# Differential equation proof

## Homework Statement

Given a function g(t)=acosωt + bsinωt, where a and b are constants, show that g(t) is the real part of the complex function: keeiωt for some k and Φ
Remark: the complex expression ke is called a phasor. If we know that g(t) has the form kcos(ωt+Φ) then we need know only the constants k and Φ-the amplitude and the phase- to know the function g. Hence we can use the phasor ke as a notation for the function g(t)=keeiωt

## Homework Equations

Euler's formula eiωt= cosωt +isinωt

## The Attempt at a Solution

Not really sure where to start here except for expanding using euler;s as the first step. any help would be greatly appreciated.

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haruspex
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Not really sure where to start here except for expanding using euler;s as the first step.
So do that, for both exponential terms.

okay so then I get k(cosΦt +isinΦt)(cosωt+isinωt)
but what should this be equal too?

haruspex
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okay so then I get k(cosΦt +isinΦt)(cosωt+isinωt)
but what should this be equal too?
How do you get φt terms?

umm I'm not sure. I didn't know we would be using that symbol. What does φ represent?

haruspex
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umm I'm not sure. I didn't know we would be using that symbol. What does φ represent?
It is a constant. You need to find the values of k and φ which make the real part match g().

haruspex
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umm I'm not sure. I didn't know we would be using that symbol. What does φ represent?
I think perhaps you did not understand my question. In your post #3, you had terms like sin(φt). I don't understand how you got those. Please post your working.

Sorry I'm kind of lost. Where did φ come from?

In my third post I only haveΦ k ω and t

no φ

haruspex
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no φ
Looks like a font issue. Are you saying Φ and φ look very different to you? In the font that comes up on my iPad they are almost the same, so I may have used the wrong one.

yes they look very different. I think thats where the confusion was.

haruspex
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yes they look very different. I think thats where the confusion was.

But in regards to your earlier quastion. I used euler's formula to obtain sin(Φt) eiΦt=cosΦt +isin(Φt)
right?

that sin (Φt) is not apart of that calculation

haruspex
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But in regards to your earlier quastion. I used euler's formula to obtain sin(Φt) eiΦt=cosΦt +isin(Φt)
right?
According to your initial post, it is e, not eiφt.

oops, my mistake, than diregard the t's

haruspex
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oops, my mistake, than diregard the t's
Ok. So multiply out the corrected version of the expression in your post #3.

k(cosΦ +isinΦ)(cosω+isinω)

haruspex
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k(cosΦ +isinΦ)(cosω+isinω)
No, the other exponential term does have a t factor: eiωt.

k(cosΦ +isinΦ)(cosωt+isinωt)

haruspex
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k(cosΦ +isinΦ)(cosωt+isinωt)
Right. Now multiply out.

kcosΦcosωt +kisinΦcosωt + kisinωtcosΦ -ksinΦsinωt

haruspex