Toronto Math Forum
MAT3342018F => MAT334Lectures & Home Assignments => Topic started by: Amarnath Parthiban on October 09, 2018, 04:32:49 PM

I can't seem to get the answer in the book.
$\int_\gamma z^2$ from $2$ to $ 3 + i$
I've attempted to substitute using $(3+i)t + 2(1t)$ but I think the problem is the $z$. From my understanding this is the magnitude of the complex number. Therefore, it should be cancelled out by the square. Since $\sqrt{x^2 + y^2} = z$. I have also tried subbing in $z\overline{z} = z^2$ to no avail.

Try using parmetrization, $ r(t) = 2 + (1+i)t, t \in [0,1] $
$ r(t) = 2 + (1 +i)t
= (2 +t)+ it \\
r'(t) = 1 + i $

You do not need to explicitly declare the components $x, y$ nor use $z\bar{z}$, just express them in terms of $t$ and solve using antiderivatives. Instead, convert $z$ into the parametric curve function $\gamma(t)$ so you have an integral in respect to $t$ quite familiar in calculus. To deal with the magnitude term, identify the real and imaginary components inside the absolute value bars.
Here, parameterize the curve $\gamma(t) = [2+t] + [it]$ where $t \in [0, 1]$. Note that the the number begins at $2 + 0i$ and both components increase by 1.
Hence, $\gamma(t) = t + 2 + it$, $f(\gamma) = \gamma^2$, and $\gamma'(t) = 1 + i$.
Then, we therefore have the integral $\int_0^1 f(\gamma(t))\gamma'(t) \:dt = \int_0^1t + 2 + it^2(1 + i) \:dt$, which will be quite familiar in earlier calculus courses. You can proceed easily from here after expressing the magnitude in terms of $t$.
Furthermore, note that this line integral is expressible as $[F(\gamma(t))]^1_0$ where $F$ is the antiderivative of $f$, since the integrand is of the form of the derivative of a function composition (that will be useful for similar line integral problems).
$\int_0^1 [t + 2] + [it]^2(1 + i) \:dt$
$= (1 + i)\int_0^1 [t + 2] + [it]^2 \:dt$, {$x=[t+2], y=[t]$}
$= (1 + i)\int_0^1 (t+2)^2 + (t)^2 \:dt$, {definition of magnitude}
$= (1 + i)\int_0^1 t^2 + 4t + 4 + t^2 \:dt$
$= (1 + i)\int_0^1 2t^2 + 4t + 4 \:dt$
$= (1 + i)[\frac{2}{3}t^3 + 2t^2 + 4t]^1_0$
$= (1 + i)[\frac{2}{3}(1) + 2(1) + 4(1)]  0$
$= (1 + i)[\frac{2}{3} + 6]$
$= (1 + i)[\frac{20}{3}]$