### Instruction: Let $S$ be the set of complex numbers $z$ such that the real part of $\frac{1}{z}$ is equal to $\frac{1}{6}.$  This set forms a curve.  Find the area of the region inside the curve.### Response: To solve this problem, I will use some basic facts about complex numbers and their geometric representation on the complex plane.
If $z = a + bi$ is a complex number, then $\frac{1}{z} = \frac{a - bi}{a^2 + b^2}$ by multiplying the numerator and denominator by the conjugate of $z$.
The real part of $\frac{1}{z}$ is then $\frac{a}{a^2 + b^2}$, and this is equal to $\frac{1}{6}$ according to the problem.
So I have the equation $\frac{a}{a^2 + b^2} = \frac{1}{6}$, which I can simplify by multiplying both sides by $a^2 + b^2$ and then by $6$.
I get $6a = a^2 + b^2$, which is a quadratic equation in $a$ and $b$.
To make it easier to see the curve, I will complete the square for both $a$ and $b$.
I have $a^2 - 6a + b^2 = 0$, which I can rewrite as $(a - 3)^2 + b^2 = 9$.
Now I recognize this as the equation of a circle with center $(3, 0)$ and radius $3$.
The curve is the boundary of this circle, and the area of the region inside the curve is just the area of the circle.
The area of a circle is $\pi$ times the radius squared, so the area of the region is $\pi \times 3^2 = 9\pi$.