# Disk of Convergence

• Jun 13th 2007, 04:59 AM
moolimanj
Disk of Convergence
(a) Determine the disk of convergence of the power series

(n^5)(z-3i)^n/4^n) for n=1 to infinity

(b) Use taylore theorem to determine the taylor series about 2i for the functio f(z) = Log(z+i), giving an expression for the general term of teh series. Also, state the largest open disk on which the function f is represented by this taylor series.

(c) Find the taylor series about 0 for each of the following functions:

(i) f(z) = (e^2z)Coshz (up to the terms in z^3)
(ii) f(z) - Log(2-cosz) (up to the terms in z^6)

(d) Let the function f be entire and suppose that

f(i/n) = -(2/n^3) for all n subset N

show that f(z) = - 2iz^3

Thankyou
• Jun 13th 2007, 06:26 AM
ThePerfectHacker
Quote:

Originally Posted by moolimanj
(a) Determine the disk of convergence of the power series

(n^5)(z-3i)^n/4^n) for n=1 to infinity

We have,
$\sum_{n=1}^{\infty} \frac{n^5(z-3i)^n}{4^n}$

Use the ratio test,
$\left| \frac{n^5 (z-3i)^n}{4^n} \right|^{1/n} = \frac{(n^{1/n})^5 |z-3i| }{4}$

To evaluate the limit of this sequence we use the famous limit $\lim n^{1/n}=1$.

Thus, (for absolute convergence we make it less than 1),
$\frac{|z-3i|}{4} <1$
Thus,
$|z-3i| < 4$
Hence the open disk is described as,
$N(3i,4)$
I do not think any points on the boundary of the disk converge.