Q14P - Mathematical Methods in Physical Sciences

Question

Find the disk on convergence for each of the following complex power series.

 $\mathbf{\sum }_{\mathbf{n}\mathbf{-}\mathbf{0}}^{\mathbf{\infty }}\mathit{n}\mathbf{(}\mathit{n}\mathbf{+}\mathbf{1}\mathbf{)}{\mathbf{(}\mathit{z}\mathbf{-}\mathbf{2}\mathit{i}\mathbf{)}}^{\mathbf{n}}$

Step-by-Step Solution

Answer

The region of convergent is $\left|z-2i\right|<1$ .

Step 1: Given data

The given series is,  $\sum _{\mathrm{n}-0}^{\infty }n(n+1){(z-2i)}^{\mathrm{n}}$

Step 2: Concept of Ration test

The ratio test is a check (or "criterion") to see if a series will eventually converge to $\mathbf{\sum }_{\mathbf{n}\mathbf{-}\mathbf{1}}^{\mathbf{\infty }}{\mathit{a}}_{\mathbf{n}}$ . Every term is a real or complex number, and when n is big,   is not zero.

Step 3: Calculation to check the series is convergent

Find $A_n$ and $A_{n+1}$ as follows:

$$\begin{gathered} A_n=n(n+1){(z-2i)}^n \\ A_n=(n+1){(n+2)(z-2i)}^{n+1} \end{gathered}$$ 

 

Apply ratio test as follows:

 $$\begin{gathered} \rho =\underset{n\to \infty }{\lim }\left|\frac{A_{n+1}}{A_n}\right| \\ \rho =\underset{n\to \infty }{\lim }\left|\frac{(n+1)(n+2){(z-2i)}^{n+1}}{\left(n\right)(n+1){(z-2i)}^n}\right| \\ \rho =\underset{n\to \infty }{\lim }\left|(z-2i).\frac{(n+2)}{n}\right| \\ \rho =\underset{n\to \infty }{\lim }\left|(z-2i).\frac{\left(1+{\displaystyle \frac{2}{n}}\right)}{\left({\displaystyle \frac{n}{n}}\right)}\right| \end{gathered}$$

Substitute limit in the above equation and solve further as follows:

$\rho =\left|z-2i\right|$ 

 

If,$\rho <1$ , then the series is convergence.

Step 4: Calculation to find the region of convergent

The region of convergence is $\left|z-2i\right|<1$ .

Let, $z=x+iy$ .

 

Therefore, calculate further as shown below.

 $$\begin{gathered} \left|x+iy-2i\right|<1 \\ \left|x+(y-2)i\right|<1 \\ \sqrt{x^2+{(y-2)}^2}<1 \\ x^2+{(y-2)}^2<1 \end{gathered}$$

 

It is an equation of disk with center (0,2) and r = 1 .

 

Hence, the region of convergent is $\left|z-2i\right|<1.|$ .