Notes: `1 - sqrt(3)i` Write the trigonometric form of the number.

Sunday, 20 September 2015

Given: $\displaystyle{1}-\sqrt{{{3}}}{i}$

$\displaystyle{r}=\sqrt{{{{1}}^{{2}}+{{\left(-\sqrt{{{3}}}\right)}}^{{2}}}}=\sqrt{{{1}+{3}}}=\sqrt{{{4}}}={2}$

$\displaystyle{\tan{\theta}}=-\frac{\sqrt{{3}}}{{1}}=-\sqrt{{3}}$

Since the angle is in quadrant 4, $\displaystyle<{b}{r}>$

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

In trigonometric form $\displaystyle{z}={2}{\left[{\cos{{\left(\frac{{{5}\pi}}{{3}}\right)}}}+{i}{\sin{{\left(\frac{{{5}\pi}}{{3}}\right)}}}\right]}$

Given: $\displaystyle{1}-\sqrt{{{3}}}{i}$

$\displaystyle{r}=\sqrt{{{{1}}^{{2}}+{{\left(-\sqrt{{{3}}}\right)}}^{{2}}}}=\sqrt{{{1}+{3}}}=\sqrt{{{4}}}={2}$

$\displaystyle{\tan{\theta}}=-\frac{\sqrt{{3}}}{{1}}=-\sqrt{{3}}$

Since the angle is in quadrant 4, $\displaystyle<{b}{r}>$

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

In trigonometric form $\displaystyle{z}={2}{\left[{\cos{{\left(\frac{{{5}\pi}}{{3}}\right)}}}+{i}{\sin{{\left(\frac{{{5}\pi}}{{3}}\right)}}}\right]}$

Notes