The complex number $z$ traces a circle centered at the origin with radius 2.  Then $z + \frac{1}{z}$ traces a:
(A) circle
(B) parabola
(C) ellipse
(D) hyperbola

Enter the letter of the correct option.
Let $z = a + bi,$ where $a$ and $b$ are real numbers.  Since $|z| = 2,$ $a^2 + b^2 = 4.$  Then
\begin{align*}
z + \frac{1}{z} &= a + bi + \frac{1}{a + bi} \\
&= a + bi + \frac{1}{a + bi} \\
&= a + bi + \frac{a - bi}{a^2 + b^2} \\
&= a + bi + \frac{a - bi}{4} \\
&= \frac{5}{4} a + \frac{3}{4} bi.
\end{align*}Let $x + yi = z + \frac{1}{z},$ so $x = \frac{5}{4} a$ and $y = \frac{3}{4} b.$  Then
\[\frac{x^2}{(5/4)^2} + \frac{y^2}{(3/4)^2} = a^2 + b^2 = 4,\]so
\[\frac{x^2}{(5/2)^2} + \frac{y^2}{(3/2)^2} = 1.\]Thus, $z + \frac{1}{z}$ traces an ellipse.  The answer is $\boxed{\text{(C)}}.$
(C)
Level 3
Intermediate Algebra
Problem: The complex number $z$ traces a circle centered at the origin with radius 2.  Then $z + \frac{1}{z}$ traces a:
(A) circle
(B) parabola
(C) ellipse
(D) hyperbola

Enter the letter of the correct option.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine the path traced by the expression $z + \frac{1}{z}$, we start by letting $z = 2e^{i\theta}$, where $\theta$ varies from $0$ to $2\pi$. This representation captures the circular motion of $z$ around the origin with radius 2.

Next, we compute $\frac{1}{z}$:
\[
\frac{1}{z} = \frac{1}{2e^{i\theta}} = \frac{1}{2} e^{-i\theta}.
\]

Now, we can express $z + \frac{1}{z}$:
\[
z + \frac{1}{z} = 2e^{i\theta} + \frac{1}{2} e^{-i\theta}.
\]

To simplify this expression, we can combine the terms:
\[
z + \frac{1}{z} = 2e^{i\theta} + \frac{1}{2} e^{-i\theta} = 2\left(\cos(\theta) + i\sin(\theta)\right) + \frac{1}{2}\left(\cos(\theta) - i\sin(\theta)\right).
\]

This can be rewritten as:
\[
z + \frac{1}{z} = \left(2 + \frac{1}{2}\right)\cos(\theta) + i\left(2 - \frac{1}{2}\right)\sin(\theta) = \frac{5}{2}\cos(\theta) + \frac{3}{2}i\sin(\theta).
\]

Letting $x = \frac{5}{2}\cos(\theta)$ and $y = \frac{3}{2}\sin(\theta)$, we can derive the relationship between $x$ and $y$. From the equations, we can express $\cos(\theta)$ and $\sin(\theta)$ in terms of $x$ and $y$:
\[
\cos(\theta) = \frac{2x}{5}, \quad \sin(\theta) = \frac{2y}{3}.
\]

Using the Pythagorean identity $\cos^2(\theta) + \sin^2(\theta) = 1$, we substitute:
\[
\left(\frac{2x}{5}\right)^2 + \left(\frac{2y}{3}\right)^2 = 1.
\]

This simplifies to:
\[
\frac{4x^2}{25} + \frac{4y^2}{9} = 1.
\]

Multiplying through by 225 to eliminate the denominators gives:
\[
36x^2 + 100y^2 = 225.
\]

This is the equation of an ellipse in standard form. Therefore, the path traced by $z + \frac{1}{z}$ is an ellipse. The final answer is $(C)$.<|END_OF_LATENT|> The final answer is $(C)$.