I have to Solve x^2 - (4-i)x + (5-5i) = 0 and
Prove/Verify/Derive de Moivre's Theorem

- anonymous

I have to Solve x^2 - (4-i)x + (5-5i) = 0 and
Prove/Verify/Derive de Moivre's Theorem

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- anonymous

quadratic formula for the first

- anonymous

for demoive's, you can prove by induction.

- anonymous

ok i will try right now thx for pointing me in the right direction

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## More answers

- anonymous

you're welcome

- anonymous

Bump having trouble with quadratic formula..
I am stuck at (4-i) +- sqrt(-4+12i+i^2)

- anonymous

* (4-i) +- sqrt(-4+12i+i^2)/(2)

- anonymous

any ideas?

- anonymous

i got to (4-i) +- 2i+i sqrt(12i)/(2) but I feel i made a mistake somewhere..

- 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

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

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

im lost I thought you just plugged in the coefficients into the formula. My teacher didnt talk about anything else like that

- Loser66

@phi it's too complicated to me, too. XD

- anonymous

lol @Loser66

- 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

sorry if this is a stupid question but how is sqrt(-4+12i+i^2) = sqrt(-5 + 12 i)

- phi

i= sqrt(-1), and i*i = sqrt(-1)*sqrt(-1) = -1

- Loser66

@phi, so why don't we convert \(\sqrt{-5+12i}\) to \((2+3i)^2\) to take off the squaroot?

- phi

that is good. How did you do that ?

- 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

loser i put it into wolfram also but...idk what they did. I thought they just plugged in

- 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

Lol how did u go backwards o.o

- Loser66

@Andysebb expand (2+3i)^2 , it leads to \(\sqrt{-5+12i}\)

- phi

|dw:1378762580974:dw|

- Loser66

@phi ohh I got you,

- 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

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

@phi Euler number and polar form relate to the problem, right?

- anonymous

so this problem has nothing to do with quadratic formula then

- 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

we used the quadratic formula. But we took a side trip to find the square root part.

- Loser66

Thanks phi. I review a lot from yours.

- anonymous

ya thanks phi im going over it still. What about proving de Moivre's theorem?

- 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

hmm

- anonymous

demoive's is only for integer exponents

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