# How do you change 10z^12 into polar coordinates where z = 2 +4i?

Dec 28, 2015

$2 + 4 i$

$\theta = \arctan \left(\frac{b}{a}\right)$

$\theta = \arctan \left(2\right)$

$r = \sqrt{{2}^{2} + {4}^{2}} = \sqrt{20} = 2 \sqrt{5}$

polar form of $2 + 4 i$$\implies$$2 \sqrt{5} \left(\cos \left(\arctan \left(2\right)\right) + i \sin \left(\arctan \left(2\right)\right)\right)$

${\left(2 \sqrt{5} \left(\cos \left(\arctan \left(2\right)\right) + i \sin \left(\arctan \left(2\right)\right)\right)\right)}^{12}$

using the moivre formula

${\left(2 \sqrt{5}\right)}^{12} {\left(\cos \left(\arctan \left(2\right)\right) + i \sin \left(\arctan \left(2\right)\right)\right)}^{12}$

$6 , 4 \cdot {10}^{7} \left(\cos \left(12 \arctan \left(2\right)\right) + i \sin \left(12 \arctan \left(2\right)\right)\right) = {z}^{12}$

$6 , 4 \cdot {10}^{8} \left(\cos \left(12 \arctan \left(2\right)\right) + i \sin \left(12 \arctan \left(2\right)\right)\right) = 10 {z}^{12}$