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Find all cubic roots of \(z = −1 + i \)

 Dec 22, 2015

Best Answer 

 #2
avatar+109519 
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Hi Gibsonj338   laugh

 

Find all cubic roots of

 

 \(z=-1+i \mbox{ distance of z to the origin }=\sqrt{1^2+1^2}=\sqrt2\\ \mbox{cubic toots so there are 3 of them so they are } \frac{2\pi}{3} radians\; apart.\\ z=\sqrt2(cis\frac{3\pi}{4})\\ \sqrt[3]{z}=\sqrt[6]{2}(cis(\frac{3\pi}{4}\div3)),\quad\sqrt[6]{2}(cis((\frac{3\pi}{4}+2\pi)\div3)),\quad \sqrt[6]{2}(cis((\frac{3\pi}{4}+4\pi)\div3))\\ \sqrt[3]{z}=\sqrt[6]{2}(cis(\frac{3\pi}{4}\div3)),\quad\sqrt[6]{2}(cis(\frac{11\pi}{4}\div3)),\quad \sqrt[6]{2}(cis(\frac{19\pi}{4}\div3))\\ \sqrt[3]{z}=\sqrt[6]{2}(cis(\frac{3\pi}{12})),\quad\sqrt[6]{2}(cis(\frac{11\pi}{12})),\quad \sqrt[6]{2}(cis(\frac{19\pi}{12}))\\ \sqrt[3]{z}=\sqrt[6]{2}(cis(\frac{\pi}{4})),\quad\sqrt[6]{2}(cis(\frac{11\pi}{12})),\quad \sqrt[6]{2}(cis(\frac{19\pi}{12}))\\ or\ \)

 

\(\sqrt[3]{z}=\frac{1+i}{\sqrt[3]{2}},\quad\sqrt[6]{2}(cos(\frac{11\pi}{12})+i\;sin(\frac{11\pi}{12})),\quad\sqrt[6]{2}(cos(\frac{19\pi}{12})+i\;sin(\frac{19\pi}{12}))\\ \sqrt[3]{z}=\frac{1+i}{\sqrt[3]{2}},\quad\sqrt[6]{2}(-cos(\frac{\pi}{12})+i\;sin(\frac{\pi}{12})),\quad\sqrt[6]{2}(cos(\frac{5\pi}{12})-i\;sin(\frac{5\pi}{12})) \)

 

 Dec 23, 2015
edited by Melody  Dec 23, 2015
 #1
avatar+1904 
+5

\(z =(−1 + i )^(1/3)\)

 

\(r=\sqrt((-1)^2+1^2)\)

 

\(r=\sqrt(1+1^2)\)

 

\(r=\sqrt(1+1)\)

 

\(r=\sqrt(2)\)

 

\(tan(\Theta)=1/-1\)

 

\(tan(\Theta)=-1\)

 

\(\Theta=tan^-1(-1)\)

 

\(\Theta=-\pi/4\)

 

\(\sqrt(2)e^(-\pi/4i)^(1/3)\)

 

\(\sqrt(2)^(1/3)e^(-\pi/12i)\)

 

\(\sqrt(2)^(1/3)*(cos(-\pi/12)+isin(-\pi/12))\)

 

\(\sqrt(2)^(1/3)*(0.965925826289 +isin(-\pi/12))\)

 

\(\sqrt(2)^(1/3)*(0.965925826289 +i*-0.258819045103)\)

 

\(\ 1.414213562373095)^(1/3)*(0.965925826289 +i*-0.258819045103)\)

 

\(\ 1.122462048309373*(0.965925826289 +i*-0.258819045103)\)

 

\(\ 1.084215081491274550528506797 +i*-0.290514555507789375384650419\)

 

\(\ 1.084215081491274550528506797 + -0.290514555507789375384650419i\)

 

Answer 1:\(\ 1.084215081491274550528506797 -0.290514555507789375384650419i\)

 

\(\sqrt(2)e^(7\pi/4i)^(1/3)\)

 

\(\sqrt(2)^(1/3)e^(7\pi/12i)\)

 

\(\sqrt(2)^(1/3)*(cos(7\pi/12)+isin(7\pi/12))\)

