How does $e^{-j\pi n}$ become $(-1)^n$

Question

For
$$e^{-j\pi n}$$

How does this become $$(-1)^n$$

or is it actually $$(-1)^{-n}$$ I have checked on calculator and values are all the same when the same n value is used

Answer 1

Consider the power series of $e^x$, that is:

$$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}=1+\frac x{1!}+\frac{x^2}{2!} +\frac{x^3}{3!}+\frac{x^4}{4!}+\ ...$$

Now consider Euler's constant $e$ raised to the power $ix$, such that $i:=\sqrt{-1}$

$$e^{ix}=\sum_{n=0}^{\infty}\frac{(xi)^n}{n!}=1+ix+\frac{(ix)^2}{2!}+\frac {(ix)^3}{3!}+\frac {(ix)^4}{4!}+\frac {(ix)^5}{5!}+\ ...$$

Noting $i^2=-1$, an equivalent expression is:

$$e^{ix}=\sum_{n=0}^{\infty}\frac{(xi)^n}{n!}=1+ix-\frac{x^2}{2!}-\frac {i(x)^3}{3!}+\frac {x^4}{4!}+\frac {i(x)^5}{5!}-\ ...$$

Factoring yields a very interesting expression:

$$=\sum_{k=0,m=1,k \ even}^{\infty}\frac {x^k(-1)^{(m+1)}}{k!}+(i)\sum_{k=1,m=1,k \ odd}^{\infty}\frac {{x^k}{(-1)^{m+1}}}{k!}$$

$$=\left(1-\frac {x^2}{2!}+\frac {x^4}{4!}-\frac {x^6}{6!} + \ ...\right)+i\left(x-\frac {x^3}{3!}+\frac {x^5}{5!} -\frac {x^7}{7!} \ + \ ...\right)=\cos x+i \sin x$$

Denoted as "Euler's Formula," various mathematical equalities can thus be proved such as the trigonometric angle identities upon considering its real and imaginary parts.

Now that we have some intuition behind Euler's formula, your above question can now be analysed:

We have:

$$e^{i\pi}=-1$$ Why? Simply set $x=\pi$ in the formula $e^{{i}{x}}= \cos x+i \sin x $.

Thus finally:

$$e^{i\pi\ n}=e^{(i\pi)n}=(-1)^n$$

Answer 2

$e^{πj}=-1$

So $e^{-πjn}=(e^{πj})^{-n}=(-1)^{-n}$.

I am assuming $j$ is the imaginary unit, or $\sqrt{-1}$.

Answer 3

Expression $\large -j \pi n$ is equivalent to:

• $\large \quad j \pi (-n)$ and
• $\large \quad j (-\pi) n$.

This leads to two possible uses of Euler's formula:

1. $\large \quad e^{j \pi} = \cos(\pi) + j sin(\pi)$
   $\large \quad e^{j \pi} = -1+0j = -1$
   $\large \quad \implies e^{-j \pi n} = [e^{j \pi}]^{-n} = -1^{-n}$

2. $\large \quad e^{j (-\pi)} = \cos(-\pi) + j sin(-\pi)$
   $\large \quad e^{j (-\pi)} = -1+0j = -1$
   $\large \quad \implies e^{-j \pi n} = [e^{j (-\pi)}]^{n} = -1^{n}$

Actually:

$\large \quad \quad -1^{-n} = \frac {1} {-1^n} = \frac {1^n} {-1^n} = {[\frac {1} {-1}]}^n = -1^{n}$

We could also use De Moivre's formula, which deals directly with angles like $n \pi$. This would lead to the same result.