Exercise 05.15 - proposed solution

Prove

(w nonzero) requires choosing

where

the specific RHS choice of

.

Dealing with the special cases where a and/or b are zero:

If a is zero then

and there's nothing to prove.

If b is zero, then

and there's nothing to prove.

So from now on w, a and b are all assumed to be nonzero.

Let

We note first off that

So

Giving

To begin, by definition

and

We require these to be equal, which means the moduli must match and the arguments must differ by a multiple of

, hence:

Equating the moduli:

{Eqn. 1}

Equating the arguments (give or take multiples of

):

Since the LHS is a complex number, and n is an integer, this equation can only be true in general if n = 0. So

Subbing into { Eqn. 1}:

And since m is an integer and

and

are real, the only solution in general is m = 0. Hence k = 0.

Hence,