Question

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

$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}{\mathbf{2}}^{\mathbf{n}}{\left(z+i-3\right)}^{\mathbf{2}\mathbf{n}}$

Short answer

The region of convergent is $\left|z-3+i\right|<\frac{1}{\sqrt{2}}$.

Step-by-step solution

Given data

The given series is, $\sum _{\mathrm{n}=1}^{\infty }{2}^{\mathrm{n}}{(\mathrm{z}+\mathrm{i}-3)}^{2\mathrm{n}}$.

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.

Calculation to check the series is convergent

Find $A_n$ and$A_{n+1}$ as follows:

$$\begin{gathered} A_n=2^n{\left(z-3+i\right)}^{2n} \\ {A}_{n+1}=2^{n+1}{\left(z-3+i\right)}^{2n+2} \end{gathered}$$

Apply ratio test as follows:

$$\begin{gathered} \rho =\underset{n\to \infty }{lim}\left|\frac{A_{n+1}}{A_n}\right| \\ \rho =\underset{n\to \infty }{lim}\left|\frac{2^{n+1}{\left(z-3+i\right)}^{2n+2}}{2^n{\left(z-3+i\right)}^{2n}}\right| \\ \rho =\left|2{\left(z-3+i\right)}^2\right| \end{gathered}$$

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

Calculation to find the region of convergent

The region of convergence is given as follows:

$$\begin{gathered} \left|2{\left(z-3+i\right)}^2\right|<1 \\ \left|{\left(z-3+i\right)}^2\right|<\frac{1}{2} \\ \left|z-3+i\right|<\frac{1}{\sqrt{2}} \end{gathered}$$

Let, $z=x+iy$

Therefore, calculate further as follows:

$$\begin{gathered} \left|x+iy-3+i\right|<\frac{1}{\sqrt{2}} \\ \left|\left(x-3\right)+\left(y+1\right)i\right|<\frac{1}{\sqrt{2}} \\ \sqrt{{\left(x-3\right)}^2+{\left(y+1\right)}^2}<\frac{1}{\sqrt{2}} \\ {\left(x-3\right)}^2+{\left(y+1\right)}^2<\frac{1}{\sqrt{2}} \end{gathered}$$

It is an equation of disk with center $\left(3,-1\right)$ and$r=\frac{1}{2}$.

Hence, the region of convergent is $\left|z-3+i\right|<\frac{1}{\sqrt{2}}$.