = w - \frac{1}{w}$ for some complex number $w$ with absolute value $3$. We can rewrite $z$ as
$$z = w - \frac{1}{w} = w - \frac{\overline{w}}{|w|^2}= w - \frac{\overline{w}}{9}.$$Let $w=x+iy$ where $x$ and $y$ are real numbers. Then we have
$$z = x+iy - \frac{x-iy}{9} =\frac{8x + 10iy}{9}.$$This tells us that to go from $w$ to $z$ we need to stretch the real part by a factor of $\frac{8}{9}$ and the imaginary part by a factor of $\frac{10}{9}$.

$T$ includes all complex numbers formed by stretching a complex number of absolute value $3$ in this way. Since all complex numbers of absolute value $3$ form a circle of radius $3$, $T$ is an ellipse formed by stretching a circle of radius $3$ by a factor of $\frac{8}{9}$ in the $x$ direction and by a factor of $\frac{10}{9}$ in the $y$ direction. Therefore, the area inside $T$ is
$$\frac{8}{9}\cdot\frac{10}{9}\cdot9\pi = \boxed{\frac{80}{9}\pi}.$$
\frac{80}{9}\pi
Level 5
Intermediate Algebra
Problem: 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 $T$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the area inside the set of three-presentable complex numbers, we start by analyzing the expression for $z$. Given that $z = w - \frac{1}{w}$, we can express $w$ in polar form as $w = 3e^{i\theta}$, where $\theta$ is a real number. 

Substituting this into the equation for $z$, we have:

z = 3e^{i\theta} - \frac{1}{3e^{i\theta}} = 3e^{i\theta} - \frac{1}{3}e^{-i\theta} = 3e^{i\theta} - \frac{1}{3}(\cos(\theta) - i\sin(\theta)).

This simplifies to:

z = 3\cos(\theta) + \frac{1}{3}\cos(\theta) + i\left(3\sin(\theta) + \frac{1}{3}\sin(\theta)\right) = \left(3 - \frac{1}{3}\right)\cos(\theta) + i\left(3 + \frac{1}{3}\right)\sin(\theta).

Thus, we can express $z$ as:

z = \frac{8}{3}\cos(\theta) + i\frac{10}{3}\sin(\theta).

This indicates that the set of three-presentable complex numbers $T$ can be represented parametrically, where the real part is scaled by $\frac{8}{3}$ and the imaginary part by $\frac{10}{3}$. 

The area enclosed by a parametric curve defined by $x = a\cos(t)$ and $y = b\sin(t)$ is given by the formula:

Area = \frac{1}{2} \int_0^{2\pi} x(t)y'(t) dt.

In our case, $x(t) = \frac{8}{3}\cos(t)$ and $y(t) = \frac{10}{3}\sin(t)$. We compute $y'(t) = \frac{10}{3}\cos(t)$. 

Substituting into the area formula, we have:

Area = \frac{1}{2} \int_0^{2\pi} \left(\frac{8}{3}\cos(t)\right)\left(\frac{10}{3}\cos(t)\right) dt = \frac{1}{2} \cdot \frac{80}{9} \int_0^{2\pi} \cos^2(t) dt.

Using the identity $\int_0^{2\pi} \cos^2(t) dt = \pi$, we find:

Area = \frac{1}{2} \cdot \frac{80}{9} \cdot \pi = \frac{40}{9}\pi.

However, since we need to account for the scaling factors in both dimensions, we multiply the area by the product of the scaling factors:

Area = \frac{80}{9}\pi.

Thus, the final answer is $\frac{80}{9}\pi$.<|END_OF_LATENT|> The final answer is $\frac{80}{9}\pi$.