Complex functions

For real variables, a function $f$ is a rule assigning a real number $f(x)$ to an element $x$ of the domain of $f$. A function of a complex variable (a complex function) assigns a complex number given by $f(z)$ to a complex number $z$.

Complex polynomials

A complex polynomial of degree $n$ is an expression of the form,

here, $c_{i}\left(i=0, \cdots, n\right.$ and $\left.c_{n} \neq 0\right)$ are complex constants and $z=x+i y$ is a complex variable. The Fundamental Theorem of Algebra asserts that every nonconstant, singlevariable polynomial with complex coefficients, $P(z)$ has at least one root. Note that this includes real polynomials since every real number is a complex number with imaginary part equal to 0 . In general, an $n$ degree polynomial will have $n$ roots (in some cases these are repeated roots so that the number of solutions is less than $n$ ). Moreover, any polynomial can be factorised as follows

where $a_{1}, a_{2}, \cdots, a_{n}$ are the (complex) roots. Even if the $c_{i}$ coefficients in Eq. (7.13) are real, the roots can still be complex. However, in this case, the roots occur in complex pairs, i.e. if $a_{1}$ is complex, $\overline{a_{1}}$ will also be a root. A simple example is

where the roots $z=i$ and $\bar{z}=-i$ are complex conjugates. It follows that if $P(z)$ is a polynomial with real coefficients of degree $n$ and $n$ is odd, that there will be at least one real root.

Proof: the roots of complex polynomials with real coefficients occur in complex conjugate pairs

Suppose that $a$ is a root of $P(z)$ as defined in Eq. (7.13) with real coefficients $c_{i}$. Then,

Taking the complex conjugate of the above equation yields

Since the coefficients $c_{i}$ are assumed to be real then

Therefore, Eq. (7.16) can be written as

comparing Eqs. (7.15) and (7.17), we see that $\bar{a}$ is also a root of $P$.

For the case of a polynomial with real coefficients therefore, any root is real or one of a complex conjugate pair of roots. And so if $a_{1}$ is a complex root, then another root, say $a_{2}$ is equal to $\overline{a_{1}}$. Then

gives a real quadratic. For example, the quartic $x^{4}+x^{2}+1$ is factorised as follows (see Example 7.3 for the factorisation),

The first two factors of Eq. (7.18) are complex conjugates of each other so their product is a real quadratic,

Similarly, $\left(x-e^{2 \pi i / 3}\right)\left(x-e^{-2 \pi i / 3}\right)=x^{2}+x+1$, therefore

In general, any real polynomial can be factorised into a product of real linear and real quadratic terms. The linear factors are associated with real roots and the quadratic factors with complex conjugate pairs of roots.

Example 7.2 Find the cube roots of 1.

Solution We are looking for roots $z$ with the property

Note that there are more than one ways to go about solving for the roots. Here, we choose to use DeMoivre's theorem [Eq. (7.10)]. We first express the RHS of (7.19) in polar form, $r e^{i \theta}$ where $r=|1|=1$ and $\arg (1)=0+2 k \pi$ where $k=0,1,2, \cdots$; this yields

Taking the cube roots of both sides

where we have used DeMoivre's theorem. In Eq. (7.21), $k$ can take any integer value (including 0). Here, we expect at most 3 roots and so it is customary to choose 3 consecutive values of $k$, e.g. 0,1 , and 2 . Any other value of $k$ yields a root that repeats one of the ones already determined. And so, for $k=0$,

for $k=1$,

and for $k=2$,

Example 7.3 Factorise the complex polynomial $P(z)=z^{4}+z^{2}+1$.

Solution Quartic polynomials can be difficult to solve, however, in this case $P(z)$ can be recast as quadratic equation for $z^{2}$,

This can be solved via the standard formula for the roots of a quadratic,

It is convenient to rewrite this in terms of complex exponentials,

Next, we want to solve

an obvious solution is $z=e^{\pi i / 3}$. Another solution is $z=e^{-2 \pi i / 3}$. To see this,

Similarly, $z=e^{-\pi i / 3}$ and $z=e^{2 \pi i / 3}$ are solutions to $z^{2}=e^{-2 \pi i / 3}$. Accordingly, the four roots of $P(z)$ are

$P(z)$ has the factorised form

or

where the complex exponentials have been converted into Cartesian form (the roots are shown in the Argand diagram in Fig. 7.1).

Complex power series

As with real numbers, powers of complex numbers are defined through multiplication,

For example, if $z=x+i y$,

Negative powers can be defined by the reciprocal,

Figure 7.1: The roots from Example 7.3 shown on the Argand diagram.

Complex power series can be defined in the same way as for real numbers

where $c_{0}, c_{1}, \cdots, c_{n}$ is an infinite list of complex numbers. The notion of absolute convergence and the ratio test carry through - here absolute values are replaced with the complex modulus. Such a series converges absolutely for $|z|<R$ and diverges for $|z|>R$ where $R$ is the radius of convergence given by a positive constant. Recall that if a power series is expanded around the point $a$ and the radius of convergence is $R$, then the set of all points $z$ such that $|z-a|=R$ is a circle called the boundary of the disk of convergence (refer to Subsec. 6.2.3). We look at some examples next.

