How many complex numbers $z$ such that $\left| z \right| < 30$ satisfy the equation
\[
  e^z = \frac{z - 1}{z + 1} \, ?
\]
Let $z = x + yi$, where $x$ and $y$ are real. Then
$$|e^z| = |e^{x+yi}| = |e^x \cdot e^{iy}| = |e^x| \cdot |e^{iy}| = e^x \cdot 1 = e^x.$$So $e^z$ is inside the unit circle if $x < 0$, is on the unit circle if $x = 0$, and is outside the unit circle if $x > 0$.

Also, note that $z$ is closer to $-1$ than to $1$ if $x < 0$, is equidistant to $1$ and $-1$ if $x = 0$, and is closer to $1$ than to $-1$ if $x > 0$. So $\frac{z-1}{z+1}$ is outside the unit circle (or undefined) if $x < 0$, is on the unit circle if $x = 0$, and is inside the unit circle if $x > 0$.

Comparing the two previous paragraphs, we see that if $  e^z = \frac{z - 1}{z + 1},$ then $x = 0$.  So $z$ is the purely imaginary number $yi$.

Also, note that $z$ satisfies the original equation if and only if $-z$ does. So at first we will assume that $y$ is positive, and at the end we will double the number of roots to account for negative $y$. (Note that $y \ne 0$, because $z = 0$ is not a root of the original equation.)

Substituting $z = yi$ into the equation $  e^z = \frac{z - 1}{z + 1}$ gives the new equation
$$  e^{iy} = \frac{iy - 1}{iy + 1}.$$By the first two paragraphs, we know that both sides of the equation are always on the unit circle. The only thing we don’t know is when the two sides are at the same point on the unit circle.

Given a nonzero complex number $w$, the angle of $w$ (often called the argument of $w$) is the angle in the interval $[0, 2\pi)$ that the segment from $0$ to $w$ makes with the positive $x$-axis. (In other words, the angle when $w$ is written in polar form.)

Let’s reason about angles. As $y$ increases from $0$ to $\infty$, the angle of $iy -1$ strictly decreases from $\pi$ to $\frac{\pi}{2}$, while the angle of $iy+1$ strictly increases from $0$ to $\frac{\pi}{2}$. So the angle of $\frac{iy - 1}{iy + 1}$ strictly decreases from $\pi$ to $0$.

Let $n$ be a nonnegative integer. We will consider $y$ in the interval from $2n\pi$ to $(2n + 2)\pi$.  As $y$ increases from $2n\pi$ to $(2n + 1)\pi$, the angle of $e^{iy}$ strictly increases from $0$ to $\pi$.  As $y$ increases from $(2n+ 1)\pi$ to just under $(2n+ 2)\pi$, the angle of $e^{iy}$ strictly increases from $\pi$ to just under $2\pi$.

Comparing the angle information for  $\frac{iy - 1}{iy + 1}$ and $e^{iy}$ above, we see that $\frac{iy - 1}{iy + 1}$ and $e^{iy}$ are equal for exactly one $y$ in $(2n\pi,(2n + 1)\pi)$, and for no $y$ in $[(2n + 1)\pi,(2n + 2)\pi]$. So we have exactly one root of $y$ in each of $(0, \pi)$, $(2\pi, 3\pi), (4\pi, 5\pi), (6\pi, 7\pi)$, and $(8\pi, 9\pi)$. That gives $5$ positive roots for $y$. We don’t have to go further because $9\pi < 30 < 10\pi$.

Because we have $5$ positive roots for $y$, by symmetry we have $5$ negative roots for $y$. Altogether, the total number of roots is $\boxed{10}$.