Let $z$ be a nonreal complex number such that $|z| = 1.$  Find the real part of $\frac{1}{1 - z}.$
Let $z = x + yi,$ where $x$ and $y$ are real numbers.  Then from the condition $|z| = 1,$ $\sqrt{x^2 + y^2} = 1,$ so $x^2 + y^2 = 1.$

Now,
\begin{align*}
\frac{1}{1 - z} &= \frac{1}{1 - x - yi} \\
&= \frac{1 - x + yi}{(1 - x - yi)(1 - x + yi)} \\
&= \frac{1 - x + yi}{(1 - x)^2 + y^2} \\
&= \frac{1 - x + yi}{1 - 2x + x^2 + y^2} \\
&= \frac{1 - x + yi}{2 - 2x}.
\end{align*}The real part of this complex number is $\frac{1 - x}{2 - 2x} = \frac{1 - x}{2(1 - x)} = \boxed{\frac{1}{2}}.$
\frac{1}{2}
Level 4
Intermediate Algebra
Problem: Let $z$ be a nonreal complex number such that $|z| = 1.$  Find the real part of $\frac{1}{1 - z}.$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the real part of \(\frac{1}{1 - z}\), we start by expressing \(z\) in terms of its polar coordinates since \(|z| = 1\). Let \(z = e^{i\theta}\) for some angle \(\theta\). This gives us \(1 - z = 1 - e^{i\theta}\).

Next, we can rewrite \(1 - e^{i\theta}\) using Euler's formula:
\[
1 - e^{i\theta} = 1 - (\cos \theta + i \sin \theta) = (1 - \cos \theta) - i \sin \theta.
\]

To find \(\frac{1}{1 - z}\), we need to compute the reciprocal of the complex number \(1 - e^{i\theta}\). The reciprocal of a complex number \(a + bi\) is given by:
\[
\frac{1}{a + bi} = \frac{a - bi}{a^2 + b^2}.
\]

Applying this to our expression:
\[
\frac{1}{1 - e^{i\theta}} = \frac{(1 - \cos \theta) + i \sin \theta}{(1 - \cos \theta)^2 + \sin^2 \theta}.
\]

Now, we simplify the denominator:
\[
(1 - \cos \theta)^2 + \sin^2 \theta = 1 - 2\cos \theta + \cos^2 \theta + \sin^2 \theta = 1 - 2\cos \theta + 1 = 2(1 - \cos \theta).
\]

Thus, we have:
\[
\frac{1}{1 - e^{i\theta}} = \frac{(1 - \cos \theta) + i \sin \theta}{2(1 - \cos \theta)}.
\]

This can be separated into real and imaginary parts:
\[
\frac{1}{1 - e^{i\theta}} = \frac{1 - \cos \theta}{2(1 - \cos \theta)} + i \frac{\sin \theta}{2(1 - \cos \theta)} = \frac{1}{2} + i \frac{\sin \theta}{2(1 - \cos \theta)}.
\]

The real part of \(\frac{1}{1 - z}\) is therefore \(\frac{1}{2}\). This leads us to conclude that the final answer is \(\frac{1}{2}\).<|END_OF_LATENT|> The final answer is $\frac{1}{2}$.