3
$\begingroup$

According to the Proof :

\begin{align} X_n &= \sum_{k=0}^{N-1}x_ke^{-j\frac{2\pi k n}{N}}\\ X_{N-n} &= \sum_{k=0}^{N-1}x_ke^{-j\frac{2\pi k (N-n)}{N}}\\ &=\sum_{k=0}^{N-1}x_k e^{-j 2\pi k}e^{j\frac{2\pi k n}{N}} \end{align}

Using $\exp(-j2\pi k) = 1 \quad \forall \ k$

\begin{align} X_{N-n} &= \sum_{k=0}^{N-1}x_k e^{j\frac{2\pi k n}{N}}\\ \end{align}

How is $\exp(-j2\pi k) = 1 \quad \forall \ k$ true ? Doesn't it mean $-j2\pi k = 0$ ? But that's not possible right?

$\endgroup$
6
$\begingroup$

Hint:

According to Euler's formula we have $$e^{-j2\pi k}=\cos(2\pi k)-j\sin(2\pi k)=\ldots$$

$\endgroup$
5
$\begingroup$

Remember that $e^z$ has a very different meaning than $e^x$ (taking $z\in\mathbb{C}$ and $x\in\mathbb{R}$).

If the exponent was real, then, as you state in your question:

$$e^x = 1 \iff x=0$$

However, when the exponent is complex, this function acquires a very different meaning. Let $z=x+jy$, where $x,y\in\mathbb{R}$ and $j$ is the imaginary unit. Then

$$e^z=e^{x+jy}=e^x \cdot e^{jy}$$

The first factor is real and, therefore, equals $1$ if and only if $x=0$. The second factor, however, has a purely imaginary exponent (such as the one in your question). Such a function can be calculated as:

$$e^{jy}=\cos(y)+j\sin(y)$$

As you can see, this function is periodic. In your case, $y=-2\pi k$, with $k\in \mathbb{Z}$. Then

$$e^{-j2\pi k}=\cos(-2\pi k)+j\sin(-2\pi k)$$

Cosine is an even function and sine is odd, so

$$\cos(-2\pi k)+j\sin(-2\pi k)=\cos(2\pi k)-j\sin(2\pi k)$$

What are the values of $\cos(2\pi k)$ and $\sin(2\pi k)$? Well... they seem to depend on $k$. But... do they? The answer is no.

The cosine equals $1$ for any multiple of $2\pi$. The sine equals $0$ for any multiple of $2\pi$. This leads to the final result:

$$e^{-j2\pi k}=\cos(2\pi k)-j\sin(2\pi k)=1-j0=1 \ \forall k$$

$\endgroup$
3
$\begingroup$

Did you ever think about where $\pi $ came from? Watch out...

Let us first draw this weird function $e^{-2j\pi t}$ for several $t\in[0,10]$ (the little blue circles joined by line segments):

Fourier phasor

On one axis, the variable $t$, on the others the real and imaginary parts, respectively. It looks like an everlasting spring. Now draw the function at integers $k\in [0,10] \cap \mathbb{Z} = \{0,1,\ldots,10\}$. These are the little red crosses, joined by red segments. They exactly have a real part equal to one, and an imaginary part to zero: $e^{-2j\pi k} = 1+0j$.

It might seem weird with respect to the behavior of the standard exponential, but the complex argument $-2j\pi k$ is a deal changer. For real $t$, $e^{ t} = 1$ has only one solution. But $e^{-2j\pi t} = 1$ has an infinity. Its maps the real axis onto the unit circle.

In fact, in some math books, like Rudin's Real and Complex analysis (very first pages), $e$ comes first, and $\pi$ subsequently.

One first defines for any complex $z$:

$$e^z=\sum_{n=0}^{+\infty} \frac{z^n}{n!}$$

which is absolutely convergent. And then you get some results, like $e^z$ is never equal to zero. But the most striking one is:

There exists a positive number $\pi$ such that $e^{\pi j/2} = j$ and such that $e^z = 1$ if and only if $z/(2\pi j )$ is an integer.

So $\pi $ first exists as the "measure" of the periodicity of the everlasting spring of the complex exponential. The site Euler's Equation provides other nice graphs.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.