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I came across a question where the following was done:

$$ (e^{jw_0t})^2 = 1. $$

After some searching I realized that there was an identity that says that $e^{jn\pi} = 1$ if $n$ is even and $-1$ if $n$ is odd.

Can someone provide a proof for this identity? I am really confused as to how $e^{jn\pi} = 1$ if $n$ is even?!

thanks.


NOTE: I am guessing w0 = (2*pi) / period because we usually use cos(w0*t) notation so

After looking at the responses below, if I now use the Euler formula:

e^(j* [(2*pi) / period]) so this equals cos((2*pi) / period) + jsin((2*pi) / period)... this does not equal 1..

am I doing something wrong?

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    Your expressions are very difficult to read. What do u mean by what you have written down?2011-10-15

2 Answers 2

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Euler's formula says that, for every real number $x$, $\mathrm e^{\mathrm i x}=\cos(x)+\mathrm i\sin(x)$. Plug $x=n\pi$ into this.

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    And mathematicians usually write $\mathrm i$ for what some physicists call $j$.2011-10-15
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    Alternatively, demonstrate the values of $\exp(\pi j)$ and $\exp(2\pi j)$, and then note that all even numbers are of the form $2n$, and odd numbers are of the form $2n+1$...2011-10-15
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    @J.M. How does one demonstrate the value of, say, $\exp(\pi j)$ without Euler?2011-10-15
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    Oh, I was just specializing your solution. To be explicit: use Euler for $x=\pi$ and then consider what happens if you raise the result of that to an odd or even power...2011-10-15
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    @J.M. Right. Except that this supposes that $[\cos(x)+\mathrm i\sin(x)]^n=\cos(nx)+\mathrm i\sin(nx)$ for every integer $n$, at least for $x=\pi$.2011-10-15
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    Oh, right. It seems that establishing de Moivre (sans Euler) is more complicated than I remembered...2011-10-15
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    e^j*w0*t is not same as having n*pi in the exponential. So how can they say that [e^(j*w0*t)]^2 = 1?2011-10-15
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    @rrazd: FWIW, 1. you never defined $w_0$; 2. you never said where you saw this; 3. "I am really confused as to how $e^{jn\pi} = 1$ if $n$ is even?!", to quote you...2011-10-15
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    I am guessing w0 = (2*pi) / period because we usually use cos(w0*t) notation so then e^(j* [(2*pi) / period]) so this equals cos((2*pi) / period) + jsin((2*pi) / period)... this does not equal 1...2011-10-15
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    Since the person who asked teh question is probably physicist, you can probably convince him/her that the Euler formula is right by replacing $e^x, \sin(x), \cos(x)$ by their power series...2011-10-15
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Let $t$ denote a real number, and start from $e^{it} = \displaystyle\lim_{n\to\infty} \left( 1+\frac{it}{n} \right)^n$. Since $1\leq \displaystyle\left(1+\frac{t^2}{n^2}\right)^n \leq \left(e^{t^2}\right)^{1/n}\to 1$, we have $|e^{it}|=1$. This $e^{it}$ lies on the unit cricle.

Define $\sin(x), \cos(x), \tan(x) $ to be the trigonometric functions defined by points on the unit circle, not power series. Draw a picture to convince yourself that $\sin(x) \leq x \leq \tan(x) $ for small positive $x$, and $\tan(x) \leq x \leq \sin(x)$ for small negative $x$. Argue via squeeze theorem that $\tan(x) \sim x $ as $x\to 0$.

Now note $\arg(e^{it}) = \displaystyle\lim_{n\to\infty} n\cdot \tan^{-1}(t/n) = t$ . Thus, $e^{it}$ is the point on the unit circle whose argument is $t$, which answers your question.

Some nice side effects:

  • This connects the power series trig functions back to geometry quite quickly. When one starts from power series, it is not obvious that they are $2\pi $ periodic, so this is a nice way to show that.
  • This allows us to avoid having to define $\pi$ as the smallest positive $x$ such that $\sin(x) =0$ or something like that - it matches with our geometric definition that $2\pi $ is the circumference of the unit circle.