We enter the topic of complex analysis through the usual lens: trying to take rational powers of real numbers. Within the domain of real numbers, \( (-1)^{1/2} \) is not defined. It turns out that if we deal with this example, we create a number system that is closed under rational powers. We define

so that \( i^2 = -1 \).

A complex number is any expression of the form

We call \( x = Re(z) \in \Reals \) the real part of \( z \) and \( y = Im(z) \in \Reals \) the imaginary part. The set of all complex numbers is denoted by \( \Complex \) and we may write \( \Complex = \Reals + i \Reals \). Thus a complex number is identified by two real numbers. Gauss realized that a complex number can be geometrically represented by a vector in \( \Reals ^2 \).

As we will see in the course, this analogy goes very far and is very useful. Of course, the above implies that we can represent complex numbers using polar coordinates in \( \Reals ^2 \) just as well. Let

Then \( z \rho (\cos \theta + i \sin \theta) \) and \( \rho = \sqrt{x^2 + y^2} \colonequals \abs{z} \), the modulus or absolute value of \( z \). Further, \( \theta \colonequals arg(z) \), the argument of \( z \). We have

where \( n \) is any integer and the quadrant has to be specified so as to be able to determine whether \( \alpha = 0 \) or \( 1 \).

This extension now means that every polynomial has a zero. Previously, we would say \( x^2 + 1 \) has no zero. But now, we can understand why the series

diverges at \( |x| = 1 \) .

If \( z \in \Complex \) then the complex conjugate \( \overline{z} \) is defined as \( \overline{z} \colonequals x - iy = \rho (\cos \theta - i \sin \theta) = \rho(\cos (-\theta) + i \sin(-\theta)) \). These equalities show that the complex conjugate \( \overline{x} \) of \( z \) is found in the Gaussian complex plane by reflecting \( z \) with respect to the real axis.

Note that

This is useful to determine the real and imaginary part of the division of two complex numbers

Division is only defined for 1, 2, and 4 dimensional complex numbers. For quaternions:

\[(a + bi + cj + k)^{-1} = \frac{a - bi - cj - dk}{a^2 + b^2 + c^2}\] where \[i^2 = j^2 = k^2 = -1 \\ ij = k \\ ji = -k \\ kj = i \\ kj = -i \\ ki = j \\ ik = -j\]

Quaternions do not have commutative multiplication.

The triangle inequality will come up very often.

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Theorem 1.1

For any \( z_1, z_2 \in \Complex \) we have

\[\abs{z_1 + z_2} \leq \abs{z_1} + \abs{z_2} \qquad (i)\] \[\abs{z_1 - z_2} \geq \abs{\abs{z_1} - \abs{z_2}} \qquad (ii)\]

Proof

We prove (i) first. Consider

\[|z_1 + z_2| ^2 - (|z_1| + |z_2|^2) = \cancel{|z_1|^2} + \cancel{|z_2|^2} + z_1 \overline{z_2} + \overline{z_1} z_2 - \cancel{|z_1|^2} - \cancel{|z_2|^2} - 2 |z_1| |z_2| \\ = z_1 \overline{z_2} + \overline{z_1 \overline{z_2}} - 2 |z_1| |\overline{z_2}| \\ = 2 (Re(z_1 \overline{z_2}) - |z_1 \overline{z_2}|) \\ \geq 0\]

It follows that \( |z_1 + z_2| - |z_1| \geq |z_2| \) . Let \( w_1 \colonequals z_1 + z_2 \) and \( w_2 \colonequals z_1 \). Then the first inequality becomes \( |w_1| - |w_2| \leq |w_1 - w_2| \). Similarly, by switching the indices, we get \( |w_2| - |w_1| \leq |w_2 - w_1| \). These results combined give (ii).

The triangle inequality is generalized to

The geometrical meaning of the triangle inequality is just that the length of any side of a triangle is shorter than the sum of the lengths of the other two sides.

