2.3 Functions

Home | Assessment | Notes | Index | Worksheets | Blackboard

Our main objects of study in this course are complex-valued functions $f$ where the inputs come from a domain $D\subset \mathbb{C}$. We will mainly be interested in differentiating and integrating such functions.

Figure 1: A function defined on a domain $D$ maps each complex number $z\in D$ to a complex number $f(z)\in \mathbb{C}$.

Example

Often, functions are defined by formulae.

1. $f(z)=z^2$ defines a function $f:\mathbb{C}\to \mathbb{C}$.
2. $f(z)={\displaystyle \frac{1}{z}}$ defines a function $f:\mathbb{C}\setminus \{0\}\to \mathbb{C}$.
3. $f(z)={\displaystyle \frac{z}{1+z^2}}$ defines a function $f:\mathbb{C}\setminus \{i,-i\}\to \mathbb{C}$.

More complicated functions such as trigonometric and exponential functions will be formally defined by power series later on in the course. A complex version of the natural logarithm will have to wait until we have discussed integration!

Given a function $f:D\to \mathbb{C}$ we can think of $D$ as a subset of ${\mathbb{R}}^2$ and of $\mathbb{C}$ as the whole plane ${\mathbb{R}}^2$. Writing $z=x+iy$ and $$f(z)=f(x+iy)=u(x,y)+iv(x,y)$$ lets us think of $f$ as being made up of two real-valued functions $u,v:D\to \mathbb{R}$.

Example

What are the real and imaginary parts of $f(z)=z^2$? Writing $z=x+iy$ we have $$f(x+iy)=(x+iy{)}^2=x^2-y^2+2xyi$$ so $u(x,y)=x^2-y^2$ and $v(x,y)=2xy$.

We cannot easily draw the graph of a function $f:D\to \mathbb{C}$. We conclude this section with a brief discussion of two alternative methods for visualizing such a function.

The first method is to plot the level curves of the real and imaginary parts $u$ and $v$ of $f:D\to \mathbb{C}$. Recall that the level curves of a function $h:D\to \mathbb{R}$ are the curves $h(x,y)=k$ for different values of $k$. We plot level curves in the domain of the function. If the input $z$ is nudged along a level curve of $u$ the real part of the output will not change. Similarly, if the input is nudged along a level curve of $v$ then the imaginary part of the output will not change.

Figure 2: The level curves of the real and imaginary parts $u(x,y)=x^2-y^2$ and $v(x,y)=2xy$ of $f(z)=z^2$ respectively.

The second method is to think of $f$ as a vector field on $D$. Indeed, for every $z\in D$ the output $f(z)$ can be thought of as a vector in ${\mathbb{R}}^2$. Placing the tail of this vector at the point $z\in D$ lets us think of $f$ as a vector field on $D$.

Figure 3: The function $f(z)=z^2$ as the vector field $⟨x^2-y^2,2xy⟩$ on ${\mathbb{R}}^2$.