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I came across a problem that says:

It is given that $\sum_{n=0}^{\infty}a_{n}z^{n}$ converges at $z=3+4i.$

Then the radius of convergence of the power series $\sum_{n=0}^{\infty}a_{n}z^{n}$ is

(a)$\leq 5$
(b)$\geq 5$
(c)$<5$
(d)$>5$.

We know if a power series $\sum_{n=0}^{\infty}a_{n}z^{n}$ converges for $z=z_{0},$ then it is absolutely convergent for every $z=z_{1},$ when $|z_{1}|<|z_{0}|.$ Using this property, i can conclude that $(a)$ is the correct choice as equality sign occurs keeping in mind that the given series converges at $|3+4i|=5.$ Am i going in the right direction? Please help. Thanks in advance for your time.

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So, can the radius of convergence be $4$? –  Thomas Andrews Dec 13 '12 at 20:06

2 Answers 2

up vote 4 down vote accepted

No, sorry, your choice of (a) is not correct.

You know that the series does converge for a point at a radius 5 from the origin, but it might also converge for points further away than 5. So, from what you know, it converges inside a circle whose radius is at least 5.

Edit:

Another way to look at it is like this: we know that the circle of convergence includes the point $3+4i$ either in its interior or on the circumference, so we know that the radius of convergence must be at least 5 ... it could be bigger than 5, but it cannot be smaller than 5 (as then it would not include $3+4i$).

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Thank you sir for the clarification. It is now crystal clear. –  user52976 Dec 13 '12 at 21:45

Wrong direction. As you say, if it converges at $3+4i$, it also converges absolutely for all $z$ with $|z| < 5$. But it might also converge for some $z$ with $|z| > 5$.

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Sir, you say " But it might also converge for some $z$ with $|z|>5.$" But it also might not. So what is the most definite answer? –  user52976 Dec 13 '12 at 20:11
    
@user52976: $\geq 5$ means "greater than $5$ OR equal to $5$". These two possibilities encapsulate "might ... but it also might not". –  Jonas Meyer Dec 13 '12 at 20:21

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