Evaluate \int_{C} tan z dz where C is |z|=2

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.

Learn how to evaluate the contour integral \(\int_{C} \tan z \, dz\) where \(C\) is the circle with radius 2 centered at the origin. This problem utilizes complex analysis concepts including the residue theorem and contour integration techniques. Suitable for advanced undergraduate students and anyone interested in complex analysis.

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