# Convert Partial Fractions & Taylor Series: Solving Complex Equations

• dykuma
In summary: For ##1<|z|<2##, we have, ##\left|\frac{1}{z}\right|<1## and ##\left|\frac{z}{2}\right|<1##. So you can now use the formula for infinite geometric series.
dykuma

## Homework Statement

and the solution (just to check my work)

## Homework Equations

None specifically. There seems to be many ways to solve these problems, but the one used in class seemed to be partial fractions and Taylor series.

## The Attempt at a Solution

The first step seems to be expanding this using partial fractions, giving me

Now, for 0 < |z| < 1, we expand each of the fractions in the parenthesis in powers of z.

This is the Laurent series for f (z) which is valid in the region 0 < |z| < 1. I then need to get the other two series, which the next one I should try to get is for the region |z| > 2. To get that, it is suggested that I write the two partial fractions as:

However I am not sure what to do with this. I have seen things saying I should expand these two functions, and then add them together, however this does not give me the answer for the region |z| > 2, (in fact, it just gives me the first series, but a degree higher, which makes sense).

dykuma said:
I then need to get the other two series, which the next one I should try to get is for the region |z| > 2. To get that, it is suggested that I write the two partial fractions as:
View attachment 109656
However I am not sure what to do with this. I have seen things saying I should expand these two functions, and then add them together, however this does not give me the answer for the region |z| > 2, (in fact, it just gives me the first series, but a degree higher, which makes sense).
The reason for writing
$$-\frac{1}{z-1} = -\frac 1z\left(\frac{1}{1-\frac 1z}\right)$$ and then expanding is because when ##|z|>1##, you have ##|1/z| < 1## so a series in positive powers of (1/z) will converge. If you look at the way you expanded 1/(z-2), you should see that you'll get a series of positive powers of (z/2), which won't converge for |z|>2.

dykuma
vela said:
The reason for writing
$$-\frac{1}{z-1} = -\frac 1z\left(\frac{1}{1-\frac 1z}\right)$$ and then expanding is because when ##|z|>1##, you have ##|1/z| < 1## so a series in positive powers of (1/z) will converge. If you look at the way you expanded 1/(z-2), you should see that you'll get a series of positive powers of (z/2), which won't converge for |z|>2.

Thanks! Right, I okay, I see what I did wrong there, and I was able to get the correct series solution for |z| > 2.

The only question I have left is how to get the 1< |z| < 2 series. I am not sure where to start there.

dykuma said:
The only question I have left is how to get the 1< |z| < 2 series. I am not sure where to start there.
$$\frac{1}{z}\left(\frac{1}{z-2} - \frac{1}{z-1}\right)= \frac{1}{z}\left(\frac{1}{-2}\cdot\frac{1}{1-\frac{z}{2}} - \frac{1}{z}\cdot\frac{1}{1-\frac{1}{z}}\right)$$
For ##1<|z|<2##, we have, ##\left|\frac{1}{z}\right|<1## and ##\left|\frac{z}{2}\right|<1##. So you can now use the formula for infinite geometric series.

## 1. What are partial fractions?

Partial fractions are a method used to break down a complex fraction into smaller, simpler fractions. This makes it easier to solve and manipulate the equation.

## 2. Why do we need to convert partial fractions?

Converting partial fractions allows us to solve complex equations more easily and accurately. It also helps us to identify patterns and relationships between the different parts of the equation.

## 3. What is a Taylor series?

A Taylor series is a mathematical representation of a function as an infinite sum of terms. It allows us to approximate complex functions using simpler polynomials.

## 4. How do we use partial fractions and Taylor series to solve complex equations?

First, we use partial fractions to break down the complex equation into simpler fractions. Then, we use the Taylor series to approximate each fraction as a polynomial. Finally, we combine the polynomials to solve for the original complex equation.

## 5. Can we use partial fractions and Taylor series for all types of equations?

Yes, partial fractions and Taylor series can be used for a wide variety of equations, including rational functions, trigonometric functions, and exponential functions. However, the process may be more complex for some equations compared to others.

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