De moivres theorem.

**johnsy123** · Mar 1st 2011, 03:18 AM

Use DMT to solve the following equations in polar form.

$z^3=-1-i$
$z^3=-4+4\sqrt{3i}$

**FernandoRevilla** · Mar 1st 2011, 03:22 AM

What have you tried so far?. These problems are just routine knowing the corresponding theory.

**Prove It** · Mar 1st 2011, 03:30 AM

I'll show you how to do one, then you try to do the other...

$\displaystyle |-1-i| = \sqrt{(-1)^2 + (-1)^2} = \sqrt{2}$ .

Note that this complex number is in the third quadrant. The angle in the first quadrant is $\displaystyle \arctan{\frac{1}{1}} = \frac{\pi}{4}$ .
So the angle in the third quadrant, $\displaystyle \arg{(-1 - i)} = \pi + \frac{\pi}{4} = \frac{5\pi}{4}$ .

$\displaystyle z^3 = \sqrt{2}\,\textrm{cis}\,{\left(\frac{5\pi}{4}\righ t)}$

$\displaystyle z = (\sqrt{2})^{\frac{1}{3}}\,\textrm{cis}\,{\left(\fr ac{1}{3}\cdot \frac{5\pi}{4}\right)}$

$\displaystyle = \sqrt[6]{2}\,\textrm{cis}\,{\left(\frac{5\pi}{12}\right)}$ .

There are always three cube roots, and they are of the same magnitude and evenly spaced about a circle. So they differ by an angle of $\displaystyle \frac{2\pi}{3}$ .

So the other solutions are $\displaystyle \sqrt[6]{2}\,\textrm{cis}\,{\left(\frac{5\pi}{12} + \frac{2\pi}{3}\right)} = \sqrt[6]{2}\,\textrm{cis}\,{\left(\frac{13\pi}{12}\right)}$

and $\displaystyle \sqrt[6]{2}\,\textrm{cis}\,{\left(\frac{13\pi}{12} + \frac{2\pi}{3}\right)} = \sqrt[6]{2}\,\textrm{cis}\,{\left(\frac{21\pi}{12}\right)}$ .

But since $\displaystyle -\pi < \arg{z} \leq \pi$ , we rewrite this as

$\displaystyle \sqrt[6]{2}\,\textrm{cis}\,{\left(-\frac{11\pi}{12}\right)}$ and $\displaystyle \sqrt[6]{2}\,{\left(-\frac{3\pi}{12}\right)}$ .

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