What is the trigonometric form of  12+4i ?

May 19, 2016

$12 + 4 i = 4 \sqrt{10} \left(\cos \alpha + i \sin \alpha\right)$, where $\tan \alpha = \frac{1}{3}$

Explanation:

$a + i b$ can be written in trigonometric form $r {e}^{i \theta} = r \cos \theta + i r \sin \theta = r \left(\cos \theta + i \sin \theta\right)$,
where $r = \sqrt{{a}^{2} + {b}^{2}}$.

Hence $12 + 4 i = \sqrt{{12}^{2} + {4}^{2}} \left[\cos \alpha + i \sin \alpha\right]$

= $\sqrt{160} \left[\cos \alpha + i \sin \alpha\right]$

= $4 \sqrt{10} \left[\cos \alpha + i \sin \alpha\right]$,

where $\cos \alpha = \frac{12}{4 \sqrt{10}} = \frac{3}{\sqrt{10}}$

and $\sin \alpha = \frac{4}{4 \sqrt{10}} = \frac{1}{\sqrt{10}}$

or $\tan \alpha = \frac{\frac{1}{\sqrt{10}}}{\frac{3}{\sqrt{10}}} = \frac{1}{3}$

May 19, 2016

$4 \sqrt{10} \left(\cos \left(0.322\right) + i \sin \left(0.322\right)\right)$

Explanation:

Given a complex number z = x + yi , this can be written in trig. form as.

$z = x + y i = r \left(\cos \theta + i \sin \theta\right)$

where $r = \sqrt{{x}^{2} + {y}^{2}}$

and $\theta = {\tan}^{-} 1 \left(\frac{y}{x}\right)$

here x = 12 and y = 4

$\Rightarrow r = \sqrt{{12}^{2} + {4}^{2}} = \sqrt{160} = 4 \sqrt{10}$

and theta=tan^-1(4/12)≈0.322" radians or"18.43^@

$\Rightarrow 12 + 4 i = 4 \sqrt{10} \left(\cos \left(0.322\right) + i \sin \left(0.322\right)\right)$or

$12 + 4 i = 4 \sqrt{10} \left(\cos {\left(18.43\right)}^{\circ} + i \sin {\left(18.43\right)}^{\circ}\right)$