# Limits of Complex exponentials

1. Apr 6, 2005

### Sanjeev-ISE

I was told that the limit of e^-iwt as t-> infinity is 1 or is at least modelled as 1. Can anyone tell me why this is?

Also, what is the limit of e^iwt as t-> infinity and likewise, the limits of e^iwt and e^-iwt as t-> 0

How do you conceptually visualise the complex exponential?

Hope someone can help.

2. Apr 6, 2005

### matt grime

The limits you were told exist do not (as t tends to infinity), and you should be able to figure out why if you know a little about the complex numbers (consider what happens when t is an even multiple of pi, and an odd multiple of pi).

When t is zero the function is well defined and is 1.

Suppose t is real, then e^{iwt] (why the w?) is cos(wt) +i sin(wt).

What do you know about the complex plane?

3. Apr 6, 2005

### dextercioby

No.They don't exist.Period.Roughly 250 yrs ago,Euler found a nice formula linking circular trigonometric functions & complex exponentials (of arbitrary complex argument)

$$e^{i\varphi}=\cos\varphi+i\sin\varphi$$

Since it's a common fact that,due to their "oscillation",the circular trigonometrical functions of real arg. do not have asymptotic limits (on the real axis),we conclude the same thing for the complex exponential of real argument (i.e.$\varphi\in \mathbb{R}$).

So to answer your first question,you've been either lied to,or someone made a bad joke.

To answer your second.Think of the complex/Gauss/Argand plane and since the complex exp.of real arg.is a multivalued function of modulus one,u can visualize the $e^{i\varphi}$ as an arrow with its top on the unit circle & its base/origin in the point $(0,0)$.

Daniel.

4. Apr 6, 2005

### HallsofIvy

Staff Emeritus
I suspect that was intended to be $$\omega$$- typically used to represent the frequency.

I wonder if the original poster wasn't told that the limit of $$e^{i\omega t}$$ was 1 as t goes to 0 rather than infinity.

5. Oct 22, 2011

### schtruklyn

Hi. Answer to original question is - yes, exponential function changes values at the point at infinity depending on the path one chooses complex variable to approach the point at infinity. This is due to fact that exponential function has Essential Singularity at infinity. You know that infinities of analytic functions are poles. All poles are of some finite order n. If singularity is pole of infinite order, then there is an essential singularity there. By Weierstrass theorem, analytic functions assume any possible complex value at any neighborhood of essential singularity. This means one gets different values as approaches essential infinity along different directions. I hope this helps someone. Peace c:

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