Math Problem Statement

Compute R |z|2dz, where γ is the straight line segment from 1 to 3 + 2i. γ

Solution

To compute the line integral γz2dz\int_{\gamma} |z|^2 \, dz, where γ\gamma is the straight line segment from 11 to 3+2i3 + 2i, we will follow these steps:

Step 1: Parametrize the line segment

The complex line segment from 11 to 3+2i3 + 2i can be parametrized as:

z(t)=1+t(3+2i1)=1+t(2+2i),t[0,1]z(t) = 1 + t(3 + 2i - 1) = 1 + t(2 + 2i), \quad t \in [0, 1]

Thus,

z(t)=1+2t+2it=(1+2t)+2itz(t) = 1 + 2t + 2it = (1 + 2t) + 2it

The derivative of z(t)z(t) with respect to tt is:

dzdt=2+2i\frac{dz}{dt} = 2 + 2i

Step 2: Express z(t)2|z(t)|^2

We now compute z(t)2|z(t)|^2. For a complex number z=x+iyz = x + iy, z2=x2+y2|z|^2 = x^2 + y^2.

For z(t)=(1+2t)+2itz(t) = (1 + 2t) + 2it, the real part x(t)=1+2tx(t) = 1 + 2t and the imaginary part y(t)=2ty(t) = 2t. Therefore:

z(t)2=(1+2t)2+(2t)2|z(t)|^2 = (1 + 2t)^2 + (2t)^2

Expanding both terms:

z(t)2=(1+4t+4t2)+4t2=1+4t+8t2|z(t)|^2 = (1 + 4t + 4t^2) + 4t^2 = 1 + 4t + 8t^2

Step 3: Set up the integral

We can now set up the integral:

γz2dz=01z(t)2dzdtdt\int_{\gamma} |z|^2 \, dz = \int_0^1 |z(t)|^2 \frac{dz}{dt} \, dt

Substitute z(t)2=1+4t+8t2|z(t)|^2 = 1 + 4t + 8t^2 and dzdt=2+2i\frac{dz}{dt} = 2 + 2i:

01(1+4t+8t2)(2+2i)dt\int_0^1 (1 + 4t + 8t^2)(2 + 2i) \, dt

Step 4: Break the integral into real and imaginary parts

We expand the integrand:

(1+4t+8t2)(2+2i)=(1+4t+8t2)(2)+(1+4t+8t2)(2i)(1 + 4t + 8t^2)(2 + 2i) = (1 + 4t + 8t^2)(2) + (1 + 4t + 8t^2)(2i)

This gives two parts:

(2+8t+16t2)+(2i+8it+16it2)(2 + 8t + 16t^2) + (2i + 8it + 16it^2)

Now, split the integral into real and imaginary parts:

Real part: 01(2+8t+16t2)dt\text{Real part: } \int_0^1 (2 + 8t + 16t^2) \, dt Imaginary part: 01(2+8t+16t2)idt\text{Imaginary part: } \int_0^1 (2 + 8t + 16t^2) \, i \, dt

Step 5: Compute the integrals

Real part:

01(2+8t+16t2)dt=012dt+018tdt+0116t2dt\int_0^1 (2 + 8t + 16t^2) \, dt = \int_0^1 2 \, dt + \int_0^1 8t \, dt + \int_0^1 16t^2 \, dt

[ = 2[1]_0^1 + 8 \left[\frac{t^2}{2}\right]_0^1 + 16 \left[\frac{t^3}{3}\right]_0^1 ]

=2+8×12+16×13=2+4+163=343= 2 + 8 \times \frac{1}{2} + 16 \times \frac{1}{3} = 2 + 4 + \frac{16}{3} = \frac{34}{3}

Imaginary part:

01(2+8t+16t2)dt=i(343)\int_0^1 (2 + 8t + 16t^2) \, dt = i \left( \frac{34}{3} \right)

Step 6: Final answer

The final answer is the sum of the real and imaginary parts:

γz2dz=343+i343=343(1+i)\int_{\gamma} |z|^2 \, dz = \frac{34}{3} + i \frac{34}{3} = \frac{34}{3}(1 + i)

Thus, the integral is:

343(1+i)\boxed{\frac{34}{3}(1 + i)}

Would you like more details on any part of this process or have additional questions?

Here are some related questions:

  1. What are some examples of line integrals in physics?
  2. How does the parametrization of a curve affect the integral?
  3. What is the geometric meaning of z2|z|^2 in this context?
  4. Can line integrals be computed over non-straight paths?
  5. How would the answer change if γ\gamma were a different path?

Tip: When parametrizing complex paths, always verify the derivative with respect to tt carefully to ensure accurate results.

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Math Problem Analysis

Mathematical Concepts

Complex Analysis
Line Integrals
Parametric Curves

Formulas

|z|² = x² + y² for a complex number z = x + iy
Line integral: ∫_γ f(z) dz = ∫_0^1 f(z(t)) (dz/dt) dt

Theorems

Parametrization of curves
Fundamental theorem of line integrals

Suitable Grade Level

Undergraduate