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- #1

- Thread starter dwsmith
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- Thread starter
- #1

- Jan 29, 2012

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Well, the first thing I would do is go ahead and do this division- $$\frac{\frac{1}{z- i}}{z+ i}= \frac{1}{z^2+ 1}$$

Then I would rewrite it as $$\frac{1}{1-(-z^2)}$$ and use the formula for the sum of a geometric series:

$$\sum r^n= \frac{1}{1- r}$$

Then I would rewrite it as $$\frac{1}{1-(-z^2)}$$ and use the formula for the sum of a geometric series:

$$\sum r^n= \frac{1}{1- r}$$

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Thanks, I was trying to do something a little different with it but that will suit the objective.Well, the first thing I would do is go ahead and do this division- $$\frac{\frac{1}{z- i}}{z+ i}= \frac{1}{z^2+ 1}$$

Then I would rewrite it as $$\frac{1}{1-(-z^2)}$$ and use the formula for the sum of a geometric series:

$$\sum r^n= \frac{1}{1- r}$$

- Jan 29, 2012

- 1,151

- Thread starter
- #5

It had to do with a contour integral and I was going to integrate it as two separate integrals, because I thought it would work out nicer that way. However, doing it in this manner is fine. I was trying to fit a different form but this was obviously easier and still fit the right form in the end.