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OpenStudy (anonymous):
OpenStudy (anonymous):
hey
OpenStudy (bahrom7893):
hi
OpenStudy (bahrom7893):
this reminds me.. e^(i*pi) = -1. What class is this? calc 2?
OpenStudy (anonymous):
yes
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OpenStudy (bahrom7893):
Pippa, just so u know.. I suck at series. I hated that part of Calc 2
OpenStudy (anonymous):
well i gotta hand in work that i never even studied yet lol
OpenStudy (anonymous):
I am sooooo behind
OpenStudy (bahrom7893):
What was the e^x Maclaurin Expansion?
1 + x + x^2/2! + x^3/3! + ... + x^n/n! right?
OpenStudy (anonymous):
lol hahaha i didnt even touch that so like i kind of dont know any of this stuff
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OpenStudy (anonymous):
eix = cos(x) + i*sin(x)
For what values of x will the above equal i?
i = cos(x) + i*sin(x)
When is cos(x) = 0 and sin(x) = 1? When x = π/2:
So,
i = eiπ/2
So,
ii = (eiπ/2)i = e-π/2
de Moivre's formula:
(cos(x) + i*sin(x))n = cos(nx) + i*sin(nx)
31+i = 3*3i = 3*ei*ln(3) = 3(cos(ln(3)) + i*sin(ln(3)))
3*cos(ln(3)) + 3*i*sin(ln(3)) = 1.364 + 2.672*i
OpenStudy (anonymous):
this is what i found on cramster. i paid two bucks for it. Do u think its correct
OpenStudy (bahrom7893):
u paid 2 bucks for it..
OpenStudy (anonymous):
lol ya
OpenStudy (anonymous):
1.99 to be exact
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OpenStudy (bahrom7893):
I could've done that myself.
OpenStudy (anonymous):
lol hahaha
OpenStudy (bahrom7893):
I mean just calculating the value for 3^(1+i)
OpenStudy (anonymous):
so u want me to pay u two bucks for every answer?????
OpenStudy (bahrom7893):
That's the easy part apparently, and yes it is correct, though i think u have a typo
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OpenStudy (bahrom7893):
No pippa i said u wasted ur 2 bucks
OpenStudy (anonymous):
lol i was just teasing u
OpenStudy (anonymous):
whats the mistake?
OpenStudy (bahrom7893):
ii = (eiπ/2)i = e(-π)/2
there's a difference between e-pi and -e*pi
OpenStudy (anonymous):
ohhhh i seee
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OpenStudy (bahrom7893):
hold on let me write this on paper, i get confused if i type
OpenStudy (anonymous):
lol ok i am waiting patiently :D
OpenStudy (bahrom7893):
do u know how in the world the answerer got to the third line?
OpenStudy (bahrom7893):
i = cos(x) + i*sin(x).. where did that come from?
OpenStudy (anonymous):
lol no
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OpenStudy (bahrom7893):
can u post a screen of cramster's solution? because what I got was not i = Cos(x)+i*Sin(x)
I got:
i = ln(Cos(x)+i*Sin(x))/x
OpenStudy (bahrom7893):
Maybe it somehow magically turns to cos(x)+i*sin(x) but that's weird.. that would mean e^ix = i
OpenStudy (anonymous):
lol
OpenStudy (anonymous):
This is the exact answer
OpenStudy (bahrom7893):
oh crap...
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