To evaluate the integral of $x \, dz$ over a circle $C$ defined by $|z| = r$, we need to understand the expression properly.

Step 1: Parametrize the contour $C$

The contour $C$ is a circle of radius $r$, which means $|z| = r$. In the complex plane, this circle can be parametrized using the exponential form:

$z(t) = r e^{it}, \quad \text{where} \quad t \in [0, 2\pi]$

Thus, $dz = \frac{d}{dt}(r e^{it}) dt = i r e^{it} dt$.

Step 2: Express $x$ in terms of $z$

In your integral, we have $x \, dz$, but it's unclear whether $x$ refers to the real part of $z$, or some other function. Assuming that $x$ is the real part of $z$, we can express it as:

$x = \text{Re}(z) = r \cos(t)$

Step 3: Set up the integral

Now, we can express the integral as:

$\int_C x \, dz = \int_0^{2\pi} \left( r \cos(t) \right) \left( i r e^{it} \right) dt$

Simplifying the expression:

$\int_0^{2\pi} r^2 \cos(t) i e^{it} dt$

Since $e^{it} = \cos(t) + i \sin(t)$, we get:

$i e^{it} = i (\cos(t) + i \sin(t)) = -\sin(t) + i \cos(t)$

Therefore, the integrand becomes:

$r^2 \cos(t) \left( -\sin(t) + i \cos(t) \right)$

Expanding this:

$r^2 \left( -\cos(t) \sin(t) + i \cos^2(t) \right)$

Step 4: Compute the integral

Now, we split the real and imaginary parts of the integral:

$\int_0^{2\pi} r^2 \left( -\cos(t) \sin(t) + i \cos^2(t) \right) dt = r^2 \left( \int_0^{2\pi} -\cos(t) \sin(t) \, dt + i \int_0^{2\pi} \cos^2(t) \, dt \right)$

1. The first integral $\int_0^{2\pi} \cos(t) \sin(t) \, dt$ evaluates to 0, since it is an odd function over a symmetric interval.

2. The second integral $\int_0^{2\pi} \cos^2(t) \, dt$ can be computed using the identity $\cos^2(t) = \frac{1 + \cos(2t)}{2}$:

$\int_0^{2\pi} \cos^2(t) \, dt = \int_0^{2\pi} \frac{1 + \cos(2t)}{2} \, dt = \frac{1}{2} \left[ \int_0^{2\pi} 1 \, dt + \int_0^{2\pi} \cos(2t) \, dt \right]$

The first integral is simply $2\pi$, and the second integral vanishes, as $\int_0^{2\pi} \cos(2t) \, dt = 0$.

Thus, we have:

$\int_0^{2\pi} \cos^2(t) \, dt = \frac{1}{2} \times 2\pi = \pi$

Step 5: Final result

Combining both integrals, we get:

$\int_C x \, dz = r^2 \left( 0 + i \pi \right) = i \pi r^2$

Conclusion:

The value of the contour integral $\int_C x \, dz$ over the circle $|z| = r$ is:

$\boxed{i \pi r^2}$

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Would you like more details on any part of this process? Or have any further questions?

Related Questions:

1. How do we evaluate integrals of complex functions over different types of contours?
2. What is the significance of using parametric forms in contour integrals?
3. How do we handle integrals involving trigonometric functions in complex analysis?
4. What is the connection between contour integrals and Cauchy’s integral theorem?
5. How can we apply Green’s Theorem to complex integrals?

Tip:

When dealing with contour integrals, it’s helpful to parametrize the curve and express all functions in terms of that parametrization, which simplifies the integration process.