Example Adopt A Group Page

(1)

\begin{align} e^{i\theta}=\cos(\theta)+i\sin(\theta) \end{align}

+ Show justification

Recall the following four facts:

(2)

\begin{align} i^2=-1;\ i^3=-i;\ i^4=1;\ i^5=i;\ i^6=-1;\ \ldots \end{align}

(3)

\begin{align} \cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots \end{align}

(4)

\begin{align} \sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots \end{align}

(5)

\begin{align} e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots \end{align}

Therefore, replacing $x$ by $i\theta$ in (5) gives

(6)

\begin{align} e^{i\theta}=1+(i\theta)+\frac{(i\theta)^2}{2!}+\frac{(i\theta)^3}{3!}+\frac{(i\theta)^4}{4!}+\cdots \end{align}

Now, using (2) we see that this means

(7)

\begin{align} e^{i\theta}=1+i\theta -\frac{\theta^2}{2!}-\frac{i(\theta)^3}{3!}+\frac{\theta^4}{4!}+\frac{i(\theta)^5}{5!}-\frac{\theta^6}{6!}-\cdots \end{align}

and so

(8)

\begin{align} e^{i\theta}=1-\frac{\theta^2}{2}+\frac{\theta^4}{4!}-\frac{\theta^6}{6!}+\cdots + i\theta-\frac{i(\theta)^3}{3!}+\frac{i(\theta)^5}{5!}+\cdots. \end{align}

Factoring the $i$ out of the second part of the previous equation gives

(9)

\begin{align} e^{i\theta}=\left(1-\frac{\theta^2}{2}+\frac{\theta^4}{4!}-\frac{\theta^6}{6!}+\cdots\right) + i\left(\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}+\cdots\right)=\cos(\theta)+i\sin(\theta). \end{align}

according to equations (3) and (4).

(10)

\begin{align} e^{i\pi}+1=0 \end{align}

+ Show justification

Replacing $\theta$ with $\pi$ in (1) above, we get that

(11)

\begin{align} e^{i\pi}=\cos(\pi)+i\sin(\pi)=-1+i\cdot 0=-1 \end{align}

and so adding 1 to both sides we get $e^{i\pi}+1=0$ as required.