Question

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

$\mathbf{\sum }_{\mathbf{n}\mathbf{-}\mathbf{0}}^{\mathbf{m}}\frac{{\left(z-2+i\right)}^{\mathbf{n}}}{{\mathbf{2}}^{\mathbf{n}}}$

Short answer

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

Step-by-step solution

Given data

The given series is, $\sum _{n=0}^{\infty }\frac{{\left(z-2+i\right)}^n}{2^n}$.

Concept of Ration test

The ratio test is a check (or "criterion") to see if a series will eventually converge to $\sum _{n-1}^{\infty }{a}_n$. 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:

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

Apply ratio test as follows:

role="math" localid="1658727517111" $$\begin{gathered} \rho =\underset{n\to \infty }{lim}\left|\frac{A_{n+1}}{A_n}\right| \\ \rho =\underset{n\to \infty }{lim}\left|\frac{{\displaystyle \frac{(2-2+i^{n+1}}{2^{n+1}}}}{{\displaystyle \frac{{\left(2-2+i\right)}^n}{2^n}}}\right| \\ \rho =\left|\frac{\left(z-2+i\right)}{2}\right| \end{gathered}$$

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

Calculation to find the region of convergent

The region of convergence is given as follows:

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

Let, $z=x+iy$.

Therefore, calculate as follows:

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

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

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