find all roots ((√6) −(√2)i)^(1/3) by using demover theorem ?


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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
	$let find z / z^2 =((√6)−i(√2))^(1/3) we have (√6)−i(√2)=2(√2){((√6)/(2(√2)))−((i(√2))/(2(√2)))} =2(√2)e^(iarctan(−((√2)/(√6)))) ⇒ ((√6)−i(√2))^(1/3) =(2(√2))^(1/3) e^(−(i/3)arctan(((√2)/(√6)))) let z =re^(iθ) (e)⇒r^2 e^(2iθ ) =(2(√2))^(1/3) e^(−(i/3)arctan(((√2)/(√6)))) ⇒r =(2(√2))^(1/6) and 2θ =−(1/3)arctan(((√2)/(√6)))+2kπ ⇒θ_k =−(1/6) arctan(((√2)/(√6)))+kπ so the roots are z_z =^(12) (√8) e^(i(−(1/6)arctan(((√2)/(√6)))+kπ)) k ∈{0,1}$
	$let find z / z^{2} = {(\sqrt{6} - i \sqrt{2})}^{\frac{1}{3}} we have$ $\sqrt{6} - i \sqrt{2} = 2 \sqrt{2} {\frac{\sqrt{6}}{2 \sqrt{2}} - \frac{i \sqrt{2}}{2 \sqrt{2}}} = 2 \sqrt{2} e^{iarctan (- \frac{\sqrt{2}}{\sqrt{6}})} \Rightarrow$ ${(\sqrt{6} - i \sqrt{2})}^{\frac{1}{3}} = {(2 \sqrt{2})}^{\frac{1}{3}} e^{- \frac{i}{3} \arctan (\frac{\sqrt{2}}{\sqrt{6}})} let z = {re}^{i θ}$ $(e) \Rightarrow r^{2} e^{2 i θ} = {(2 \sqrt{2})}^{\frac{1}{3}} e^{- \frac{i}{3} \arctan (\frac{\sqrt{2}}{\sqrt{6}})} \Rightarrow r = {(2 \sqrt{2})}^{\frac{1}{6}} and$ $2 θ = - \frac{1}{3} \arctan (\frac{\sqrt{2}}{\sqrt{6}}) + 2 k π \Rightarrow θ_{k} = - \frac{1}{6} \arctan (\frac{\sqrt{2}}{\sqrt{6}}) + k π$ $so the roots are z_{z} =^{12} \sqrt{8} e^{i (- \frac{1}{6} \arctan (\frac{\sqrt{2}}{\sqrt{6}}) + k π)}$ $k \in {0, 1}$
	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

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