anonymous
  • anonymous
I have to Solve x^2 - (4-i)x + (5-5i) = 0 and Prove/Verify/Derive de Moivre's Theorem
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
quadratic formula for the first
anonymous
  • anonymous
for demoive's, you can prove by induction.
anonymous
  • anonymous
ok i will try right now thx for pointing me in the right direction

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anonymous
  • anonymous
you're welcome
anonymous
  • anonymous
Bump having trouble with quadratic formula.. I am stuck at (4-i) +- sqrt(-4+12i+i^2)
anonymous
  • anonymous
* (4-i) +- sqrt(-4+12i+i^2)/(2)
anonymous
  • anonymous
any ideas?
anonymous
  • anonymous
i got to (4-i) +- 2i+i sqrt(12i)/(2) but I feel i made a mistake somewhere..
phi
  • phi
sqrt(-4+12i+i^2)= sqrt(-5 + 12 i) if we write this in polar form, we have sqrt ( 13 * exp( i * A) ) = (13 * (exp(i*A)) ^ (1/2) where A = arctan(12/-5) or \[ \sqrt{13} e^{ i \frac{A}{2}}= \sqrt{13}(\cos A/2 + i \sin A/2) \] cos A= -5/12, sin A= 12/13 cos A/2 = sqrt( (1/2) (1 + cos A)) sin A/2 = sqrt ( (1/2) (1 - cos A)) it's a bit complicated, but eventually you can get to an answer
anonymous
  • anonymous
Do I have to write it in polar form? I can't pick A to be the coefficent of x^2 and so on for B and C. So A = 1, B= -(4-i), and C=(5-5i) Then plug into quadratic formula: x=-b +- sqrt(b^2-4ac)/(2)
phi
  • phi
First, my A is just the angle, not related to your A,B,C coefficients I am just trying to find the square root of -5 + 12i Normally I use polar coordinates. You could use this approach http://www.1728.org/compnum2.htm (scroll down to square root)
anonymous
  • anonymous
im lost I thought you just plugged in the coefficients into the formula. My teacher didnt talk about anything else like that
Loser66
  • Loser66
@phi it's too complicated to me, too. XD
anonymous
  • anonymous
lol @Loser66
phi
  • phi
As I said, the easiest way to take the square root of a complex number is write it in polar form, and then raise to the 1/2 power.
anonymous
  • anonymous
sorry if this is a stupid question but how is sqrt(-4+12i+i^2) = sqrt(-5 + 12 i)
phi
  • phi
i= sqrt(-1), and i*i = sqrt(-1)*sqrt(-1) = -1
Loser66
  • Loser66
@phi, so why don't we convert \(\sqrt{-5+12i}\) to \((2+3i)^2\) to take off the squaroot?
phi
  • phi
that is good. How did you do that ?
Loser66
  • Loser66
to be honest, I didn't know and I got stuck like what the asker did until I put it into worfram. To me, it's not good at aaaaall. Because I didn't know how to do. :)
anonymous
  • anonymous
loser i put it into wolfram also but...idk what they did. I thought they just plugged in
Loser66
  • Loser66
@Andysebb if I go backward, its answer makes sense to me. the problem is I don't know how to go forward hehehehe, shame on me
anonymous
  • anonymous
Lol how did u go backwards o.o
Loser66
  • Loser66
@Andysebb expand (2+3i)^2 , it leads to \(\sqrt{-5+12i}\)
phi
  • phi
|dw:1378762580974:dw|
Loser66
  • Loser66
@phi ohh I got you,
phi
  • phi
\[ \left(13 e^{iA}\right)^{\frac{1}{2}} =13^{\frac{1}{2}}e^{i {\frac{A}{2}}}\\ = \sqrt{13} \left( \cos\frac{A}{2} + i \sin \frac{A}{2} \right) \] From the figure we see cos A= -5/13 and sin A = 12/13 and using the 1/2 angle formula we have \[ \cos A/2 = \sqrt{ \frac{1}{2}(1 + \cos A)} = \sqrt{ \frac{1}{2}\left(1 - \frac{5}{13} \right)} = \sqrt{\frac{4}{13} } = \frac{2}{\sqrt{13}} \] working through it, you find \[ \sin A/2 = \frac{3}{\sqrt{13}}\]
phi
  • phi
finally we get, for the square root \[ ±\sqrt{13} \left( \cos\frac{A}{2} + i \sin \frac{A}{2} \right) = \\ \sqrt{13} \left( \frac{2}{ \sqrt{13}}+ i \frac{3}{ \sqrt{13}}\right) \\ = ±(2+3i) \]
Loser66
  • Loser66
@phi Euler number and polar form relate to the problem, right?
anonymous
  • anonymous
so this problem has nothing to do with quadratic formula then
phi
  • phi
and going back to the problem \[x= ( 4 - i + 2 + 3i)/2 = (6+2i)/2 = 3 + i \] and \[x= ( 4 - i - 2 - 3i)/2 = (2 -4i)/2= 1 - 2i\]
phi
  • phi
we used the quadratic formula. But we took a side trip to find the square root part.
Loser66
  • Loser66
Thanks phi. I review a lot from yours.
anonymous
  • anonymous
ya thanks phi im going over it still. What about proving de Moivre's theorem?
phi
  • phi
I am wondering about how to prove de Moivre's theorem from this. It seems that we used de Moivre's (to take the square root). I may be missing something, as the solution x= 1-2i and x= 3+i does not suggest de Moivre's
anonymous
  • anonymous
hmm
anonymous
  • anonymous
demoive's is only for integer exponents

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