The integral to evaluate is $\int_{C} \tan z \, dz$, where $C$ is the contour defined by $|z| = 2$. This is a closed contour integral, and we'll apply the residue theorem to solve it.

Step 1: Identify Singularities

The function $\tan z$ can be written as:

$\tan z = \frac{\sin z}{\cos z}$

The singularities occur where the denominator is zero, i.e., where $\cos z = 0$.

The cosine function is zero at $z = \frac{\pi}{2} + n\pi$ for any integer $n$. Hence, the singularities of $\tan z$ are at:

$z = \frac{\pi}{2} + n\pi \quad \text{for } n \in \mathbb{Z}$

Step 2: Determine Which Singularities Are Inside the Contour

The contour $C$ is the circle $|z| = 2$. We need to identify which singularities lie within this circle.

For $n = 0$, we have $z = \frac{\pi}{2}$ and for $n = -1$, we have $z = -\frac{\pi}{2}$. Both of these values are inside the circle $|z| = 2$, as:

$\left|\frac{\pi}{2}\right| \approx 1.57 < 2$ $\left|-\frac{\pi}{2}\right| \approx 1.57 < 2$

So, the singularities within the contour are $z = \frac{\pi}{2}$ and $z = -\frac{\pi}{2}$.

Step 3: Calculate the Residues

The residues of $\tan z$ at these singularities can be computed using:

$\text{Residue}\left( \tan z, z_0 \right) = \lim_{z \to z_0} (z - z_0) \tan z$

For $z = \frac{\pi}{2}$ and $z = -\frac{\pi}{2}$, these residues are:

1. Residue at $z = \frac{\pi}{2}$:

$\text{Residue}\left(\tan z, \frac{\pi}{2}\right) = \lim_{z \to \frac{\pi}{2}} \left(z - \frac{\pi}{2}\right) \tan z = \lim_{z \to \frac{\pi}{2}} \frac{z - \frac{\pi}{2}}{\cos z} = \frac{1}{\cos'(\frac{\pi}{2})}$ Since $\cos'(\frac{\pi}{2}) = -\sin(\frac{\pi}{2}) = -1$, $\text{Residue}\left(\tan z, \frac{\pi}{2}\right) = \frac{1}{-1} = -1$

2. Residue at $z = -\frac{\pi}{2}$:

$\text{Residue}\left(\tan z, -\frac{\pi}{2}\right) = \lim_{z \to -\frac{\pi}{2}} \left(z + \frac{\pi}{2}\right) \tan z = \lim_{z \to -\frac{\pi}{2}} \frac{z + \frac{\pi}{2}}{\cos z} = \frac{1}{\cos'(-\frac{\pi}{2})}$ Since $\cos'(-\frac{\pi}{2}) = -\sin(-\frac{\pi}{2}) = 1$, $\text{Residue}\left(\tan z, -\frac{\pi}{2}\right) = \frac{1}{1} = 1$

Step 4: Apply the Residue Theorem

The residue theorem states:

$\int_{C} \tan z \, dz = 2\pi i \times \sum \text{Residues inside } C$

Thus, the integral is:

$\int_{C} \tan z \, dz = 2\pi i \times \left( -1 + 1 \right) = 2\pi i \times 0 = 0$

Final Answer:

$\int_{C} \tan z \, dz = 0$

Would you like any further explanation on any part of this solution?

Here are some related questions you might find interesting:

1. How do you evaluate the contour integral $\int_{C} \frac{1}{z} dz$ where $C$ is $|z| = 1$?
2. What are the general steps to apply the residue theorem?
3. How do you compute the residue of a pole of order higher than 1?
4. Can the residue theorem be applied to contours that are not circles?
5. How does the result change if the contour was $|z| = 1$ instead?

Tip: When applying the residue theorem, always carefully identify the singularities inside the contour and calculate their residues accurately.