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(Solved) Using Euler's Formula

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I'm trying to fully understand why:
Use euler's formula (e^ix = cosx + isinx) to put pi^(-1+2i) in the form a+ib

The answer should be... pi^-1 * (cos[2ln(pi)]) + pi^-1 * (i * sin[2ln(pi)])

- I do know that you can separate the pi^-1 and put it in the form pi^-1 * (cos mx + i * sinmx)
- The two comes from the m, which is 2.

Why is it ln(pi) though?... if you have pi^-1 * pi^(m*ln(cosx+isinx)), how do you convert that into the answer?

Solved: Thanks dullard!
 
Start with pi^(-1+2i)

Take the natural log and then the exponent of that (this doesn't change the answer at all).

pi^(-1+2i) = e^ln(pi^(-1+2i))

With logs, you can move the exponent in front of the log:

e^ln(pi^(-1+2i)) = e^[(-1+2i)*ln(pi)]

Simplify:

e^[(-1+2i)*ln(pi)] = e^[-1*ln(pi)]*e^[2i*ln(pi)]

Rearrange by putting the exponent BACK on the first log:

e^[-1*ln(pi)]*e^[2i*ln(pi)] = e^[ln(pi^-1)]*e^[2i*ln(pi)]

Simplify:

e^[ln(pi^-1)]*e^[2i*ln(pi)] = pi^-1*e^[2i*ln(pi)]

Let x = 2*ln(pi):

pi^-1*e^[2i*ln(pi)] = pi^-1*e^[i*x]

Do the rest of your homework for yourself. It is just one or two last steps depending on how much math you like to do in one step. Plus, what does this have to do with a differential equation (your thread title)? Nothing that I can see.
 
Originally posted by: dullard
Do the rest of your homework for yourself. It is just one or two last steps depending on how much math you like to do in one step. Plus, what does this have to do with a differential equation (your thread title)? Nothing that I can see.

Gotcha. I didn't think about manipulating the equation into e^lnx.
To answer your question, we need to utilize euler's formula for complex second degree equations. This one they threw out as one of the problems, while the others were exp[<insert here>]. So in the end, I suppose it may have some relevance to DE? Maybe not. In the end though, an opinion is an opinion

I greatly appreciate the help though 🙂.
Thanks :cookie:
 
Originally posted by: dullard
Start with pi^(-1+2i)

Take the natural log and then the exponent of that (this doesn't change the answer at all).

pi^(-1+2i) = e^ln(pi^(-1+2i))

With logs, you can move the exponent in front of the log:

e^ln(pi^(-1+2i)) = e^[(-1+2i)*ln(pi)]

Simplify:

e^[(-1+2i)*ln(pi)] = e^[-1*ln(pi)]*e^[2i*ln(pi)]

Rearrange by putting the exponent BACK on the first log:

e^[-1*ln(pi)]*e^[2i*ln(pi)] = e^[ln(pi^-1)]*e^[2i*ln(pi)]

Simplify:

e^[ln(pi^-1)]*e^[2i*ln(pi)] = pi^-1*e^[2i*ln(pi)]

Let x = 2*ln(pi):

pi^-1*e^[2i*ln(pi)] = pi^-1*e^[i*x]


I made peepee in my poopoo place
 
Originally posted by: Ricemarine
I greatly appreciate the help though 🙂.
Thanks :cookie:
You are welcome.
Originally posted by: Ocguy31
I made peepee in my poopoo place
Hmm, maybe I should have made it more simple. Start with:

pi^(-1+2i) = pi^(-1)*pi^(2i)

That would skip a few steps. I wasn't thinking clearly due to my jet lag (37 hours of travel time yesterday, yes 37).
 
Originally posted by: Ricemarine
Now where did you go? lol.
Kommetjie, South Africa -> Cape Town South Africa airport, ~1 hr + ~2 hours waiting for plane
Cape Town -> Johannesburg, South Africa ~2 hr + ~6 hr wait for next plane
Johannesburg -> Dakar, Senegal ~ 8 hr + ~2 hr wait for security check of plane
Dakar -> Atlanta ~ 9 hr + ~3 hr wait for next plane
Atlanta -> Omaha ~2.5 hr
Omaha airport -> Car across town ~1 hr including wait for baggage
Omaha -> home ~1 hr

That adds up to 37 hours.
 
Originally posted by: zoiks
Originally posted by: Ocguy31
I made peepee in my poopoo place

So basically yellow urine is coming out of your anus instead of a turd?

Well it looks like Urine after I drink Mexican beer or eat Taco Bell......I could shit through a screen-door and not hit a wire.
 
Originally posted by: Ocguy31
Originally posted by: zoiks
Originally posted by: Ocguy31
I made peepee in my poopoo place

So basically yellow urine is coming out of your anus instead of a turd?

Well it looks like Urine after I drink Mexican beer or eat Taco Bell......I could shit through a screen-door and not hit a wire.

Woah... TMI.
 
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