Answer to Question #168969 in Math for EUGINE HAWEZA

Question

a.      Express the roots of (−14+3i)^−2/5 complex number in polar form. 

Answer 1

$(-14+3i)^{-2}=\dfrac{187}{42025}+\dfrac{84}{42025}i$

$r=\dfrac{1}{205}, \theta=\tan^{-1}(\dfrac{84}{187})$

According to the De Moivre's Formula, all $n$-th roots of a complex number $r(\cos \theta+i\sin \theta)$ are given by $\sqrt[n]{r}\bigg(\cos (\dfrac{\theta+2\pi k}{n})+i\sin (\dfrac{\theta+2\pi k}{n})\bigg),$

$k=0, 1, 2, ... n-1$

$k=0:$

$\dfrac{1}{\sqrt[5]{205}}(\cos (\dfrac{\tan^{-1}(\dfrac{84}{187})}{5})+i\sin (\dfrac{\tan^{-1}(\dfrac{84}{187})}{5}))$

$k=1:$

$\dfrac{1}{\sqrt[5]{205}}(\cos (\dfrac{\tan^{-1}(\dfrac{84}{187})+2\pi}{5})+i\sin (\dfrac{\tan^{-1}(\dfrac{84}{187})+2\pi}{5}))$

$k=2:$

$\dfrac{1}{\sqrt[5]{205}}(\cos (\dfrac{\tan^{-1}(\dfrac{84}{187})+4\pi}{5})+i\sin (\dfrac{\tan^{-1}(\dfrac{84}{187})+4\pi}{5}))$

$k=3:$

$\dfrac{1}{\sqrt[5]{205}}(\cos (\dfrac{\tan^{-1}(\dfrac{84}{187})+6\pi}{5})+i\sin (\dfrac{\tan^{-1}(\dfrac{84}{187})+6\pi}{5}))$

$k=4:$

$\dfrac{1}{\sqrt[5]{205}}(\cos (\dfrac{\tan^{-1}(\dfrac{84}{187})+8\pi}{5})+i\sin (\dfrac{\tan^{-1}(\dfrac{84}{187})+8\pi}{5}))$