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Find the Taylor Series expansion of the given analytic function $f(z)$, centered at point $z_0$; find the disk of convergence.

a) $f(z)=\frac{1}{-2+3i-z}$ $z_0=3$

b) $f(z)=(2-z)\cos{(3z^2)}$ $z_o=0$


I know that I have to start off by finding the derivatives of each function until they start to repeat. From there I need to use the following formula to find the Taylor coefficients: $a_n=\frac{f^{(n)}(z_0)}{n!}$

For part (b) I want to confirm that I should be using the product rule to compute the derivatives.

For both part (a) and (b) I also get a little confused at the very end when it is time to actually write out the Taylor series expansion. How do I know where to find the disk of convergence?

Any help would be appreciated. Maybe if I can see a solution of just one of them I can continue with the other?

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  • $\begingroup$ The prime in problem (a) is probably a typo, meant to be a comma before the $z_0 = 3$ statement. $\endgroup$ Apr 1, 2015 at 23:55
  • $\begingroup$ And for both these problems you can find the radius of convergence using the ratio test. $\endgroup$ Apr 1, 2015 at 23:57
  • $\begingroup$ Mark you're right -- it is a comma. Thank you! $\endgroup$
    – Kristin
    Apr 2, 2015 at 14:07

1 Answer 1

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a) $f(z) = \frac{1}{-2+3i-z} = \frac{1}{-5+3i-(z-3)}$ converges in $|z-3| \lt |-5+3i|=\sqrt{34}$
also using geometric series expansion we have $$ f(z) = \frac{1}{-5+3i-(z-3)} = \frac{1}{-5+3i}\cdot\frac{1}{1-\frac{z-3}{-5+3i}} = \sum_{n=0}^{\infty}\frac{(z-3)^n}{(-5+3i)^{n+1}} $$

b) $f$ is entire and plugging $3z^2$ in the Taylor series of $\cos$ we have $$ f(z) = 2\cos(3z^2)-z\cos(3z^2) = \sum_{n=0}^{\infty}\frac{2\cdot 9^n\cdot(-1)^n}{(2n)!}z^{4n} - \sum_{n=0}^{\infty}\frac{9^n\cdot (-1)^n}{(2n)!}z^{4n+1} = $$

$$ = 2-z-9z^4+\frac{9}{2}z^5+\dots $$

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  • $\begingroup$ Thank you for your explanation! Did you include the point $z_0$ in the denominator so that it could be of the form $z-z_0$? Is this the only way to go about it? I was thinking of treating point $z$ as point $z_0$. Also I'm a little confused about why you didn't need to find the derivatives. I appreciate your help! $\endgroup$
    – Kristin
    Apr 2, 2015 at 14:10
  • $\begingroup$ Notice: benji did not do this by taking repeated derivatives. Instead he used series he already knew (geometric and cosine). $\endgroup$
    – GEdgar
    Apr 2, 2015 at 14:56
  • $\begingroup$ Oh I see thank you! I went back and worked through part (a) by taking repeated derivatives and I got almost the same answer. The only difference is that in the denominator I have $(n-1)!(-5+3i)$. Is this incorrect? $\endgroup$
    – Kristin
    Apr 2, 2015 at 15:12
  • $\begingroup$ @Kristin yes, I believe this is incorrect. please double check your work. As for your previous question, if you want a Taylor series around $z_0$ you need a power series of $(z-z_0)$. Since I used the geometric series expansion I had to put it in that form. $\endgroup$
    – benji
    Apr 3, 2015 at 2:27

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