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I'm really rusty on my series convergence, and I guess I was more asking for a bit of clarification on a question I'm working on.

I've been asked to find the power series expansion for $\frac{1}{3 - z}$ about the point $4i$, and to then give the radius of convergence.

After a bit of tedium, I got the power series expansion as follows;

$$\frac{1}{3 - z} = \frac{1}{3 - 4i - (z - 4i)}$$ $$ = \frac{1}{3-4i} \frac{1}{1 - \frac{z-4i}{3-4i}}$$ $$ = \sum_{n = 0}^{\infty} \frac{1}{3-4i} (\frac{z-4i}{3-4i})^n$$ $$ = \sum_{n = 0}^{\infty} \frac{1}{(3-4i)^{n+1}} (z-4i)^n$$

Now, I know this will hold true for $\frac{|z - 4i|}{5} < 1$, which is equivalent to $|z-4i| < 5$, so is it correct to say that the radius of convergence for this function is 5 (ie; the ball of radius 5, centred at 4i)??

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  • $\begingroup$ Yes...and the work looks just fine. Nice. +1 $\endgroup$
    – DonAntonio
    May 31, 2014 at 12:35

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Your answer is correct and also the method is reasonable. Let me just add a result that would tell you right way what radius of covergence to expect. You could use this as a check.

For any rational function (and this holds in more generality) the radius of convergence of the series development around a point is the distance to the nearest pole of the function.

The only pole of the function being at $3$ and the distance of $3$ and $4i$ being $5$ you can confirm the radius of convergence.

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