At this point, let’s skip ahead a bit to Euler’s equation. We define the complex exponential function

For real \( z \) this coincides with the real exponential that we know. For imaginary \( z \), we have

As a proof, let \( f(\theta) = \cos \theta + i \sin \theta \). Since both \( \sin \theta \) and \( \cos \theta \) satisfy the second order equation

Then

With the initial condition

The exponential function should have the property

So \( e^{i \theta} \) satisfies \( \left[ \pdv{^2}{\theta ^2} + 1 \right] e^{i \theta} = 0 \). Since this is a solution to the same equation with the same initial conditions, via the uniqueness theorem for 2nd order linear PDEs \( f(\theta) = e^{i \theta} \)

Limits, continuity, and differentiation #

To recap, we consider \( w = f(z) \) defined in a neighborhood of \( z_0 \) (except maybe at \( z_0 \) itself). We say that

if for all \( \epsilon > 0 \) sufficiently small, there is a \( \delta > 0 \) such that

If \( z_0 \) is a boundary point of the domain of definition, we impose that the above definition only holds for paths of approach that lie entirely in the domain. This is a generalization of the idea of a one-sided limit.

As an example, let’s verify that

We have to show that for a given \( \epsilon \) there is a \( \delta \) such that

this implies

It follows that if we choose \( \delta < \epsilon / 3 \) then the condition is satisfied and we have verified the limit statement.

We can extend the limit concept to the point at infinity by saying that

if for sufficiently small \( \epsilon > 0 \) there is a \( \delta > 0 \) such that

A function is called continuous at \( z_0 \) if

In terms of \( \epsilon \) and \( \delta \) we have that \( f(z) \) is continuous at \( z = z_0 \) if \[\forall \epsilon > 0 \quad \exist \delta > 0 : \quad |z - z_0| < \delta \rightarrow |f(x) - f(z_0)| < \epsilon\]

The same algebraic properties that hold for limits also hold for continuous functions. The sum of continuous functions is continuous, the product of continuous functions is continuous, etc.

It follows immediately from the definition that if \( f(z) \) is continuous at \( z_0 \) then so is \( f(z) \). So if \( f(z) \) is continuous at \( z_0 \), so are \( f(z) \) , \( Re(f(z)) \), \( Im(f(z)) \), and \( |f(z)|^2 \), since all of these are sums and products of continuous functions in this case.

We say that a function is continuous in a region if it is continuous at every point in the region. Typically, \( \delta \) depends on which point \( z_0 \) in the region is being considered, e.g. \( \delta = \delta(\epsilon, z_0) \). If \( \delta \) can be chosen to be independent of \( z_0 \), we say that \( f(z) \) is uniformly continuous in the region.

Let \( f(z) \) be defined in \( \Reals \), which contains a neighborhood of \( z_0 \). Then

is called the derivative of \( f(z) \) at \( z_0 \) if this limit exists. If so, then also

The funciton \( f(z) \) is differentiable in a region if its derivative at every point in the region is defined.

\[\]

Theorem

If \( f(z) \) is differentiable at \( z_0 \) then \( f(z) \) is continuous at \( z_0 \)

Proof

\[\lim_{z \rightarrow z_0} (f(z) - f(z_0)) = \lim_{z \rightarrow z_0} \left( \frac{f(z) - f(z_0)}{z - z_0} \right) (z - z_0) \\ = \lim_{z \rightarrow z_0} \left( \frac{f(z) - f(z_0)}{z - z_0} \right) \lim_{z \rightarrow z_0} (z - z_0) \\ = f'(z_0) \cdot 0 \\ = 0\]

For many features of complex functions, it suffices to repeat the proof in the real case in \( \Reals \). But there are some important consequences of being in two dimensions. As an example, consider \( f(z) = \overline{z} \). We know that this function is continuous for every \( z_0 \in \Complex \). However

If we let \( h = \epsilon e^{i \theta} \). Then

This limit is not defined, since the result depends on the direction along which z is approached. It follows that \( f(z) = \overline{z} \) is nowhere differentiable. This is reminiscent of the case of a function of two variables like

Here the limit as \( (x, y) \rightarrow (0, 0) \) is undefined. Reminiscent but not quite in the same league: the above example \( f(x) = \overline{z} \) is nowhere differentiable.

It’s easy to show that

And the regular differentiation formulae work for the elementary functions we defined before: