6.2: Complex Functions
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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)≡z Perhaps the next simplest function assigns to each number its square, i.e., f(z)≡z2. As we decomposed the argument of f, namely z, value of f, z2 in this case, into its real and imaginary parts. In general, we write
f(x+iy)=u(x,y)+iv(x,y)
where u and v are both real-valued functions of two real variables. In the case that f(z)≡z2 we find
u(x,y)=x2−y2
and
v(x,y)=2xy
With the tools of complex numbers, we may produce complex polynomials
f(z)=zm+cm−1zm−1+⋯+c1z+c0
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−λ1)d1(z−λ2)d2⋯(z−λh)dh
Each λj is a root of degree dj. Here h is the number of distinct roots of f. We call λj a simple root when dj=1 rational functions. Suppose
q(z)=f(z)g(z)
in rational, that f is of order at most m−1 while g is of order m with the simple roots {λ1,⋯,λm}. It should come as no surprise that such q should admit a Partial Fraction Expansion
q(z)=m∑j=1qjz−λj
One uncovers the qj by first multiplying each side by z−λj and then letting z tend to λj. For example, if
1z2+1=q1z+i+q2z−i
then multiplying each side by z+i produces
1z−i=q1+q2(z+i)z−i
Now, in order to isolate q1 it is clear that we should set z=−i. So doing we find that q1=i2. In order to find q2 we multiply Equation by z−i and then set z=i. So doing we find q2=−i2, and so
1z2+i=i2z+i+−i2z−i
Returning to the general case, we encode the above in the simple formula
qj=limzZ→λj(z−λj)q(z)
You should be able to use this to confirm that
zz2+1=1/2z+i+1/2z−i
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=(0−110)
is
(zI−B)−1=1z2+1(z1−1z)=1z+i(12i2−i212)+1z−i(12−i2i212)
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
ez≡∞∑n=0znn!
and find that
ee=1+iθ+(iθ)22+(iθ)33!+(iθ)44!+⋯=1−θ22+θ44−⋯+i(θ−θ33+θ55−⋯)=cosθ+isinθ
With this observation, the polar form is now simply z=|z|eiθ
One may just as easily verify that
cos(θ)=eiθ+e(−i)θ2
and
sin(θ)=eiθ−e(−i)θ2i
These suggest the definitions, for complex z
cos(z)≡eiz+e(−i)z2
and
sin(z)≡eiz−e(−i)z2i
As in the real case the exponential enjoys the property that
ez1+z2=ez1ez2
and in particular
ex+iy=exeiy=excos(y)+iexsin(y)
Finally, the inverse of the complex exponential is the complex logarithm,
ln(z)≡ln(|z|)+iθ
for z=|z|eiθ. One finds that ln(−1+i)=ln(√2)+i3π4.