Solving the Simple Equation exp(ikL) - exp(-ikL) = 0

[kasse]

Homework Statement

What does it take for

exp(ikL) - exp(-ikL) = 0

?

The Attempt at a Solution

cos(kL) + i sin(kL) - cos(kL) - i sin(kL) = 0

0 = 0

This suggests that kL can be anything, but this is not the right answer.

 

[HallsofIvy]

Homework Statement

What does it take for

exp(ikL) - exp(-ikL) = 0

?

The Attempt at a Solution

cos(kL) + i sin(kL) - cos(kL) - i sin(kL) = 0

You are using exp(ikL)= cos(kl)+ i sin(kL), right?

So exp(-ikL)= cos(kL)- i sin(kL) because cosine is an even function and sine is an odd function.

However, cos(kL)+ i sin(kL)- (cos(kL)- sin(kL)) is NOT 0.

0 = 0

This suggests that kL can be anything, but this is not the right answer.

 

[Mark44]

Homework Statement

What does it take for

exp(ikL) - exp(-ikL) = 0

?

The Attempt at a Solution

cos(kL) + i sin(kL) - cos(kL) - i sin(kL) = 0

The equation above should be:
cos(kL) + i sin(kL) - cos(-kL) - i sin(-kL) = 0
or
cos(kL) + i sin(kL) - cos(kL) + i sin(kL) = 0
So 2 sin(kL) = 0, which says that kL = n*pi, where n is an integer.

0 = 0

This suggests that kL can be anything, but this is not the right answer.

 

[kasse]

Here's the whole problem:

There are two harmonic waves on a string:

y1=Aexp(i(kx-wt))
y2=Bexp(i(-kx-wt))

(1) y(x=0)=0
(2) y(=L) = 0

The total wave is

(3) y=exp(-iwt)[Aexp(ikx) + Bexp(-ikx)]

(1) gives A = -B, so that (3) can be written

y = exp(-iwt)[Aexp(ikL) - Aexp(-ikL)] = 0

For (2) to be true, we must have exp(ikL) - exp(-ikL) = 0, that is

cos(kL) + i sin(kL) - cos(-kL) - i sin (-kL) = 0 ---> 2sin(kL) = 0 ---> kL = n*pi

My book is wrong, then?

 

[Mark44]

Books have been known to have wrong answers. Your equation is satisfied if kL = pi, or 3*pi, or 5*pi, and so on, values that your book's solution doesn't include.