Question

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

${\mathbf{e}}^{\mathbf{z}}\mathbf{=}\mathbf{1}\mathbf{+}\mathbf{z}\mathbf{+}\frac{{\mathbf{z}}^{\mathbf{2}}}{\mathbf{2}\mathbf{!}}\mathbf{+}\frac{{\mathbf{z}}^{\mathbf{3}}}{\mathbf{3}\mathbf{!}}\mathbf{·}\mathbf{·}\mathbf{·}\left[\mathrm{equation}(8.1)\right]$

Short answer

The entire complex plane is the radius of convergence.

Step-by-step solution

Given Information.

The given power series, i.e.,${\mathrm{S}}_{\mathrm{n}}={\mathrm{e}}^{\mathrm{z}}$

Definition of Radius of convergence.

The interior of the set of points of convergence of a power series is called the disc of convergence. Its radius is known as the series' convergence radius.

Find the value of the general term.

Expand the series to find the general term.

$$\begin{gathered} S_n=e^z=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+......\left(1\right) \\ {S}_n=\sum _{}\frac{{\mathrm{z}}^{\mathrm{n}}}{\mathrm{n}!} \end{gathered}$$

Find the value of ${\mathrm{a}}_{\mathrm{n}}$and${\mathrm{a}}_{\mathrm{n}+1}$

$\begin{array}{l}{\mathrm{a}}_{\mathrm{n}}=\frac{{\mathrm{z}}^{\mathrm{n}}}{\mathrm{n}!}\\ {\mathrm{a}}_{\mathrm{n}+1}=\frac{{\mathrm{z}}^{\mathrm{n}+1}}{(\mathrm{n}+1)!}\end{array}$

Find the radius of convergence.

Use the ratio test.

$$\begin{gathered} {\mathrm{\rho }}_{\mathrm{n}}=\frac{{\mathrm{a}}_{\mathrm{n}+1}}{{\mathrm{a}}_{\mathrm{n}}} \\ {\mathrm{\rho }}_{\mathrm{n}}=\left|\frac{{\displaystyle \frac{{\mathrm{z}}^{\mathrm{n}+1}}{\left(\mathrm{n}+1\right)!}}}{{\displaystyle \frac{{\mathrm{z}}^{\mathrm{n}}}{\mathrm{n}!}}}\right| \\ =\left|\frac{\mathrm{z}}{\mathrm{n}+1}\right| \end{gathered}$$

Calculate the value of localid="1658729145101" $\mathrm{\rho }$,i.e.,

$\begin{array}{l}\mathrm{p}=\underset{\mathrm{n}\to \infty }{\lim }\left|{\mathrm{\rho }}_{\mathrm{n}}\right|\\ =\underset{\mathrm{n}\to \infty }{\lim }\left|\frac{\mathrm{z}}{\mathrm{n}+1}\right|\\ =0\end{array}$

${\mathrm{S}}_{\mathrm{n}}={\mathrm{e}}^{\mathrm{z}}$ is convergent for all values of z.

Hence, the entire complex plane is the radius of convergence.