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[SOLVED] Taylor Series

dwsmith

Well-known member
Feb 1, 2012
1,673
I am trying to find the Taylor series for
$$\displaystyle
\dfrac{\left(\dfrac{1}{z-i}\right)}{z+i}
$$
where z is a complex number.


There is a reason it is set up as a fraction over the denominator so lets not move it down.
 
Last edited:

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
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:
$$\sum r^n= \frac{1}{1- r}$$
 
Last edited:

dwsmith

Well-known member
Feb 1, 2012
1,673
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:
$$\sum r^n= \frac{1}{1- r}$$
Thanks, I was trying to do something a little different with it but that will suit the objective.
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
It was after posting that I noticed your "There is a reason it is set up as a fraction over the denominator so lets not move it down". That seems strange. What was your reason? Of course, the Taylor's series (about 0) will be the same however you write the function.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
It was after posting that I noticed your "There is a reason it is set up as a fraction over the denominator so lets not move it down". That seems strange. What was your reason? Of course, the Taylor's series (about 0) will be the same however you write the function.
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.