Transforming Complex Solutions into Polar Form

[w3390]

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]

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 ​

 

[w3390]

Why isn't x real?

 

[tiny-tim]

it's the product of two complex numbers … it's very unlikely to be real

 

[w3390]

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]

dunno

maybe ​

 

[w3390]

What I'm saying is:

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

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

Then taking only the real part of this:

x(t) = Gcos(\omegat - \delta).

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

Does this make sense?

 

[tiny-tim]

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(\omegat - \delta) + i*sin(\omegat - \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).