Question

Find the disk of convergence of the series$\mathbf{\sum }_{\mathbf{}}{\left(z-2i\right)}^{\mathbf{n}}\mathbf{/}\mathit{n}$.

Short answer

The equation $\left|z-2i\right|<1$represents the equation of a disk having $\left(0,2\right)$as a center.

Step-by-step solution

Given Information

The given expression is $\sum _{}\frac{{\left(z-2i\right)}^n}{n}$.

Definition of Complex Numbers

The numbers that are presented in the form of $\mathit{x}\mathbf{+}\mathit{i}\mathit{y}$where, $\mathbf{\text{'}}\mathit{x}\mathbf{\text{'}}$is real numbers and $\mathbf{\text{'}}\mathit{i}\mathit{y}\mathbf{\text{'}}$ is an imaginary number, those numbers are referred to as called Complex numbers.

Step 3:Find the ratio of the series.

Consider $s=\sum _{}\frac{\left(z-2i\right)}{n}$be a series

Let $\frac{{\left(z-2\right)}^n}{n}$be $a_n$

$a_n=\frac{{\left(z-2i\right)}^n}{n}$ …(1)

Replace $n$by $n+1$in equation (1).

$a_{n+1}=\frac{{\left(z-2i\right)}^{n+1}}{\left(n+1\right)}$ …(2)

Find the ratio of $p_n$.

$$\begin{gathered} p_n=\frac{\left({\displaystyle \frac{{\left(z-2i\right)}^{n+1}}{\left(n+1\right)}}\right)}{\left({\displaystyle \frac{{\left(z-2i\right)}^n}{n}}\right)} \\ =\left(\frac{{\left(z-2i\right)}^{n+1}}{n+1}\right)\times \left(\frac{n}{{\left(z-2i\right)}^n}\right) \\ {p}_n=\left(z-2i\right)\times \left(\frac{n}{n+1}\right) \end{gathered}$$

Finding the Limit

Find the limit of $p_n$.

role="math" localid="1658906723093" $$\begin{gathered} p=\underset{\mathrm{n}\to \infty }{\lim }\left|\left(z-2i\right)\times \frac{n}{n+1}\right| \\ =\underset{\mathrm{n}\to \infty }{\lim }\left|\left(z-2i\right)\right|\left|\frac{n}{n+1}\right| \\ =\text{'}\left|\left(z-2i\right)\right|\underset{n\to \infty }{lim}\left|\frac{1}{1+{\displaystyle \frac{1}{n}}}\right| \\ =\left|\left(z-2i\right)\right| \end{gathered}$$

The condition $p=\underset{n\to \infty }{lim}\left|p_n\right|<1$must be valid for this series to converge.

$\left|z-2i\right|<1$

Hence the equation $\left|z-2i\right|<1$represents the equation of a disk having as a center.