We have
\frac{2+8i}{7+i}
=\frac{\sqrt{68}(\cos(\tan^{-1}(4))+i\sin(\tan^{-1}(4)))}{\sqrt{50}(\cos(\tan^{-1}(1/7))+i\sin(\tan^{-1}(1/7)))}
=\sqrt{\frac{68}{50}}(\cos(\tan^{-1}(4)-\tan^{-1}(1/7))+i\sin(\tan^{-1}(4)-\tan^{-1}(1/7)))
=\sqrt{\frac{34}{25}}(\cos(\tan^{-1}(27/11))+i\sin(\tan^{-1}(27/11)))
=\sqrt{\frac{34}{25}}(11/\sqrt850+i27/\sqrt{850})
=\sqrt{\frac{34}{25\times 850}}(11+27i)
=1/25(11+27i)