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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

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    $\begingroup$ Check Euler's Formula (en.wikipedia.org/wiki/Euler%27s_formula) $\endgroup$
    – amcalde
    Apr 8 '17 at 2:15
  • $\begingroup$ so the negative doesnt matter for the final answer for the exponential? $\endgroup$ Apr 8 '17 at 2:19
  • $\begingroup$ Not for the base $(-1)$. If $n$ is even, $-n$ is even, and if $n$ is odd, $-n$ is odd. $(-1)^n$ is 1 for even $n$ and $-1$ for odd $n$. $\endgroup$ Apr 8 '17 at 3:19
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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:

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

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  • $\begingroup$ in your third line after you substitute $i^2 = -1$, I think you meant: $1 + ix -\frac{x^2}{2!} - \frac{i(x^3)}{3!} + \frac{x^4}{4!} + \frac{i(x^5)}{5!} - \dots$ $\endgroup$ Apr 8 '17 at 3:09
  • $\begingroup$ Ah yes, my bad, a simple typing error. Thanks for the notice, all fixed! $\endgroup$ Apr 8 '17 at 3:15
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$e^{πj}=-1$

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

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

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  • $\begingroup$ yes j is what you are assuming, so why doesnt the negative matter for the exponential (e^-<---)? $\endgroup$ Apr 8 '17 at 2:21
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    $\begingroup$ @Ihavenoclue If you turn 180 degrees ($\pi$), it doesn't matter whether you turned right ($+$) or left ($-$). $\endgroup$
    – Neal
    Apr 8 '17 at 2:37
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    $\begingroup$ @Ihavenoclue, $(-1)^{-n} = \dfrac1{(-1)^n} = \left(\dfrac1{-1}\right)^n = (-1)^n$ $\endgroup$
    – user307169
    Apr 8 '17 at 2:39
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It's best to simply sketch an Argand diagram for this. You will soon see that the value of $e^{n \pi j}$ just oscillates between $-1$ and $1$ depending on whether $n$ is odd or even hence it is equal to $(-1)^n$.

To see this explicitly just use the fact that $e^{j \pi n}=\cos nx +j \sin nx$ and noting that the sine of any integer multiple of $\pi$ is zero. If you do the same for $-\pi$ you will get the same answer (equivalent to $-n$). It just means that the complex vector is rotating around the unit circle in the opposite direction now.

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