don't know where should i post this.

Find the real part of the complex number (i-1)^3 by using De Moivre's formula

My lecture notes is too brief need help !

thanks ")

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- Dec 17th 2010, 05:03 AMsaccharineComplex number with De Moivre's formula
don't know where should i post this.

Find the real part of the complex number (i-1)^3 by using De Moivre's formula

My lecture notes is too brief need help !

thanks ") - Dec 17th 2010, 05:13 AMTheCoffeeMachine
Write it in the polar form:

$\displaystyle \displaystyle (i-1)^3 = \bigg[\sqrt{2}\left(\cos\left(\frac{3\pi}{4}\right)+i\si n\left(\frac{3\pi}{4}\right)\right)\bigg]^3$.

De Moivre's theorem tells you that, for any arbitrary $\displaystyle x$ and integer $\displaystyle n$, we have:

$\displaystyle \left(\cos{x}+i\sin{x}\right)^n = \cos\left(nx\right)+i\sin\left(nx\right)$. - Dec 17th 2010, 05:37 AMPlato
Expand $\displaystyle \left( { - 1 + i} \right)^3 = \sum\limits_{k = 0}^3 {\dbinom{3}{k}\left( { - 1} \right)^{3 - k} \left( i \right)^k }. $

The real parts occur when $\displaystyle k=0~\&~2$.

What do they add up to?