 

\(\sqrt(2)^(1/3)*(-0.258819045103 +isin(7\pi/12))\)

 

\(\sqrt(2)^(1/3)*(-0.258819045103 +i* 0.965925826289)\)

 

\(\ 1.414213562373095)^(1/3)*(-0.258819045103 +i* 0.965925826289)\)

 

\(\ 1.122462048309373*(-0.258819045103 +i* 0.965925826289)\)

 

\(\ -0.290514555507789375384650419 +i* 1.084215081491274550528506797\)

 

Answer 2:\(\ -0.290514555507789375384650419 +1.084215081491274550528506797i\)

 

\(\sqrt(2)e^(15\pi/4i)^(1/3)\)

 

\(\sqrt(2)^(1/3)e^(15\pi/12i)\)

 

\(\sqrt(2)^(1/3)*(cos(15\pi/12)+isin(15\pi/12))\)

 

\(\sqrt(2)^(1/3)*(-0.707106781187 +isin(15\pi/12))\)

 

\(\sqrt(2)^(1/3)*(-0.707106781187 +i* -0.707106781187)\)

 

\(\ 1.414213562373095)^(1/3)*(-0.707106781187 +i* -0.707106781187)\)

 

\(\ 1.122462048309373*(-0.793700525984607637192165751 +i* -0.793700525984607637192165751)\)

 

\(\ -0.793700525984607637192165751 +i*-0.793700525984607637192165751\)

 

\(\ -0.793700525984607637192165751 +-0.793700525984607637192165751i\)

 

Answer 3:\(\ -0.793700525984607637192165751 -0.793700525984607637192165751i\)

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 Dec 22, 2015
edited by gibsonj338  Dec 22, 2015
 #2
avatar+109519 
+10
Best Answer

Hi Gibsonj338   laugh

 

Find all cubic roots of

 

 \(z=-1+i \mbox{ distance of z to the origin }=\sqrt{1^2+1^2}=\sqrt2\\ \mbox{cubic toots so there are 3 of them so they are } \frac{2\pi}{3} radians\; apart.\\ z=\sqrt2(cis\frac{3\pi}{4})\\ \sqrt[3]{z}=\sqrt[6]{2}(cis(\frac{3\pi}{4}\div3)),\quad\sqrt[6]{2}(cis((\frac{3\pi}{4}+2\pi)\div3)),\quad \sqrt[6]{2}(cis((\frac{3\pi}{4}+4\pi)\div3))\\ \sqrt[3]{z}=\sqrt[6]{2}(cis(\frac{3\pi}{4}\div3)),\quad\sqrt[6]{2}(cis(\frac{11\pi}{4}\div3)),\quad \sqrt[6]{2}(cis(\frac{19\pi}{4}\div3))\\ \sqrt[3]{z}=\sqrt[6]{2}(cis(\frac{3\pi}{12})),\quad\sqrt[6]{2}(cis(\frac{11\pi}{12})),\quad \sqrt[6]{2}(cis(\frac{19\pi}{12}))\\ \sqrt[3]{z}=\sqrt[6]{2}(cis(\frac{\pi}{4})),\quad\sqrt[6]{2}(cis(\frac{11\pi}{12})),\quad \sqrt[6]{2}(cis(\frac{19\pi}{12}))\\ or\ \)

 

\(\sqrt[3]{z}=\frac{1+i}{\sqrt[3]{2}},\quad\sqrt[6]{2}(cos(\frac{11\pi}{12})+i\;sin(\frac{11\pi}{12})),\quad\sqrt[6]{2}(cos(\frac{19\pi}{12})+i\;sin(\frac{19\pi}{12}))\\ \sqrt[3]{z}=\frac{1+i}{\sqrt[3]{2}},\quad\sqrt[6]{2}(-cos(\frac{\pi}{12})+i\;sin(\frac{\pi}{12})),\quad\sqrt[6]{2}(cos(\frac{5\pi}{12})-i\;sin(\frac{5\pi}{12})) \)

 

Melody Dec 23, 2015
edited by Melody  Dec 23, 2015

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