Using the representation for ${\mathit{e}}^{\mathbf{(}\mathit{\alpha }\mathbf{+}\mathit{i}\mathit{\beta }\mathbf{)}\mathit{t}}$ in6, verify the differentiation formula7.

When Euler's formula $e^{i\theta }=cos\theta +isin\theta$(with $\theta =\beta t$) is used in equation $e^{(\alpha +i\beta )t}=e^{\alpha t+i\beta t}=e^{\alpha t}{e}^{i\beta t}$, we find $e^{(\alpha +i\beta )t}=e^{\alpha t}cos\beta t+isin\beta t$, which expresses the complex function $e^{(\alpha +i\beta )t}$in terms of familiar real functions. Having made sense out of $e^{(\alpha +i\beta )t}$, we can now show that $\left(7\right)\frac{d}{dt}{e}^{(\alpha +i\beta )t}=(\alpha +i\beta )e^{(\alpha +i\beta )t}$,

$e^{(\alpha +i\beta )t}=e^{\alpha t}(cos\beta t+isin\beta t)$given from (6) in the chapter.

$\frac{d}{dt}{e}^{\alpha t}(cos\beta t+isin\beta t)...\left(a\right)$

Differentiate by parts,

$$\begin{gathered} \frac{d}{dt}{e}^{\alpha t}cos\beta t=\left(e^{\alpha t}(-\beta sin\beta t)+\alpha e^{\alpha t}\left(cos\beta t\right)\right) \\ \frac{d}{dt}{e}^{\alpha t}isin\beta t=i\left(e^{\alpha t}\left(\beta cos\beta t\right)+\alpha e^{\alpha t}\left(sin\beta t\right)\right) \end{gathered}$$

Now substitute the both into the equation (a),

$\frac{d}{dt}{e}^{\alpha t}cos\beta t+\frac{d}{dt}{e}^{\alpha t}isin\beta t=\left(e^{\alpha t}(-\beta sin\beta t)+\alpha e^{\alpha t}\left(cos\beta t\right)\right)+i\left(e^{\alpha t}\left(\beta cos\beta t\right)+\alpha e^{\alpha t}\left(sin\beta t\right)\right)$.

Expand$-\beta e^{\alpha t}\left(sin\beta t\right)+\alpha e^{\alpha t}\left(cos\beta t\right)+i\beta e^{\alpha t}\left(cos\beta t\right)+i\alpha e^{\alpha t}\left(sin\beta t\right)$

Collect like terms of $cos\beta t$and $sin\beta t$

$(\alpha +i\beta )e^{\alpha t}\left(cos\beta t\right)+(i\alpha -\beta )e^{\alpha t}\left(sin\beta t\right)$

Factor out $e^{\alpha t}$

$\frac{d}{dt}{e}^{\alpha t}cos\beta t+\frac{d}{dt}{e}^{\alpha t}isin\beta t=e^{\alpha t}\left(\right(\alpha +i\beta \left)\right(cos\beta t)+(i\alpha -\beta \left)\right(sin\beta t\left)\right)$

Multiply $1=-(i\times i)$ to $(i\alpha -\beta )$

$$\begin{gathered} i-\left(i\right(i\alpha -\beta \left)\right)=-i\left(i^2\alpha -i\beta \right)=-i(-\alpha -i\beta )=i(\alpha -i\beta ) \\ {e}^{\alpha t}\left(\right(\alpha +i\beta \left)\right(cos\beta t)+i(\alpha +i\beta \left)\right(sin\beta t\left)\right) \end{gathered}$$

Factor out $(\alpha +i\beta )$

$(\alpha +i\beta )\times e^{\alpha t}\left(\right(cos\beta t)+i(sin\beta t\left)\right)$

Now it's in the format of $e^{(\alpha +i\beta )t}=e^{\alpha t}(cos\beta t+isin\beta t)$, so use it to get the correct solution $\frac{d}{dt}{e}^{\alpha t}(cos\beta t+isin\beta t)=(\alpha +i\beta )\times e^{(\alpha +i\beta )t}$.