e^inx = 0 ?

aequinoctium

I have come up with this recently and i am unable to determine if it is true, or even possible!

It can be shown that a linear combination of sines and cosines can be represented purely as a cosine function:
Acos(nx) + Bsin(nx) = Dcos(nx - d) **
where A, B are constants, D is square root(A^2 + B^2) and tan d = B/A

consider f(x) = e^inx
where i is square root(-1), n is a constant

using the maclaurin series it can be shown that
e^inx = cos(nx) + i.sin(nx) ***

equation ** then says that *** is equivalent to
Dcos(nx - d)
where tan d = i and D = square root(1^2 + i^2)
this implies that D = square root[1 + (-1)] = 0

this implies that:
e^inx = 0.cos(nx - d) = 0




however, i do not know if eq. ** only applies to real numbers A & B or even if d can be evaluated [tan d = i] (or even if this d matters since it is all being multiplied by zero!)
can anyone clarify this situation for me?

seandoiler

this is what i think...so assume you can and proceed as follows
well if you take tan[d]=i, this then evaluates d to be a complex infinity, and hence the term cos(nx-d) becomes infinity...and so D.Cos(nx-d) is this case is undefined, so by contradiction i guess you can't write e^inx as DCos(nx-d).....

Precision

Euler formula

e^(i.y) = cos y + i.sin y

(where y is an angle in units of radians).

pi = 3.1415926535897932384626433832795

Letting y = pi

e^(i.pi) = cos pi + i.sin pi = -1 + i.0

therefore

e^(i.pi) = -1

Also

e^(i.pi/2) = i

e^(-i.pi/2) = -i

e^(2.i.pi) = 1

e^(-i.pi) = -1

p to the e

also to reiterate what seandoiler said if you take the natural log of both sides it tends towards infinity.