Question

The value of \( \int_0^{1+i} (x^2 - iy) dz \), along the path \( y = x^2 \) is:

• A. \( \frac{5}{6} - \frac{1}{6}i \)
• B. \( \frac{5}{6} + \frac{1}{6}i \)
• C. \( \frac{1}{6} - \frac{5}{6}i \)
• D. \( \frac{1}{6} + \frac{5}{6}i \)

Hint:

For complex line integrals, parameterization is key. Convert \(z, dz, f(z)\) into functions of a single real parameter (like \(x\) or \(t\)). Be very careful with complex number multiplication, especially with signs and \(i^2 = -1\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem requires evaluating a complex line integral along a specified path. The standard method is to parameterize the path and convert the complex integral into an integral of a single real variable.

Step 2: Key Formula or Approach:
1. Parameterize the path. The path is \( y = x^2 \). We can use \(x\) as the parameter. The path starts at \(z=0\) (i.e., \(x=0, y=0\)) and ends at \(z=1+i\) (i.e., \(x=1, y=1\)). So, \(x\) goes from 0 to 1. 2. Express everything in terms of the parameter \(x\). - \( z(x) = x + iy = x + ix^2 \) - \( dz = \frac{dz}{dx} dx = (1 + 2ix) dx \) - The integrand is \( f(z) = x^2 - iy = x^2 - ix^2 = x^2(1-i) \) 3. Substitute into the integral and evaluate. \[ \int_C f(z) dz = \int_{0}^{1} x^2(1-i) \cdot (1+2ix) dx \]
Step 3: Detailed Explanation:
We evaluate the integral by first expanding the product inside: \[ \int_{0}^{1} x^2(1-i)(1+2ix) dx = (1-i) \int_{0}^{1} x^2(1+2ix) dx \] \[ = (1-i) \int_{0}^{1} (x^2 + 2ix^3) dx \] Now integrate term by term: \[ = (1-i) \left[ \frac{x^3}{3} + 2i\frac{x^4}{4} \right]_{0}^{1} = (1-i) \left[ \frac{x^3}{3} + i\frac{x^4}{2} \right]_{0}^{1} \] \[ = (1-i) \left( \left(\frac{1}{3} + i\frac{1}{2}\right) - (0) \right) \] Now, multiply the complex numbers: \[ = (1-i) \left(\frac{1}{3} + \frac{i}{2}\right) = 1\left(\frac{1}{3} + \frac{i}{2}\right) - i\left(\frac{1}{3} + \frac{i}{2}\right) \] \[ = \frac{1}{3} + \frac{i}{2} - \frac{i}{3} - \frac{i^2}{2} = \frac{1}{3} + \frac{1}{2} + i\left(\frac{1}{2} - \frac{1}{3}\right) \] \[ = \frac{2+3}{6} + i\left(\frac{3-2}{6}\right) = \frac{5}{6} + \frac{1}{6}i \]
Step 4: Final Answer:
The value of the integral is \( \frac{5}{6} + \frac{1}{6}i \). This corresponds to option (B). (Note: If the path were \(y=x\), the answer would be \( \frac{5}{6} - \frac{1}{6}i \), which is option A. It's possible the question had a typo.)