find-all-roots-6-2-i-1-3-by-using-demover-theorem-

Question Number 95259 by mhmd last updated on 24/May/20

f i n d a l l r o o t s {(\sqrt{6} - \sqrt{2} i)}^{\frac{1}{3}} b y u s i n g d e m o v e r t h e o r e m ?

Commented by mhmd last updated on 24/May/20

h e l p m e ?

Answered by mathmax by abdo last updated on 24/May/20

let find z / z^{2} = {(\sqrt{6} - i \sqrt{2})}^{\frac{1}{3}} we have

\sqrt{6} - i \sqrt{2} = 2 \sqrt{2} {\frac{\sqrt{6}}{2 \sqrt{2}} - \frac{i \sqrt{2}}{2 \sqrt{2}}} = 2 \sqrt{2} e^{iarctan (- \frac{\sqrt{2}}{\sqrt{6}})} \Rightarrow

{(\sqrt{6} - i \sqrt{2})}^{\frac{1}{3}} = {(2 \sqrt{2})}^{\frac{1}{3}} e^{- \frac{i}{3} \arctan (\frac{\sqrt{2}}{\sqrt{6}})} let z = {re}^{i θ}

(e) \Rightarrow r^{2} e^{2 i θ} = {(2 \sqrt{2})}^{\frac{1}{3}} e^{- \frac{i}{3} \arctan (\frac{\sqrt{2}}{\sqrt{6}})} \Rightarrow r = {(2 \sqrt{2})}^{\frac{1}{6}} and

2 θ = - \frac{1}{3} \arctan (\frac{\sqrt{2}}{\sqrt{6}}) + 2 k π \Rightarrow θ_{k} = - \frac{1}{6} \arctan (\frac{\sqrt{2}}{\sqrt{6}}) + k π

so the roots are z_{z} =^{12} \sqrt{8} e^{i (- \frac{1}{6} \arctan (\frac{\sqrt{2}}{\sqrt{6}}) + k π)}

k \in {0, 1}

Commented by mhmd last updated on 24/May/20

t h a n k y o u s i r

Commented by turbo msup by abdo last updated on 24/May/20

y o u a r e w e l c o m e