# Complex to polar form

## Homework Statement

Show that the solution x(t) = Ge^(iwt), where G is in general complex, can be written in the form x(t) = Dcos(wt - $$\delta$$).

D(w) and $$\delta$$(w) are real functions of w.

## Homework Equations

z = Ae^(i$$\phi$$)

## The Attempt at a Solution

So I know I should start by writing G in polar form. I am confused though as to how to go to polar form with just the G. Is it simply just Ge^(i$$\phi$$). Then, I could use Euler's formula to write:

Ge^(i$$\phi$$) = Gcos($$\phi$$) + iGsin($$\phi$$).

I am not sure where this gets me. Any help on where to go from here or if this is even correct would be much appreciated.

tiny-tim
Homework Helper
hi w3390!

(have a delta: δ and a rho: ρ and a phi: φ and an omega: ω )

## Homework Statement

Show that the solution x(t) = Ge^(iwt), where G is in general complex, can be written in the form x(t) = Dcos(wt - $$\delta$$).

D(w) and $$\delta$$(w) are real functions of w.

but that's obviously not true …

the RHS is real, but x isn't

Why isn't x real?

tiny-tim
Homework Helper
it's the product of two complex numbers … it's very unlikely to be real

I am confused then because my question was from a test prep sheet from my professor. Should I perhaps only consider the real part of x(t)?

tiny-tim
Homework Helper
dunno

maybe

What I'm saying is:

x(t) = Ge^(i$$\phi$$)

x(t) = G[cos($$\omega$$t - $$\delta$$) + i*sin($$\omega$$t - $$\delta$$)

Then taking only the real part of this:

x(t) = Gcos($$\omega$$t - $$\delta$$).

From here, I can compare to the given solution of x(t) = Dcos($$\omega$$t - $$\delta$$) and say that G = D.

Does this make sense?

tiny-tim
Homework Helper
Hi w3390!

(what happened to that δ φ and ω i gave you? )

I don't understand where your second line came from …

x(t) = G[cos($$\omega$$t - $$\delta$$) + i*sin($$\omega$$t - $$\delta$$)

HallsofIvy
Write $G= re^{i\theta}$. Then $Ge^{i\omega t}= r e^{i(\omega t+ \theta)}= r (cos(\omega t+ \theta)+ i sin(\omega t+ \theta)$.