Write the series for${\mathbf{e}}^{\mathbf{x}\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{i}\mathbf{)}}$. Write$\mathbf{1}\mathbf{+}\mathit{i}\mathbf{}$in theform${\mathbf{re}}^{\mathbf{i\theta }}$ and so obtain (easily) the powers of ($\mathbf{1}\mathbf{+}\mathit{i}$). Thus show, for example, that the${\mathbf{e}}^{\mathbf{x}}\mathbf{cos}\mathbf{}\mathbf{x}$series has no${\mathbf{x}}^{\mathbf{2}}$term, no${\mathbf{x}}^{\mathbf{6}}$term, etc., and a similar result for the${\mathbf{e}}^{\mathbf{x}}\mathbf{sinx}$series. Find (easily) a formula for the general term for each series.

The given expression is$e^{x(1+i)}$ .

Represent the complex number in rectangular form means writing the given complex number in the form of in$\mathit{x}\mathbf{+}\mathit{i}\mathit{y}$ which x is the real part and y is the imaginary part.

Consider be a complex number.

$$\begin{gathered} e^{\left(z\right)}=e^{\left(x\right)}\left|e^{\left(xi\right)}\right| \\ {e}^{\left(z\right)}=e^{\left(x\right)}\left[cos\right(x)+isin(x\left)\right].......\left(1\right) \end{gathered}$$

Write in series form.

$$\begin{gathered} {\mathrm{e}}^{\left(\mathrm{z}\right)}=\sum \frac{{\mathrm{z}}^{\mathrm{n}}}{\mathrm{n}!} \\ {\mathrm{e}}^{\left(\mathrm{z}\right)}=\sum \frac{(\mathrm{x}+\mathrm{ix}{)}^{\mathrm{n}}}{\mathrm{n}!} \\ {\mathrm{e}}^{\left(\mathrm{z}\right)}=\sum \frac{{\mathrm{x}}^{\mathrm{n}}(1+\mathrm{i}{)}^{\mathrm{n}}}{\mathrm{n}!} \\ {\mathrm{e}}^{\left(\mathrm{z}\right)}=\sum \frac{{\mathrm{x}}^{\mathrm{n}}[\sqrt{2}{\mathrm{e}}^{(\mathrm{\pi i}/4)}{]}^{\mathrm{n}}}{\mathrm{n}!} \end{gathered}$$

$$\begin{gathered} {\mathrm{e}}^{\left(\mathrm{z}\right)}==\sum \frac{{\mathrm{x}}^{\mathrm{n}}[\sqrt{2}{]}^{\mathrm{n}}{\mathrm{e}}^{(\mathrm{\pi ni}/4)}}{\mathrm{n}!} \\ {\mathrm{e}}^{\left(\mathrm{z}\right)}=\sum \frac{{\mathrm{x}}^{\mathrm{n}}\left[\sqrt{2}{]}^{\mathrm{n}}\right[\cos (\mathrm{n\pi }/4)+\mathrm{isin}(\mathrm{n\pi }/4)]}{\mathrm{n}!} \\ {\mathrm{e}}^{\left(\mathrm{z}\right)}=\sum \frac{{\mathrm{x}}^{\mathrm{n}}\left[\sqrt{2}{]}^{\mathrm{n}}\cos \right(\mathrm{n\pi }/4)}{\mathrm{n}!}+\mathrm{i}\frac{{\mathrm{x}}^{\mathrm{n}}\left[\sqrt{2}{]}^{\mathrm{n}}\sin \right(\mathrm{n\pi }/4)}{\mathrm{n}!}......\left(2\right) \\ {\mathrm{e}}^{\left(\mathrm{z}\right)}=\frac{\sum {\mathrm{x}}^{\mathrm{n}}{\left(2\right)}^{\mathrm{n}/2}\cos (\mathrm{n\pi }/4)}{\mathrm{n}!}+\mathrm{i}\sum \frac{{\mathrm{x}}^{\mathrm{n}}(2{)}^{\mathrm{n}/2}\sin (\mathrm{n\pi }/4)}{\mathrm{n}!} \end{gathered}$$

From equation (1) and (2), take the real and imaginary part differently.

$$\begin{gathered} e^{\left(x\right)}\cos \left(x\right)=\underset{}{\sum \frac{x^n{\left(2\right)}^{n/2}\cos \left(n\pi /4\right)}{n!}} \\ {e}^{\left(x\right)}\sin \left(x\right)=\underset{}{\sum \frac{x^n{\left(2\right)}^{n/2}\sin \left(n\pi /4\right)}{n!}} \end{gathered}$$

Hence the values of the given question are:

$$\begin{gathered} {\mathrm{e}}^{\left(\mathrm{x}\right)}\cos \left(\mathrm{x}\right)=\underset{}{\sum \frac{{\mathrm{x}}^{\mathrm{n}}{\left(2\right)}^{\mathrm{n}/2}\cos \left(\mathrm{n\pi }/4\right)}{\mathrm{n}!}} \\ {\mathrm{e}}^{\left(\mathrm{x}\right)}\sin \left(\mathrm{x}\right)=\underset{}{\sum \frac{{\mathrm{x}}^{\mathrm{n}}{\left(2\right)}^{\mathrm{n}/2}\sin \left(\mathrm{n\pi }/4\right)}{\mathrm{n}!}} \end{gathered}$$.