Is exp(i.x) with x=-infinity equal to zero?

[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 doesn't converge to anything!

regards,

 

[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.

 

[matt grime]

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

 

[marlon]

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 doesn't converge to anything!

regards,

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

 

[sam2]

Got it.

Many thanks for the replies.

Regards,
Sam

 

[Hurkyl]

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.