Lecture 21 - Integration, Antiderivatives, Homotopies

This lecture will cover how to integrate complex functions in the complex plane. It will be similar to the notion of line integral in $R^2$. In the next lecture, Cauchy’s Theorem will be proved, which is one of the cornerstones of complex analysis.

Integration in the Complex Plane

Definition of the Integral

${\int }_a^{b}f(t)dt:={\int }_a^b\mathrm{Re}f(t)dt+i{\int }_a^b\mathrm{Im}f(t)dt$
A more detailed definition of an integral is provided here.

Example of a Complex Integral

Consider the complex-valued function $f(t)=t+i{t}^2$. Then
${\int }_0^{1}f(t)dt={\int }_0^{1}tdt+i{\int }_0^1{t}^{2}dt=\frac{1}{2}+\frac{i}{3}$

Integral along a Curve

${\int }_{C}f(z)dz:={\int }_a^{b}f(\gamma (t)){\gamma }^{\prime }(t)dt.$
The full definition can be found here.

This notion is actually analogous to that of line integral of a integral of a vector field in ${\mathbb{R}}^2$. The reason is that we can identify complex numbers with vectors in $R^2$ by $z=x+iy\mapsto (x,y)$.

The meaning of ${\int }_C$

This is used to denote the integral along the curve. which has the parameters $[a,b]$, hence it gets simplified that way. It is the same as ${\int }_C\equiv {\int }_a^b$.

Note on future notation

The integral ${\int }_{C}f(z)dz$ will also be denoted simply by ${\int }_{C}f$.

Example of Integrating a Complex Curve

Consider the complex curve $C$ parametrized by $\gamma (t)=1+it$ with $0\le t\le 1$. Given the complex function $f(z)=z$, we compute
${\int }_{C}f={\int }_0^1(1+it)idt=-{\int }_0^{1}tdt+i{\int }_0^{1}dt=-\frac{1}{2}+i.$

Note

The integral ${\int }_{C}f$ depends on the curve $C$, in general.

Reparametrizations

The complex integral ${\int }_{C}f$ was defined by choosing a smooth parametrization $\gamma :[a,b]\to \mathbb{C}$ of $C$. But the question is how does its value depend on such a choice? To figure this out, we need to introduce some terminology regarding curves.

Definitions

• Reparametrization
• Same Orientation
• Opposite Orientation

Example of Reparametrization

Let $C$ be the line segment with endpoints $0$ and $4i$. An obvious parametrization for this curve is given by ${\gamma }_1:[0,4]\to \mathbb{C}$ defined by ${\gamma }_1(t)=it$.

Now consider ${\gamma }_2:[0,2]\to \mathbb{C}$ defined by ${\gamma }_2(t)=i{t}^2$.
We claim that ${\gamma }_2$ is a reparamatrization of ${\gamma }_1$ and with the same orientation. To show this, we observe that we have ${\gamma }_2(t)={\gamma }_1(\alpha (t))$ with $\alpha (t)=t^2$ and moreover ${\alpha }^{\prime }(t)=2t\ge 0$ for all $t\in [0,2]$.

On the other hand consider the parametrization ${\gamma }_3:[0,4]\to \mathbb{C}$ defined by ${\gamma }_3(t)=i(4-t)$.
We have ${\gamma }_3(t)={\gamma }_1(\beta (t))$ with $\beta (t)=4-t$ and ${\beta }^{\prime }(t)=-1$.
We conclude that ${\gamma }_3$ is a reparamatrization of ${\gamma }_1$ with opposite orientation.

Proposition

Let $C$ be a curve with a smooth parametrization $\gamma :[a,b]\to \mathbb{C}$. If $\sigma :[c,d]\to \mathbb{C}$ is a reparametrization of $\gamma$ with the same orientation we have
${\int }_a^{b}f(\gamma (t)){\gamma }^{\prime }(t)dt={\int }_c^{d}f(\sigma (t)){\sigma }^{\prime }(t)dt.$
If instead $\sigma$ has opposite orientation, the two integrals are equal up to minus sign..

Let ${\int }_{C}f$ be defined with respect to a parametrization $\gamma$. If we write ${\int }_{-C}f$ for the integral computed with, to a reparametrization $\sigma$ with opposite orientation as above, we have
${\int }_{-C}f=-{\int }_{C}f.$
Notice that this is analogous to what we have for the line integral of a vector field in $R^2$.

Example

Let $\gamma :[a,b]\to \mathbb{C}$ be a smooth parametrization of $C$. We define the opposite parametrization ${\gamma }_{op}:[-b,-a]\to \mathbb{C}$ by setting ${\gamma }_{op}(t)=\gamma (-t)$.

