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# 6.2: Complex Functions

• • Contributed by Steve Cox
• Emeritus Professor (Computational and Applied Mathematics) at Rice University
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## Complex Functions

A complex function is merely a rule for assigning certain complex numbers to other complex numbers. The simplest (nonconstant) assignment is the identity function $$f(z) \equiv z$$ Perhaps the next simplest function assigns to each number its square, i.e., $$f(z) \equiv z^2$$. As we decomposed the argument of $$f$$, namely $$z$$, value of $$f$$, $$z^2$$ in this case, into its real and imaginary parts. In general, we write

$f(x+iy) = u(x,y)+iv(x,y) \nonumber$

where $$u$$ and $$v$$ are both real-valued functions of two real variables. In the case that $$f(z) \equiv z^{2}$$ we find

$u(x,y) = x^{2}-y^{2} \nonumber$

and

$v(x,y) = 2xy \nonumber$

With the tools of complex numbers, we may produce complex polynomials

$f(z) = z^{m}+c_{m-1}z^{m-1}+\cdots+c_{1}z+c_{0} \nonumber$

We say that such an $$f$$ is order $$m$$. We shall often find it convenient to represent polynomials as the product of their factors, namel

$f(z) = (z- \lambda_{1})^{d^{1}}(z-\lambda_{2})^{d^{2}} \cdots (z-\lambda_{h})^{d^{h}} \nonumber$

Each $$\lambda_{j}$$ is a root of degree $$d_{j}$$. Here $$h$$ is the number of distinct roots of $$f$$. We call $$\lambda_{j}$$ a simple root when $$d_{j} = 1$$ rational functions. Suppose

$q(z) = \frac{f(z)}{g(z)} \nonumber$

in rational, that $$f$$ is of order at most $$m-1$$ while $$g$$ is of order $$m$$ with the simple roots $$\{\lambda_{1}, \cdots, \lambda_{m}\}$$. It should come as no surprise that such $$q$$ should admit a Partial Fraction Expansion

$q(z) = \sum_{j = 1}^{m} \frac{q_{j}}{z-\lambda_{j}} \nonumber$

One uncovers the $$q_{j}$$ by first multiplying each side by $$z-\lambda_{j}$$ and then letting $$z$$ tend to $$\lambda_{j}$$. For example, if

$\frac{1}{z^{2}+1} = \frac{q_{1}}{z+i}+\frac{q_{2}}{z-i} \nonumber$

then multiplying each side by $$z+i$$ produces

$\frac{1}{z-i} = q_{1}+\frac{q_{2}(z+i)}{z-i} \nonumber$

Now, in order to isolate $$q_{1}$$ it is clear that we should set $$z = -i$$. So doing we find that $$q_{1} = \frac{i}{2}$$. In order to find $$q_{2}$$ we multiply Equation by $$z-i$$ and then set $$z = i$$. So doing we find $$q_{2} = -\frac{i}{2}$$, and so

$\frac{1}{z^2+i} = \frac{\frac{i}{2}}{z+i}+\frac{\frac{-i}{2}}{z-i} \nonumber$

Returning to the general case, we encode the above in the simple formula

$q_{j} = \lim_{zZ \rightarrow \lambda_{j}} (z-\lambda_{j})q(z) \nonumber$

You should be able to use this to confirm that

$\frac{z}{z^2+1} = \frac{1/2}{z+i}+\frac{1/2}{z-i} \nonumber$

Recall that the transfer function we met in The Laplace Transform module was in fact a matrix of rational functions. Now, the partial fraction expansion of a matrix of rational functions is simply the matrix of partial fraction expansions of each of its elements. This is easier done than said. For example, the transfer function of

$B = \begin{pmatrix} {0}&{-1}\\ {1}&{0} \end{pmatrix}$

is

\begin{align*} (zI-B)^{-1} &= \frac{1}{z^{2}+1} \begin{pmatrix} {z}&{1}\\ {-1}&{z} \end{pmatrix} \\[4pt] &= \frac{1}{z+i} \begin{pmatrix} {\frac{1}{2}}&{\frac{i}{2}}\\ {\frac{-i}{2}}&{\frac{1}{2}} \end{pmatrix}+\frac{1}{z-i} \begin{pmatrix} {\frac{1}{2}}&{\frac{-i}{2}}\\ {\frac{i}{2}}&{\frac{1}{2}} \end{pmatrix} \end{align*}

The first line comes form either Gauss-Jordan by hand or via the symbolic toolbox in Matlab. More importantly, the second line is simply an amalgamation of Equation and Equation. Complex matrices have finally entered the picture. We shall devote all of Chapter 10 to uncovering the remarkable properties enjoyed by the matrices that appear in the partial fraction expansion of $$(zI-B)^{-1}$$ Have you noticed that, in our example, the two matrices are each projections, and they sum to I. and that their product is 0? Could this be an accident?

In The Laplace Transform module we were confronted with the complex exponential. By analogy to the real exponential we define

$e^{z} \equiv \sum_{n = 0}^{\infty} \frac{z^n}{n!} \nonumber$

and find that

\begin{align*} e^{e} &= 1+i \theta+\frac{(i \theta)^2}{2}+\frac{(i \theta)^{3}}{3!}+\frac{(i \theta)^{4}}{4!}+ \cdots \\[4pt] &= 1-\frac{\theta^{2}}{2}+\frac{\theta^{4}}{4}-\cdots+i(\theta-\frac{\theta^{3}}{3}+\frac{\theta^{5}}{5}-\cdots) \\[4pt] &= \cos \theta+i \sin \theta \end{align*}

With this observation, the polar form is now simply $$z = |z|e^{i \theta}$$

One may just as easily verify that

$\cos(\theta) = \frac{e^{i \theta}+e^{(-i) \theta}}{2} \nonumber$

and

$\sin(\theta) = \frac{e^{i \theta}-e^{(-i) \theta}}{2i} \nonumber$

These suggest the definitions, for complex $$z$$

$\cos(z) \equiv \frac{e^{iz}+e^{(-i)z}}{2} \nonumber$

and

$\sin(z) \equiv \frac{e^{iz}-e^{(-i)z}}{2i} \nonumber$

As in the real case the exponential enjoys the property that

$e^{z_{1}+z_{2}} = e^{z_{1}}e^{z_{2}} \nonumber$

and in particular

\begin{align*} e^{x+iy} &= e^{x}e^{iy} \\[4pt] &= e^{x} \cos(y)+ie^{x} \sin(y) \end{align*}

Finally, the inverse of the complex exponential is the complex logarithm,

$\ln (z) \equiv \ln(|z|)+i \theta \nonumber$

for $$z = |z|e^{i \theta}$$. One finds that $$\ln(-1+i) = \ln(\sqrt{2})+i \frac{3\pi}{4}$$.