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Taylor Series for Functions of a Complex Variable

Taylor Series for Functions of a Complex Variable
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Instructor: Gerald Lemay

Gerald has taught engineering, math and science and has a doctorate in electrical engineering.

This lesson explores how analytic functions can be expressed as Taylor series. We use the concepts of complex differentiable functions and Cauchy-Riemann equations.

Taylor Series: Complex Variables

You probably know the formula for finding the Taylor series of a function like f(x) = 1/(1 - x). Using the Taylor series formula we get:


1/(1-x)=1+x+x^2+x^3+...=sum_n=0,infty_of_x^n


For clarity, x is a real variable, meaning it can have values on the number line, and the summation formula on the right-hand side is a compact way of writing 1 + x + x2 + ... where n takes values from 0 to ∞. The more terms we add, the better our approximation to the actual function. But there is a catch! We need more information.

For example, with 1/(1 - x), the summation formula only works for |x| < 1. The last part, |x| < 1, tells us where on the x-axis we can pick values (like x = 0.9) and still be able to use the summation expression. This is called the region of convergence. If we pick any value for x outside of the region of convergence (like x = 1.1), the summation will not represent the function, f(x), no matter how many terms we add together.

What if instead of a real variable, x, we have a complex variable, z ?

In this lesson we explore this question using the ideas of complex differentiability, Cauchy-Riemann conditions, analytic functions, and a region of convergence defined by a circle instead of a number line.

Complex Differentiability

If you can do basic manipulations of complex numbers and calculate partial derivatives, you can test for complex differentiability. Here are the steps:

First, write the function in the form:


f(x,y)=u(x,y)+iv(x,y)

identifying u(x,y) and v(x,y).

Then, calculate 4 partial derivatives:


du/dx,dv/dy,dv/dx,du/dy

Let's do an example with f(z) = 1/(1 - z).

Step 1: Write f(z) in the form f(x,y) = u(x,y) + i v(x,y)

We have f(z) = 1/(1 - z). But z is complex, meaning z = x + iy.

Substitute x + iy for z:


f(x,y)=1/[1-(x+iy)]


Which can be written as:


=1/[1-x-iy]


The denominator has a real part, 1 - x, and an imaginary part, -y.

Next, multiply the numerator and denominator by 1 - x + iy (this is sometimes called ''rationalizing the denominator''):


=[1-x+ij]/[(1-x)^2+y^2]


Separate the real parts and the imaginary parts:


=[1-x]/[(1-x)^2+y^2]+i[y]/[(1-x)^2+y^2]


Comparing this result to:


f(x,y)=u(x,y)+iv(x,y)


tells us:


u(x,y)=[1-x]/[(1-x)^2+y^2]


and


v(x,y)= [y]/[(1-x)^2+y^2]


We've done the first step!

Step 2: Calculate partial derivatives.

The partial derivative of u(x,y) with respect to x means x is the variable and everything else is a constant. When we differentiate with respect to y, x is held constant. Using the quotient rule for differentiation:


du/dx=-[(1-x)^2+y^2]-(1-x)2(1-x)(-1)_all_divided_by_[(1-x)^2+y^2]^2


This simplifies to:


du/dx=(1-x)^2-y^2]_all_divided_by_[(1-x)^2+y^2]^2


Now, differentiate v(x,y) with respect to y:


dv/dy=[(1-x)^2+y^2-y2y]_all_divided_by_[(1-x)^2+y^2]^2


which simplifies to:


dv/dy=[(1-x)^2-y^2]_all_divided_by_[(1-x)^2+y^2]^2


Two more to go. Partial derivative of v(x,y) with respect to x:


dv/dx=-y2(1-x)(-1)_all_divided_by_[(1-x)^2+y^2]^2


which simplifies to:


dv/dx=2y(1-x)_all_divided_by_[(1-x)^2+y^2]^2


Last one: differentiate u(x,y) with respect to y:


du/dy=-(1-x)2y_all_divided_by_[(1-x)^2+y^2]^2


Simplifying:


du/dy=-2y(1-x)_all_divided_by_[(1-x)^2+y^2]^2


This completes step 2.

Now we compare these results. If:


du/dx=dv/dy


and:


dv/dx=-du/dy


then, the function f(z) = 1/(1 - z) is an analytic function.

These two conditions are true, and indeed, our function is an analytic function. By the way, these two test equations are called the Cauchy-Riemann equations, which are used when exploring analytic functions.

Taylor Series: Analytic Functions

The formal statement goes like this:

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