Evaluate the following complex integrals using parameterisation $\int \limits_{\beta} f(z) dz$

Question

Evaluate the following complex integrals using parameterisation

$$\int \limits_{\beta} f(z) dz$$

where $\beta$ represents the line segment from $-i$ to $2+5i$, followed by the line segment from $2+5i$ to $5i$, and $f(x+yi)=iy+x^2$

What i tried

I tried using parameterisation such as letting $z=t+it$ but i couldnt find a suitable one. Also for the line segment from $-i$ to $2+5i$ can i split it to become $-i$ to $5i$ and then from $5i$ to $2+5i$ . And then parameterise each one individually? Is there a general approach to solving this type of questions. Could anyone please explain. Thanks

Answer 1

So you have an integral $\int_\beta f(z)\, dz$ where $\beta$ represents two line segments. First, let's parameterize each one:

$C_1$ goes from $-i$, which means the point is $(0,-1)$, to $2+5i$, which means the point is $(2,5)$, so:

\begin{align} C_1(t) &= (0,1) + t[(2, 5)-(0,-1)] \\ C_1(t) &= (2t, 6t-1). \end{align}

$C_2$ goes from $2+5i$, which means the point is $(2,-5)$, to $5i$, which means the point is $(0,5)$, so:

\begin{align} C_2(t) &= (2,5) + t[(0, 5)-(2,5)] \\ C_2(t) &= (2-2t, 5). \end{align}

Perfect, you have parametrized $\beta$, now the question is: how do you include this in the integral? You have this function $f(z) = x+iy$ and you have to convert it to $z$ notation, such as this:

\begin{align} f(z) &= x^2 +iy \\ f(z) &= x -x + x^2 + iy \\ f(z) &= x + iy -x + x^2 \\ f(z) &= z - \dfrac{z + \bar z}{2} + \left(\dfrac{z + \bar z}{2} \right)^2. \end{align}

And now the integral is:

\begin{align} \int_\beta f(z) \, dz &= \int_\beta \left[z - \dfrac{z + \bar z}{2} + \left(\dfrac{z + \bar z}{2} \right)^2 \right] \,dz \\ &= \int_{C_1} \left[z - \dfrac{z + \bar z}{2} + \left(\dfrac{z + \bar z}{2} \right)^2 \right] \,dz \\ &+ \int_{C_2} \left[z - \dfrac{z + \bar z}{2} + \left(\dfrac{z + \bar z}{2} \right)^2 \right] \,dz. \end{align}

Before you put $C_1(t)$ and $C_2(t)$, you have to convert them to $z$ notation, which might be strange but it is easy:

\begin{align} C_1(t) &= z_1(t) = 2t + i(6t-1) \\ C_2(t) &= z_2(t) = 2 -2t + 5i. \end{align}

Let's do the first intregal:

\begin{align} \int_{C_1} \left[z - \dfrac{z + \bar z}{2} + \left(\dfrac{z + \bar z}{2} \right)^2 \right] \,dz \\ &= \int_0^1 \left(2t + i(6t-1) -2t + 4t^2 \right) (2+i6)\,dt \\ &= -\dfrac{28}{3} + 12i. \end{align}

By the way, $dz_1 = (2+6i)dt$. The second integral is:

\begin{align} \int_{C_2} \left[z - \dfrac{z + \bar z}{2} + \left(\dfrac{z + \bar z}{2} \right)^2 \right] \,dz \\ &= \int_0^1 \left(2 -2t + 5i -(2-2t) + (2-2t)^2 \right) (-2)\,dt \\ &= -\dfrac{8}{3} - 10i. \end{align}

By the way, $dz_2 = -2 dt$. Therefore, the result is $-12 +2i$.

Answer 2

General: the line segment from $a$ to $b$ has the parametrisation

$C(t)=a+t(b-a)$ with $t \in [0,1]$.

Answer 3

As real vector spaces, $\mathbb C$ and $\mathbb R^2$ are identical. Finding parameterizations of curves in $\mathbb C$ is thus the same as finding them in $\mathbb R^2.$ I point this out to emphasize that this part of complex analysis is no different from what you probably saw when studying vector calculus.

Let $p,q\in \mathbb R^2.$ Then the line in $\mathbb R^2$ through $p$ and $q$ is the set of points $\{p+t(q-p): t\in \mathbb R\}.$ The line segment from $p$ to $q$ is the part of the line corresponding to $t\in [0,1].$ Thus the "standard parameterization" of the line segment $[p,q]$ is the curve $\gamma :[0,1]\to \mathbb R^2$ defined by $\gamma (t) = p +t(q-p).$ It's very important for you to know this.

It should be noted that there are other ways to parameterize $[p,q].$ For example $[(0,0),(1,1)]$ can be parameterized as above, but also as $\gamma(t) = (t^2,t^2), t\in [0,1].$ But the the standard parameterization is the most basic and easiest of them in most cases.

So with $p=-i=(0,-1), q = 2+5i = (2,5), r = 5i = (0,5),$ you're set up to find the value of the given contour integral.