# Proof of complex conjugate symmetry property of DFT

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?

Hint:

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

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:

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

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): 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.