Example 1: Consider a complex series example which corresponds to the real example given in Eq. (6.49) in Chapter 6. The power series for the function $f(z)=1 /(1-z)$, expanded around $z=0$, is simply

and has radius of convergence 1 (use the ratio test). The series diverges at every point on the boundary $z=e^{2 \pi i t}$ with $t \in[0,1]$ as $\lim _{n} z^{n}=\lim _{n} e^{2 \pi i n t}$ is never zero.

Example 2: Recall the power series for $f(x)=-\ln (1-x)$ given by Eq. (6.50) in Chapter 6. The power series for $g(z)=-\ln (1-z)$, expanded around $z=0$ is

and has radius of convergence 1 (use the ratio test) and diverges for $z=1$ (as can be seen by inspection) but converges for all other points on the boundary. This is non-trivial to see. For any $z$ in the unit circle, $z=e^{2 \pi i t}$ with $t \in[0,1]$. Let us apply the Dirichlet test (refer to Subsec. 6.2.2) which can be extended to complex numbers in the following form:

Theorem (Dirichlet test)

If $\left\{b_{n}\right\}$ is a sequence of real numbers satisfying $b_{n} \downarrow 0$ as $n \rightarrow \infty$ and $\left\{a_{n}\right\}$ a sequence of complex numbers that has bounded partial sums, then the series $\sum a_{n} b_{n}$ converges.

Here $b_{n}=1 / n$ and $a_{n}=z^{n}$. Clearly the first condition is satisfied. For the second one we need to have

for all $n$. But,

which is satisfied for every $z \neq 1$ in the circle!

In conclusion, the series converges for every $z$ with $|z| \leq 1$ other than $z=1$, and it diverges for $|z|>1$.

Example 3: The power series

has radius of convergence 1 (use ratio test) and converges everywhere on the boundary absolutely.

Complex exponential

Recall that the Maclaurin series for the exponential [see Eq. (6.20)] is valid for all real $x$, i.e. $R=\infty$. Define the exponential of a complex number $z$ via the complex series,

It follows that $f(z)=e^{z}$ defines a complex function. This converges for any complex $z$, i.e. $R=\infty$. Writing $z=x+i y$ gives

using Euler's formula. Note also that the exponential obeys the addition property

as for real numbers. And as for real numbers, we will use $\exp (z)$ and $e^{z}$ interchangeably. The exponential is a periodic function with period $2 \pi i$; from the periodicity of trigonometric functions, $e^{2 \pi i}=1$ and using Eq. (7.26), we find

It follows that the complex exponential, in sharp contrast with the real exponential, is not 1 - 1. It also follows from the definition of the exponential function

The periodicity of the exponential function results in the complex logarithm (see below) being multivalued.

Complex sine and cosine

We have already seen in Eqs. (7.11) and (7.12) how the cosine and sine functions can be expressed in terms of exponentials. In Eqs. (7.11) and (7.12), $\theta$ is taken to be real. The definitions can be extended to any complex number $z$ :

On the multivaluedness of the complex logarithm

We have seen that for real numbers, the natural logarithm is the inverse function of the exponential so that $\exp (\ln x)=x$. In contrast with the real logarithm function which is only defined for positive real numbers, the complex logarithm is defined for all nonzero complex numbers, but at a price: the function is not single-valued. This has to do with the periodicity (7.27) or, equivalently, with the multivaluedness of the $\operatorname{argument}, \arg (z)$ (recall that unlike $\operatorname{Arg}(z)$ which is single-valued, $\arg (z)$ is not because for a given complex number $z$, the number $\arg (z)$ represents an infinite number of possible values).

By analogy with the real natural logarithm, we define the complex logarithm as an inverse to the complex exponential function, $\exp (\ln z)=z$. In other words, we say that a logarithm of a nonzero complex number $z$, is any complex number $w$ such that $e^{w}=z$. We define the function $\ln z$ by

From the periodicity (7.27) of the exponential function it follows that if $w=\ln z$ so is

for any integer $k$, i.e. if $w=\ln z$ we also have $w+2 \pi i k=\ln z$. Therefore we see that $\ln z$ is a multivalued function ${ }^{15}$.

Consider now the polar form $z=r e^{i \theta}$. This can be rewritten as $z=\exp \left[\ln \left(r e^{i \theta}\right)\right]$ by direct substitution in Eq. (7.29) which gives,

