How do you convert -3+1i to polar form?

Aug 9, 2018

Explanation:

Let ,

$z = x + i y = - 3 + 1 i \implies x = - 3 \mathmr{and} y = 1$

Now ,

$| z | = r = \sqrt{{x}^{2} + {y}^{2}} = \sqrt{{\left(- 3\right)}^{2} + {\left(1\right)}^{2}} = \sqrt{9 + 1} = \sqrt{10}$

We have ,

$\cos \theta = \frac{x}{r} = \frac{- 3}{\sqrt{10}} < 0 \mathmr{and} \sin \theta = \frac{y}{r} = \frac{1}{\sqrt{10}} > 0$

$\therefore \cos \theta < 0 \mathmr{and} \sin \theta > 0 \implies {2}^{n d} Q u a \mathrm{dr} a n t$

$\therefore \tan \theta = \sin \frac{\theta}{\cos} \theta = \left(\frac{\frac{1}{\sqrt{10}}}{- \frac{3}{\sqrt{10}}}\right) = - \frac{1}{3}$

$\therefore \theta = \arctan \left(- \frac{1}{3}\right) = - a r c \tan \left(\frac{1}{3}\right) \approx {\left(- 18.43\right)}^{\circ}$

So, the polar form :

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

$\implies z = \sqrt{10} \left(\cos \theta + i \sin \theta\right)$

,where ,

$\theta = - a r c \tan \left(\frac{1}{3}\right) , \cos \theta = - \frac{3}{\sqrt{10}} \mathmr{and} \sin \theta = \frac{1}{\sqrt{10}} .$