Find the power series for ${\mathit{e}}^{\mathbf{x}}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{x}$and for ${\mathit{e}}^{\mathbf{x}}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathit{x}$from the series for $\mathit{e}\mathit{z}$in the following way: Write the series for ${\mathit{e}}^{\mathbf{z}}$; put $\mathit{z}\mathbf{=}\mathit{x}\mathbf{+}\mathit{i}\mathit{y}$. Show that ${\mathit{e}}^{\mathbf{z}}\mathbf{=}{\mathit{e}}^{\mathbf{x}}\left(cosy+isiny\right)$; take real and imaginary parts of the equation, and put $\mathit{y}\mathbf{=}\mathit{x}$.

The power series (8.1) is$\sum _{n=0}^{\infty }{a}_n{\left(x-c\right)}^n=a_0+a_1\left(x-c\right)+a_2{\left(x-c\right)}^2+\boxed{\xcancel{}}$.

A power series (8.1) is an infinite series that is shown in step 1 here ${\mathit{a}}_{\mathbf{n}}$represents the coefficient of the $\mathit{n}\mathit{t}\mathit{h}$term and c as a constant.

It is known that $e^z=\sum _{n=0}^{\infty }\frac{z^n}{n!}$.

Expand the series as:

$e^z=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+.........$

Substitute$z=x+iy$into the obtained series.

$$\begin{gathered} e^{x+iy}=1+\left(x+iy\right)+\frac{{\left(x+iy\right)}^2}{2!}+\frac{{\left(x+iy\right)}^3}{3!}+........... \\ =\sum _{n=0}^{\infty }\frac{{\left(x+iy\right)}^n}{n!} \end{gathered}$$

Use Euler theorem:

$e^x\left(\cos y+i\sin y\right)=\sum _{n=0}^{\infty }\frac{{\left(x+iy\right)}^n}{n!}$

The formula states that $Re\left(z\right)=\frac{1}{2}\left(z+\overline{z}\right)$and $Im\left(z\right)=\frac{1}{2!}\left(z-\overline{z}\right)$.

$$\begin{gathered} e^x\cos y=\frac{1}{2}\left(e^x\left(\cos y+i\sin y\right)+e^x\left(\cos y-i\sin y\right)\right) \\ =\frac{1}{2}\left[\sum _{n=0}^{\infty }\frac{{\left(x+iy\right)}^n}{n!}+\sum _{n=0}^{\infty }\frac{{\left(x+iy\right)}^n}{n!}\right] \\ =\frac{1}{2}\sum _{n=0}^{\infty }\frac{1}{n!}\left({\left(x+iy\right)}^n+{\left(x-iy\right)}^n\right) \end{gathered}$$

Substitute$y=x$into the obtained expression in step 3.

$$\begin{gathered} e^x\cos x=\frac{1}{2}\sum _{n=0}^{\infty }\frac{1}{n!}\left({\left(x+ix\right)}^n+{\left(x-ix\right)}^n\right) \\ =\frac{1}{2}\sum _{n=0}^{\infty }\frac{1}{n!}{x}^n\left({\left(1+i\right)}^n+{\left(1-i\right)}^n\right) \\ =\frac{1}{2}\sum _{n=0}^{\infty }\frac{1}{n!}\left({\left(\sqrt{2}\right)}^n{\left(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)}^n+{\left(\sqrt{2}\right)}^n{\left(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)}^n\right) \\ =\frac{1}{2}\sum _{n=0}^{\infty }\frac{1}{n!}{\left(\sqrt{2}\right)}^n\left({\left(e^{i\frac{\mathrm{\pi }}{4}}\right)}^n+{\left(e^{-i\frac{\mathrm{\pi }}{4}}\right)}^n\right) \end{gathered}$$

Simplify the expression further:

$$\begin{gathered} e^x\cos x=\frac{1}{2}\sum _{n=0}^{\infty }\frac{x^n}{n!}{2}^{\frac{n}{2}}\left(e^{i\frac{n\mathrm{\pi }}{4}}+e^{-i\frac{nx}{4}}\right) \\ =\frac{1}{2}\sum _{n=0}^{\infty }\frac{x^n}{n!}{2}^{\frac{n}{2}}\left(\frac{e^{{\displaystyle \frac{in\mathrm{\pi }}{4}}}+e^{-{\displaystyle \frac{n\mathrm{\pi }}{4}}}}{2}\right) \end{gathered}$$

The power series is $e^x\cos x=\sum _{n=0}^{\infty }\frac{x^n}{n!}{2}^{\frac{n}{2}}\cos \left(\frac{n\mathrm{\pi }}{4}\right)$

The second expression becomes:

$$\begin{gathered} e^x\sin y=\frac{1}{2i}\left(e^x\left(\cos y+i\sin y\right)-e^x\left(\cos y-i\sin y\right)\right) \\ =\frac{1}{2i}\left(\sum _{n=0}^{\infty }\frac{{\left(x+iy\right)}^n}{n!}-\sum _{n=0}^{\infty }\frac{{\left(x-iy\right)}^n}{n!}\right) \end{gathered}$$

Substituteinto the obtained expression.

$$\begin{gathered} e^x\sin x=\frac{1}{2i}\left(\sum _{n=0}^{\infty }\frac{{\left(x+ix\right)}^n}{n!}-\sum _{n=0}^{\infty }\frac{{\left(x+ix\right)}^n}{n!}\right) \\ =\sum _{n=0}^{\infty }\frac{x^n}{n!}\left({\left(1+i\right)}^n-{\left(1-i\right)}^n\right) \\ =\sum _{n=0}^{\infty }\frac{x^n}{n!}\left({\left(\sqrt{2}\right)}^n{\left(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)}^n-{\left(\sqrt{2}\right)}^n{\left(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}\right)}^n\right) \\ =\sum _{n=0}^{\infty }\frac{x^n}{n!}\left({\left(\sqrt{2}\right)}^n{\left(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)}^n-{\left(\sqrt{2}\right)}^n{\left(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}\right)}^n\right) \end{gathered}$$

Simplify the expression further.

$$\begin{gathered} e^x\sin x=\frac{1}{2i}=\sum _{n=0}^{\infty }\frac{x^n}{n!}{2}^{\frac{n}{2}}\left({\left(e^{-\frac{\mathrm{\pi }}{4}}\right)}^n-{\left(e^{-\frac{-\mathrm{\pi }}{4}}\right)}^n\right) \\ =\sum _{n=0}^{\infty }\frac{x^n}{n!}{2}^{\frac{n}{2}}\left(\frac{e^{\frac{in\mathrm{\pi }}{4}}-e^{\frac{-in\mathrm{\pi }}{4}}}{2i}\right) \end{gathered}$$

The power series is $e^x\sin x=\sum _{n=0}^{\infty }\frac{x^n}{n!}{2}^{\frac{n}{2}}\sin \left(\frac{n\mathrm{\pi }}{4}\right)$.