It starts at ${\gamma }_{op}(-b)=\gamma (b)$ and ends at ${\gamma }_{op}(-a)=\gamma (a)$, hence it goes along the curve £C£ in the opposite direction of $\gamma$. We write the integral with respect to this parametrization as
${\int }_{C}f={\int }_{-b}^{-a}f({\gamma }_{op}(t)){\gamma }_{op}^{\prime }(t)dt.$
Now we introduce the variable $u=-t$ and observe that
${\gamma }_{op}^{\prime }=-{\gamma }^{\prime }(u).$
Then change the of variable gives
${\int }_{-C}f={\int }_b^{a}f({\gamma }_{op}(-u))(-{\gamma }^{\prime }(u))(-du)$
$=-{\int }_a^{b}f(\gamma (u)){\gamma }^{\prime }(u)du=-{\int }_{C}f.$
Hence integrals with respect to $\gamma$ and ${\gamma }_{op}$ differ by a minus sign.

Antiderivatives

Regarding the rest of this page

All the following curves will be considered as smooth, and that they will be simple, meaning that the curve does not cross itself.

tl;dr: “smooth simple curve” will be referred as “curve” (or “path”).

Recall that in real analysis, antiderivatives can be used to conveniently compute integrals. This notion also makes sense for complex functions.

Definition

Let $F:D\to \mathbb{C}$ be a holomorphic function. Suppose $F^{\prime }(z)=f(z)$ for all $z\in D$. Then $F$ is an antiderivative of $f$ on $D$.

We have the following analogue of the fundamental theorem of calculus.

Theorem

Let $f:D\to \mathbb{C}$ be continuous. Suppose that $F$ is an antiderivative of $f$ on $D$. Let $\gamma :[a,b]\to \mathbb{C}$ be a parametrization of a curve $C$ in the region $D$. Then
${\int }_{C}f=F(\gamma (b))-F(\gamma (a)).$

Proof

Using the chain rule we compute
$\frac{dF(\gamma (t))}{dt}=F^{\prime }(\gamma (t)){\gamma }^{\prime }(t)=f(\gamma (t)){\gamma }^{\prime }(t),$
where we used that $F$ is an antiderivative of $f$, that is $F^{\prime }(z)=f(z)$. Hence
${\int }_{C}f={\int }_a^{b}f(\gamma (t)){\gamma }^{\prime }(t)dt={\int }_a^b\frac{dF(\gamma (t))}{dt}dt.$Now we can use the fundamental theorem of calculus (strictly speaking after splitting $F$ into its real and complex parts), the chain rule, to conclude that
${\int }_{C}f=F(\gamma (b)-F(\gamma (a))).$

Remark

Continuing our comparison between complex numbers and $R^2$, we see that functions admitting an antiderivative are the analogue of conservative vector fields.

Example

Consider $f(z)=z$. It is easy to check that $F(z)=\frac{z^2}{2}$ is an antiderivative of $f$ (for any $z$). Let $C$ be any curve starting at $1$ and ending at $1+i$. Then we get
${\int }_{C}f=\frac{(1+i{)}^2}{2}-\frac{1^2}{2}=-\frac{1}{2}+i.$
In particular, this gives another way to do the computation in Example of a Complex Integral.

Homotopies

This will be used for proving Cauchy’s Theorem.

Definition of a Homotopy

Let ${\gamma }_1(t)$ and ${\gamma }_2(t)$ be two closed (parametrized) curves in a region $D$, with parameter $0\le t\le 1$.
They are called homotopic if there exists a continuous function $h:[0,1]\times [0,1]\to D$ such that, for all $s,t\in [0,1]$, we have
$h(t,0)={\gamma }_1(t),h(t,1)={\gamma }_2(t),$
$h(0,s)=h(1,s).$
In this case we use the notation ${\gamma }_1\sim {\gamma }_2$.

Conditions

The first two conditions mean that the curve $h$ interpolates between the two curves ${\gamma }_1$ and ${\gamma }_2$.
The third condition means that for fixed $s$ in the curve $h(t,s)$ is closed.

Let us also note that the notion of homotopy makes sense for any paths, not necessarily closed. Here we insist that the interpolating curves are also closed.

Example of Homotopy

a circle in the complex plane of radius $R$ (and centred at zero) is given by those $z\in \mathbb{C}$ such that $|z|=R$.
We can parametrize this by
$\gamma (t)=R{e}^{2\pi it},0\le t\le 1.$
consider the curves ${\gamma }_1(t)=e^{2\pi it}$ and ${\gamma }_2(t)=2e^{2\pi it}$, corresponding to radii $1$ and $2$ respectively.
A homotopy between these two closed curves is given by
$h(t,s)=(1+s)e^{2\pi it}.$
Indeed it is continuous, we have $h(t,0)={\gamma }_1(t)$ and $h(t,1)={\gamma }_2(t)$, and for any fixed $s$ we have that $h(t,s)$ parametrizes a circle (of radius $1+s$).