Taking logarithms on both sides

${ }^{15}$ Be careful here as there could be a misconception: we keep on adding $2 \pi$ to $w$ and we get the same $\ln z$ so if we were to view $\ln z$ as a function of $w$ it appears to be single-valued, much like $y=\sin x$ is single-valued: we can add $2 \pi$ to $x$ and we come back to the same $y$. The problem with this reasoning is viewing $\ln z$ as a function of $w$. We have to look at $f(z)=\ln z$ as a function of $z$. This function maps $z$ to $\ln z$, much like $f(x)=\ln x \operatorname{maps} x$ to $\ln x$. Let us try again: $z=e^{w}$ is single-valued and behaves like $y=\sin x$, i.e. adding $2 \pi, 4 \pi, \ldots$ to $w$ gives the same $z$. But precisely for this reason, the inverse function $w=\ln z$ is not single-valued. Now for the same $z$ we have infinite many $w$ 's! So let us now look at $\ln z=\ln r+i \theta$. If we view $\ln z$ as a function of $\theta$ it would appear that $\ln z$ is single-valued. But in fact we have to look at $\ln z$ as a function of $z$. Adding $2 \pi k$ to $\theta$ brings us back to the same $z$ in the complex plane (after all $z=r e^{i \theta}$ is not affected by adding $2 \pi k$ to $\theta$ ), but it changes the value of $\ln z$ : for the same $z$ we have infinitely many possible values of $\ln z$ ! The function is multivalued. where $r$ and $\theta$ are real. As $\theta$ is ambiguous (adding $2 \pi$ to $\theta$ does not change $z=r e^{i \theta}$ ), so is $\ln z$. Using the modulus and argument of $z$, the complex logarithm can also be written in the form

where the ambiguity in the imaginary part is manifest. Clearly if $\theta=\arg z$ then so is $\theta+2 \pi k$ for any integer $k$. From (7.31), we now see that $\ln z$ is multivalued because its imaginary part is multivalued ${ }^{16}$. As with $\arg z$, the ambiguity may be removed by restricting the range of $\theta$. One can also define the principal value of the logarithm (denoted by Ln) by

with $-\pi<\operatorname{Arg}(z) \leq \pi$ as we saw in Subsec. 7.1.1. Note that Ln 0 is left undefined since there is no complex number $z$ satisfying $e^{z}=0$. Further, the function $\operatorname{Ln} z$ is discontinuous on the negative real axis; such discontinuities are called branch cuts. A branch cut is a curve in the complex plane across which a multivalued function is discontinuous. We explain the notion of branch cuts next by considering the complex logarithm.

Going back to Eq. (7.30), we note that the definition is fine for every point except 0 since the ln function has a singularity at that point. This is not the main concern here though. Consider going in a loop around the origin as in the unit circle below.

We see that,

Now, we start coming back around the circle toward -1 . If we get very close to -1 , the complex logarithm gets very close to

and so the function, as defined, is not continuous. Indeed, any time we go in a loop around the origin, anticlockwise, we change the value by $2 \pi i$ (going in a loop around the origin clockwise, we change the value by $-2 \pi i$ ). To deal with this, we cut the plane along a line starting at the origin (this can be the negative real axis as shown with the red line in the figure below). The origin is known as the branch point.

${ }^{16}$ Recall the point we made earlier: the multivaluedness of $\ln z$ has to do with the periodicity the exponential or, equivalently, with the multivaluedness of the argument.

We can then define the complex logarithm everywhere else apart from the points on that line; this is known as a branch cut. With branch cuts, we can now define the complex logarithm on any suitably cut plane. The definition is unique up to a factor of $2 \pi i k(k \in$ $\mathbb{Z})$. For two functions $f(z)$ and $g(z)$ we have

i.e. they are inverses of the exponential function where $f(z)=g(z)+2 \pi i k$. For different integer $k$, we obtain various functions $f$, known as the branches of the complex logarithm. It follows that we can make the logarithm function single-valued in various regions of the complex plane by choosing a different branch of the argument function. For instance, we may choose $0 \leq \operatorname{Arg}(z)<2 \pi$ as the principal value of the argument of $z$ instead of $-\pi<\operatorname{Arg}(z) \leq \pi$ chosen in Subsec. 7.1.1. Again, the principal argument is single-valued but it is now discontinuous along the positive real axis. With this definition, the branch cut for the complex logarithm is along the real positive axis and, again, the origin is a branch point.

The multivaluedness discussed above raises of course some questions. In particular, whether or not earlier examples of complex functions we have seen involved multivaluedness and hence there is a need for branch cuts too. The answer is: they did not. As representative, consider $f(z)=z^{2}=r^{2} e^{i 2 \theta}$. In the absence of a cut it is true that $\theta$ is multivalued. But the crucial point is that this does not lead to multivaluedness of the function, since

for all $n$ 's as seen earlier in this section. The same is true for $\sin z, e^{z}, \cos z / z$, and so on. The only cases that we encounter where the function is multivalued, and hence a branch cut is needed, are $\ln (z-a)$, non-integer powers of $(z-a)$ (see below), and inverse complex trigonometric functions.

Powers

If $z=r e^{i \theta}$ we have $\ln z=\ln r+i \theta$ which is defined up to integer values of $2 \pi i$. Therefore,

where we have used Eq. (7.30). For integer $n$, the ambiguity in $\ln z$ drops out since

Now, suppose $n=p$ where $p$ is not an integer; then

but like $\ln z$ itself, this is ambiguous. The ambiguity in $z^{p}$ can be removed by fixing the ambiguity in $\ln z$. For example, choose

This, like $\operatorname{Ln} z$ is discontinuous on the negative real axis.