In calculus you learned that \(e^{u}, \cos u,\) and \(\sin u\) can be represented
by the infinite series
$$
\begin{array}{c}
e^{u}=\sum_{n=0}^{\infty} \frac{u^{n}}{n !}=1+\frac{u}{1 !}+\frac{u^{2}}{2
!}+\frac{u^{3}}{3 !}+\cdots+\frac{u^{n}}{n !}+\cdots \\
\cos u=\sum_{n=0}^{\infty}(-1)^{n} \frac{u^{2 n}}{(2 n) !}=1-\frac{u^{2}}{2
!}+\frac{u^{4}}{4 !}+\cdots+(-1)^{n} \frac{u^{2 n}}{(2 n) !}+\cdots,
\end{array}
$$
and
$$
\sin u=\sum_{n=0}^{\infty}(-1)^{n} \frac{u^{2 n+1}}{(2 n+1)
!}=u-\frac{u^{3}}{3 !}+\frac{u^{5}}{5 !}+\cdots+(-1)^{n} \frac{u^{2 n+1}}{(2
n+1) !}+\cdots
$$
for all real values of \(u\). Even though you have previously considered (A)
only for real values of \(u\), we can set \(u=i \theta,\) where \(\theta\) is real,
to obtain
$$
e^{i \theta}=\sum_{n=0}^{\infty} \frac{(i \theta)^{n}}{n !}
$$
Given the proper background in the theory of infinite series with complex
terms, it can be shown that the series in (D) converges for all real \(\theta\).
(a) Recalling that \(i^{2}=-1,\) write enough terms of the sequence
\(\left\\{i^{n}\right\\}\) to convince yourself that the sequence is repetitive:
$$
1, i,-1,-i, 1, i,-1,-i, 1, i,-1,-i, 1, i,-1,-i, \cdots
$$
Use this to group the terms in (D) as
$$
\begin{aligned}
e^{i \theta}
&=\left(1-\frac{\theta^{2}}{2}+\frac{\theta^{4}}{4}+\cdots\right)+i\left(\theta-\frac{\theta^{3}}{3
!}+\frac{\theta^{5}}{5 !}+\cdots\right) \\
&=\sum_{n=0}^{\infty}(-1)^{n} \frac{\theta^{2 n}}{(2 n) !}+i
\sum_{n=0}^{\infty}(-1)^{n} \frac{\theta^{2 n+1}}{(2 n+1) !}
\end{aligned}
$$
In calculus you learned that \(e^{u}, \cos u,\) and \(\sin u\) can be represented
by the infinite series
$$
\begin{array}{c}
e^{u}=\sum_{n=0}^{\infty} \frac{u^{n}}{n !}=1+\frac{u}{1 !}+\frac{u^{2}}{2
!}+\frac{u^{3}}{3 !}+\cdots+\frac{u^{n}}{n !}+\cdots \\
\cos u=\sum_{n=0}^{\infty}(-1)^{n} \frac{u^{2 n}}{(2 n) !}=1-\frac{u^{2}}{2
!}+\frac{u^{4}}{4 !}+\cdots+(-1)^{n} \frac{u^{2 n}}{(2 n) !}+\cdots,
\end{array}
$$
and
$$
\sin u=\sum_{n=0}^{\infty}(-1)^{n} \frac{u^{2 n+1}}{(2 n+1)
!}=u-\frac{u^{3}}{3 !}+\frac{u^{5}}{5 !}+\cdots+(-1)^{n} \frac{u^{2 n+1}}{(2
n+1) !}+\cdots
$$
for all real values of \(u\). Even though you have previously considered (A)
only for real values of \(u\), we can set \(u=i \theta,\) where \(\theta\) is real,
to obtain
$$
e^{i \theta}=\sum_{n=0}^{\infty} \frac{(i \theta)^{n}}{n !}
$$
Given the proper background in the theory of infinite series with complex
terms, it can be shown that the series in (D) converges for all real \(\theta\).
(a) Recalling that \(i^{2}=-1,\) write enough terms of the sequence
\(\left\\{i^{n}\right\\}\) to convince yourself that the sequence is repetitive:
$$
1, i,-1,-i, 1, i,-1,-i, 1, i,-1,-i, 1, i,-1,-i, \cdots
$$
Use this to group the terms in (D) as
$$
\begin{aligned}
e^{i \theta}
&=\left(1-\frac{\theta^{2}}{2}+\frac{\theta^{4}}{4}+\cdots\right)+i\left(\theta-\frac{\theta^{3}}{3
!}+\frac{\theta^{5}}{5 !}+\cdots\right) \\
&=\sum_{n=0}^{\infty}(-1)^{n} \frac{\theta^{2 n}}{(2 n) !}+i
\sum_{n=0}^{\infty}(-1)^{n} \frac{\theta^{2 n+1}}{(2 n+1) !}
\end{aligned}
$$
(c) If \(\alpha\) and \(\beta\) are real numbers, define
$$
e^{\alpha+i \beta}=e^{\alpha} e^{i \beta}=e^{\alpha}(\cos \beta+i \sin \beta)
$$
Show that if \(z_{1}=\alpha_{1}+i \beta_{1}\) and \(z_{2}=\alpha_{2}+i \beta_{2}\)
then
$$
e^{z_{1}+z_{2}}=e^{z_{1}} e^{z_{2}}
$$
(d) Let \(a, b,\) and \(c\) be real numbers, with \(a \neq 0 .\) Let \(z=u+i v\) where
\(u\) and \(v\) are real-valued functions of \(x .\) Then we say that \(z\) is a
solution of
$$
a y^{\prime \prime}+b y^{\prime}+c y=0
$$
if \(u\) and \(v\) are both solutions of (G). Use Theorem \(5.2 .1(\mathbf{c})\) to
verify that if the characteristic equation of (G) has complex conjugate roots
\(\lambda \pm i \omega\) then \(z_{1}=e^{(\lambda+i \omega) x}\) and
\(z_{2}=e^{(\lambda-i \omega) x}\) are both solutions of (G).
