Notes: `(-1 + i)^6` Use DeMoivre's Theorem to find the indicated power of the complex number. Write the result in standard form.

Thursday, 8 December 2016

$\displaystyle{{\left(-{1}+{i}\right)}}^{{6}}$ Use DeMoivre's Theorem to find the indicated power of the complex number. Write the result in standard form.

$\displaystyle{{\left(-{1}+{i}\right)}}^{{6}}$

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

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

To be able to apply it, convert the complex number z=-1+i to trigonometric form.Take note that that the trigonometric form of

$\displaystyle{z}={x}+{y}{i}$

$\displaystyle{z}={r}{\left({\cos{\theta}}+{i}{\sin{\theta}}\right)}$

where

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

Applying these two formulas, the values of r and theta of z=-1+i are:

$\displaystyle{r}=\sqrt{{{{\left(-{1}\right)}}^{{2}}+{{1}}^{{2}}}}=\sqrt{{2}}$

$\displaystyle\theta\ldots$

$\displaystyle{{\left(-{1}+{i}\right)}}^{{6}}$

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

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

To be able to apply it, convert the complex number z=-1+i to trigonometric form.Take note that that the trigonometric form of

$\displaystyle{z}={x}+{y}{i}$

$\displaystyle{z}={r}{\left({\cos{\theta}}+{i}{\sin{\theta}}\right)}$

where

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

Applying these two formulas, the values of r and theta of z=-1+i are:

$\displaystyle{r}=\sqrt{{{{\left(-{1}\right)}}^{{2}}+{{1}}^{{2}}}}=\sqrt{{2}}$

$\displaystyle\theta={{\tan}}^{{-{1}}}{\left(\frac{{1}}{{-{1}}}\right)}={{\tan}}^{{-{1}}}{\left(-{1}\right)}=-{{45}}^{{o}}$

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

$\displaystyle\theta={{180}}^{{o}}+{\left(-{{45}}^{{o}}\right)}={{135}}^{{o}}$

Then, plug-in the values of r and theta to the trigonometric form

$\displaystyle{z}={r}{\left({\cos{\theta}}+{i}{\sin{\theta}}\right)}$

$\displaystyle{z}=\sqrt{{2}}{\left({{\cos{{135}}}}^{{o}}+{i}{{\sin{{135}}}}^{{o}}\right)}$

Now that z=-1+i is in trigonometric form, proceed to compute z^6 .

$\displaystyle{{z}}^{{6}}={{\left(-{1}+{i}\right)}}^{{6}}$

$\displaystyle={{\left[\sqrt{{2}}{\left({{\cos{{135}}}}^{{o}}+{i}{{\sin{{135}}}}^{{o}}\right)}\right]}}^{{6}}$

$\displaystyle={{\left(\sqrt{{2}}\right)}}^{{6}}{\left({\cos{{\left({6}\times{{135}}^{{o}}\right)}}}+{i}{\sin{{\left({6}\times{{135}}^{{o}}\right)}}}\right.}$