# Complex Number

1. Jan 11, 2005

### sam2

Hi all,

Looking for some help on the following problem. Any replies much appreciated.

I have the complex number

exp(i.x)

If x = - infinity,

is this zero?? Is there any intuitive/straightforward value that it should be? I decomposed the expression into cos and sin and it looks like the number doesnt converge to anything!

regards,

2. Jan 11, 2005

### dextercioby

Of course it doesn't converge.U need a complex number of different modulus.The one u've chosen has modulus one,i.e. 1.U cannot make it's argument go to infinity,coz that would mean rotating the unit vector (which is represented by e^(ix)) an ininite amount of times and the phase could not be determined.

Daniel.

3. Jan 11, 2005

### matt grime

It is undefined, and there is no reason why it should be defined.

4. Jan 11, 2005

### marlon

Well, it doesn't go to infinity because of the very reason that you quoted. But don't worry : there is NO problem because this function is NOT DEFINED in the infinity. So, there is no problem and why should there be any ???

marlon

5. Jan 11, 2005

### sam2

Got it.

Many thanks for the replies.

Regards,
Sam

6. Jan 11, 2005

### Hurkyl

Staff Emeritus
A comment: a number is a number, it doesn't make sense to say it "converges" to anything. While the distinction is often blurred, it is still important to remember that the concept of limit is different than that of a number.

Last edited: Jan 11, 2005