# ECE301 Fall 2008, Professor C.C. Wang

## Use this area to post your questions

I am trying to figure out how to compute the norm of the DT signal $x[n]= e^{j 2 \pi n}$. According to the solutions, the answer is

$\left| e^{j 2 \pi n} \right| = 1$.

I don't get it. Should'nt the answer be a function of n???

### Response

Since we are dealing with complex numbers, we can make the following substitution:

$\left| e^{j 2 \pi n} \right| = \left| \cos{2 \pi n} + j \sin{2 \pi n} \right|$

The magnitude of this expression will always be one since $\cos^2\theta+\sin^2\theta\equiv 1$.

### Related Question

I see your point. But if what you say is true, then we also have

$e^{j 2 \pi t} = \left( e^{j 2 \pi} \right)^t= \left( cos{2 \pi } + j sin{2 \pi } \right)^t= 1^t =1,$

which is clearly wrong, because my high school teacher showed us the graph of $e^{j 2 \pi t}$ and it was oscillating. So what am I doing wrong?

### MA181 to the rescue!

The identity $(e^a)^t=e^{ta}$ is true for real numbers, but it is not always true for complex numbers, even when $a$ is complex and $t$ is real. When $z$ is complex and $t$ is real, $z^t$ stands for $e^{t\log z}$, where $\log z$ is the complex logarithm of $z$, which has infinitely many possible values.

By the way, the constant function is a function. The modulus of $e^{j\theta}$ is equal to one for any value of $\theta$ because that complex number represents a point on the unit circle.

I saw the plea on our page and thought I'd check it out. While I can't claim to have much understanding of what is going on, it seems like that solution is valid because $n$ represents either the natural numbers or integrers, not the real numbers. Therefore, it would be like taking the value of the sine function at whole-number products of $2\pi$. Even though the function oscillates, the sequence is constant. Hope I haven't made a fool of myself. --Jmason

## Alumni Liaison

Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett