Blog: `(3 - 2i)^8` Use DeMoivre's Theorem to find the indicated power of the complex number. Write the result in standard form.

Tuesday, April 27, 2010

$\displaystyle{{\left({3}-{2}{i}\right)}}^{{8}}$ Use DeMoivre's Theorem to find the indicated power of the complex number. Write the result in standard form.

$\displaystyle{{\left({3}-{2}{i}\right)}}^{{8}}$

Take note that De Moivre's Theorem is used to compute the powers and roots of a complex number. The formula is:

$\displaystyle{{\left[{r}{\left({\cos{\theta}}+{i}{\sin{\theta}}\right)}\right]}}^{{n}}={{r}}^{{n}}{\left({\cos{{\left({n}\theta\right)}}}+{i}{\sin{{\left({n}\theta\right)}}}\right)}$

Notice that its formula is in trigonometric form. So to compute $\displaystyle{{\left({3}-{2}{i}\right)}}^{{8}}$ , it is necessary to convert the complex number $\displaystyle{z}={3}-{2}{i}$ to trigonometric form $\displaystyle{z}={r}{\left({\cos{\theta}}+{i}{\sin{\theta}}\right.}$ ).

To convert $\displaystyle{z}={x}+{y}{i}$ to $\displaystyle{z}={r}{\left({\cos{\theta}}+{i}{\sin{\theta}}\right)}$ , apply the formula

$\displaystyle{r}=\sqrt{{{{x}}^{{2}}+{{y}}^{{2}}}}$ and $\displaystyle\theta={{\tan}}^{{-{1}}}\frac{{y}}{{x}}$

So,

$\displaystyle{r}=\sqrt{{{{3}}^{{2}}+{{\left(-{2}\right)}}^{{2}}}}=\sqrt{{{9}+{4}}}=\sqrt{{13}}$

$\displaystyle\theta={{\tan}}^{{-{1}}}\frac{{-{2}}}{{3}}=-{{33.69007}}^{{o}}$

Since x is positive and y is negative, theta is located at the fourth quadrant. So the equivalent positive angle of theta is:

$\displaystyle\theta={{360}}^{{o}}+{\left(-{{33.69007}}^{{o}}\right)}={{326.30993}}^{{o}}$

Hence, the trigonometric form of the complex number

$\displaystyle{z}={3}-{2}{i}$

$\displaystyle{z}=\sqrt{{13}}{\left({{\cos{{326.30993}}}}^{{o}}+{i}{{\sin{{326.30993}}}}^{{o}}\right)}$

Now that it is in trigonometric form, proceed to apply the formula of De Moivre's Theorem to compute $\displaystyle{{z}}^{{8}}$ .

$\displaystyle{{z}}^{{8}}={{\left({3}-{2}{i}\right)}}^{{8}}$

$\displaystyle={{\left[\sqrt{{13}}{\left({{\cos{{326.30993}}}}^{{o}}+{i}{{\sin{{326.30993}}}}^{{o}}\right)}\right]}}^{{8}}$

$\displaystyle={{\left(\sqrt{{13}}\right)}}^{{8}}{\left({\cos{{\left({8}\times{{326.30993}}^{{o}}\right)}}}+{i}{\sin{{\left({8}\times{{326.30993}}^{{o}}\right)}}}\right)}$

$\displaystyle={28561}{\left({\cos{{\left({8}\times{{326.30993}}^{{o}}\right)}}}+{i}{\sin{{\left({8}\times{{326.30993}}^{{o}}\right)}}}\right)}$

$\displaystyle=-{239}+{28560}{i}$

Therefore, $\displaystyle{{\left({3}-{2}{i}\right)}}^{{8}}=-{239}+{28560}{i}$ .

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Tuesday, April 27, 2010

$\displaystyle{{\left({3}-{2}{i}\right)}}^{{8}}$ Use DeMoivre's Theorem to find the indicated power of the complex number. Write the result in standard form.

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