Definition

A *function* \(f\) defined on a set \(S \subseteq \mathbb C\) is a *rule* that assigns to each \(z \in S\) a complex number \(w\), called the *value* of \(f\) at \(z\) and is denote by \(f(z)\).

Definition

We call the set \(S\) the *domain of definition* of \(f\). Note we can't use the term "domain" because we've already defined it previously to mean an open, nonempty, and connected set.

Note that both the domain of definition and the rule should be specified for a complete definition of a function. If the domain of definition is *not* given, it is assumed to be the largest possible set.

Example

For example, the function \(w = f(z) = \frac{1}{z} = z^{-1}\) is well defined except where \(z = 0\), so the domain of definition is \(\mathbb C \setminus \{0\}\).

Example

Suppose we have a function \(w = f(z)\) for \(z = x + iy\) and \(w = u + iv\). Then we have \(u + iv = f(x + iy)\), so we can express the complex-valued function \(f\) in terms of a pair of real-valued function of \(x\) and \(y\).

\[f(z) = u(x, y) + iv(x, y)\]

And if we use polar coordinates

\[f(re^{i\theta}) = u + iv\]

we then have

\[f(re^{i\theta}) = u(r, \theta) + iv(r, \theta)\]

Example

For example, take the function \(f(z) = z^2\). Then we have \(f(x + iy) = (x + iy)^2 = x^2 - y^2 + 2ixy\). So then

\[u(x, y) = x^2 - y^2\]

and

\[v(x, y) = 2xy\]

If we use polar coordinates,

\[f(re^{i\theta}) = (re^{i\theta})^2 = r^2 e^{i2\theta} = r^2 \cos{2\theta} + ir^2 \sin{2\theta}\]

So

\[u(r, \theta) = r^2 \cos{2\theta}\]

and

\[v(r, \theta) = r^2 \sin{2\theta}\]

Definition

If \(v\) every *always* equals 0, we call \(f\) a *real-valued function* of a complex variable.

Definition

A *polynomial of degree \(n\)* is the function

\[P(z) = a_0 + a_1 z + a_2 z^2 + \dots + a_n z^n\]

for \(a_i \in \mathbb C, n \in \mathbb Z^+, \text{ and } a_n \neq 0\). Note that the requirement of \(a_n \neq 0\) simply means that the \(n\)th degreen polynomial *actually has an \(n\)th term*.

Definition

Quotients \(\displaystyle \frac{P(z)}{Q(z)}\) are called *rational functions* and are defined for all \(z\) such that \(Q(z) \neq 0\).

Definition

The *image* of a point \(z_0\) in the domain of definition \(S\) of a function \(f\) is the point \(w_0 = f(z_0)\). The term *image* of \(z\) is equivalent to the term *value* of \(f\) at \(z\).

We can also take images of *sets*. Let \(T \subseteq S\), then the *image* of \(T\) is the set of images of all the points in \(T\). We denote this by \(f(T)\).

Definition

The *range* of a function \(f\) is the image of the domain of definition \(S\). We denote this by \(f(S)\).

Definition

The *inverse image*, or *preimage* of a point \(w\) is the set of all \(z \in S\) that have an image of \(w\). Note that the preimage can have one, or many points in \(S\). My professor denotes this as \(f^{-1}(\{w\})\) to emphasize that this is *not* the inverse function of \(f\), it's just a set. Because the range is defined as the image of the domain of definition, the preimage is guaranteed to be nonempty.

Definition

A function \(f\) is called *injective*, or *one-to-one* if whenever \(f(z_1) = f(z_2)\), then \(z_1 = z_2\).

Definition

A function \(f\) is called *surjective*, or *onto* if for every \(w\) in the range of \(f\) there exists a \(z\) in the domain of definition of \(f\) such that \(f(z) = w\).

Note that how we have defined the range of a function causes every function to be surjective. This is in conflict with most textbooks that also define something called a *codomain*, or which the range of \(f\) is a subset. However, this is what my particular textbook uses, so it's what I will stick with here.

Example

The image \(w\) of a point \(z\) in the domain of definition \(S\) of a function \(f\) with range \(f(S)\).

Example

The following is an example of a function that is *not* one-to-one (injective). Note that \(f(z_1) = f(z_2)\) yet \(z_1 \neq z_2\).

Example

When we take the inverse image (preimage) of a point in the range of a function, we *don't change the direction of the arrows* we aren't finding an inverse function, only tracing the line(s) back to where they came from.

Note that the preimage is a *set* that could possibly contain one or more points in the domain of definition.

Graphing complex functions is *hard*. We have 2 dimensional inputs, *and* 2 dimensional outputs. That means we're working with a 4 dimensional function — *even for the simple ones.* Since \(w = f(z)\) is a complex number, both \(z\) and \(w\) can be interpreted on points in a plane. Therefore, we can draw the \(z\) and \(w\) planes separately. We can therefore view a function \(w = f(z)\) as a *transformation*, or a *mapping* from points in the \(z\)-plane to points in the \(w\)-plane.

We can describe these transformations with the terms *translation*, *rotation*, *reflection*, and *scaling*, or *stretching*.

Example

\(w = f(z) = z + 1 = (x + 1) + iy\) is a function that *shifts*, or *translates* points one unit to the right.

Example

Since we know that \(i = e^{i\frac{\pi}{2}}\), the function \(f(z) = iz = ire^{i\theta} = e^{i\frac{\pi}{2}}re^{i\theta} = re^{i\theta + \frac{\pi}{2}}\) actually rotates a point \(z\) \(\frac{\pi}{2}\) radians -- that is, rotates numbers 90 degrees counterclockwise.

Example

The function \(f(z) = \bar z = x - iy\) is a simple reflection about the real axis.

Example

Now consider again the function \(f(z) = z^2\). We can rewrite \(f\) in terms of real-valued functions

\[u = x^2 - y^2\]

and

\[v = 2xy\]

And think of \(f\) as a transformation from the \(x\)-\(y\) plane to the \(u\)-\(v\) plane. Because we're so unfamiliar with complex functions, even for a function so simple as \(z^2\), we'll ask ourselves a series of questions about it.

*I wonder what maps to a constant?* Just for the heck of it, we'll set \(u = c\) for some \(c > 0\) and see what falls out.

That means we have points on the family of hyperbolas \(x^2 - y^2 = c\), or \(\displaystyle \frac{x^2}{c} - \frac{y^2}{c} = 1\) in the \(x\)-\(y\) plane being mapped to points along the straight line \(u = c\) in the \(u\)-\(v\) plane.

Now what happens if we let \(u = -c\)? Then we have points on the family of hyperbolas \(x^2 - y^2 = -c\), or \(\displaystyle \frac{y^2}{c} - \frac{x^2}{c} = 1\)

That all is if we fix \(u\) to be some constant, what happens if we fix \(v\) to be some positive constant? Then we have points along the \(2xy = c\), or \(\displaystyle y = \frac{c}{2x}\) family of hyperbolas in the \(x\)-\(y\) plane being mapped to some positive \(v = c\) in the \(u\)-\(v\) plane.

Again, what if we pick a negative constant? Then we have \(\displaystyle y = \frac{-c}{2x}\) being mapped to some negative constant \(v\) in the \(u\)-\(v\) plane.

We asked ourselves what maps to a constant, now we ask, what does a constant map to? That is, let \(x = c\) and \(y\) vary. Then \(u = c^2 - y^2\) and \(v = 2cy\) which combine to make \(\displaystyle u = c^2 - \frac{v^2}{4c^2}\)