Notes: `4 - 4sqrt(3)i` Write the trigonometric form of the number.

$\displaystyle{z}={4}-{4}\sqrt{{3}}{i}$

$\displaystyle{r}=\sqrt{{{{4}}^{{2}}+{{\left(-{4}\sqrt{{3}}\right)}}^{{2}}}}=\sqrt{{{16}+{48}}}=\sqrt{{{64}}}={8}$

$\displaystyle\theta={\arctan{{\left(\frac{{-{4}\sqrt{{3}}}}{{4}}\right)}}}={\arctan{{\left(-\sqrt{{3}}\right)}}}=\frac{{{5}\pi}}{{3}}$

$\displaystyle{z}={8}{\left({\cos{{\left(\frac{{{5}\pi}}{{3}}\right)}}}+{i}{\sin{{\left(\frac{{{5}\pi}}{{3}}\right)}}}\right)}$

$\displaystyle{z}={4}-{4}\sqrt{{3}}{i}$

$\displaystyle{r}=\sqrt{{{{4}}^{{2}}+{{\left(-{4}\sqrt{{3}}\right)}}^{{2}}}}=\sqrt{{{16}+{48}}}=\sqrt{{{64}}}={8}$

$\displaystyle\theta={\arctan{{\left(\frac{{-{4}\sqrt{{3}}}}{{4}}\right)}}}={\arctan{{\left(-\sqrt{{3}}\right)}}}=\frac{{{5}\pi}}{{3}}$

$\displaystyle{z}={8}{\left({\cos{{\left(\frac{{{5}\pi}}{{3}}\right)}}}+{i}{\sin{{\left(\frac{{{5}\pi}}{{3}}\right)}}}\right)}$

Notes