Question

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

$\mathbf{\sum }_{\mathbf{n}\mathbf{-}\mathbf{1}}^{\mathbf{\infty }}\frac{{\mathbf{(}\mathbf{z}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{n}}}{\mathbf{n}}$

Short answer

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

Step-by-step solution

Given data

The given series is, $\sum _{\mathrm{n}-1}^{\infty }\frac{{(\mathrm{z}-1)}^{\mathrm{n}}}{\mathrm{n}}$.

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:

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

Apply ratio test as follows:

localid="1658740575358" role="math" $$\begin{gathered} \rho =\underset{n\to \infty }{\lim }\left|\frac{A_{n+1}}{A_n}\right| \\ \rho =\underset{n\to \infty }{\lim }\left|\frac{{\displaystyle \frac{{(z-i)}^{n+1}}{n+1}}}{{\displaystyle \frac{{(z-i)}^n}{n}}}\right| \\ \rho =\underset{n\to \infty }{\lim }\left|(z-i).\frac{n}{(n+1)}\right| \\ \rho =\underset{n\to \infty }{\lim }\left|(z-i).\frac{\left({\displaystyle \frac{n}{n}}\right)}{\left(1+{\displaystyle \frac{1}{n}}\right)}\right| \end{gathered}$$

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

$\rho =\left|z-i\right|$

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

Calculation to find the region of convergent

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

Let, $z=x+iy$.

Therefore, calculate further as follows:

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

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

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