What is the trigonometric form of  (3+5i) ?

Dec 27, 2015

$3 + 5 i = \sqrt{34} \left(\cos \left({\tan}^{- 1} \left(\frac{5}{3}\right)\right) + i \sin \left({\tan}^{- 1} \left(\frac{5}{3}\right)\right)\right)$

$\approx \sqrt{34} \left(\cos \left({59.04}^{\circ}\right) + i \sin \left({59.04}^{\circ}\right)\right)$

Explanation:

Any complex number $z = a + b i$ has a trigonometric form

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

where $r = | z | = \sqrt{{a}^{2} + {b}^{2}}$ and $\theta = {\tan}^{- 1} \left(\frac{b}{a}\right)$

For the given complex number, we have $a = 3$ and $b = 5$. Thus

$r = \sqrt{{3}^{2} + {5}^{2}} = \sqrt{9 + 25} = \sqrt{34}$

and

$\theta = {\tan}^{- 1} \left(\frac{5}{3}\right) \approx {59.04}^{\circ}$

So we have the trigonometric form

$3 + 5 i = \sqrt{34} \left(\cos \left({\tan}^{- 1} \left(\frac{5}{3}\right)\right) + i \sin \left({\tan}^{- 1} \left(\frac{5}{3}\right)\right)\right)$

$\approx \sqrt{34} \left(\cos \left({59.04}^{\circ}\right) + i \sin \left({59.04}^{\circ}\right)\right)$