6 distinct roots of complex number

• Sep 13th 2012, 06:46 PM
timeforchg
6 distinct roots of complex number
I am trying to find the z0 to z6 roots of this equation but I am stuck here. Anyone care to show the step by step on how to procced?
• Sep 13th 2012, 07:03 PM
Prove It
Re: 6 distinct roots of complex number
Quote:

Originally Posted by timeforchg
I am trying to find the z0 to z6 roots of this equation but I am stuck here. Anyone care to show the step by step on how to procced?

Try converting to polars...

Edit: We answered this exact question here.
• Sep 13th 2012, 07:36 PM
MaxJasper
Re: 6 distinct roots of complex number
Quote:

Originally Posted by timeforchg
I am trying to find the z0 to z6 roots of this equation but I am stuck here. Anyone care to show the step by step on how to procced?

$\displaystyle {z6}=1+\left(\frac{64 i (1-i)}{1+2 i}\right)^6$

$\displaystyle z6 = ((64 i (1 - i))/(1 + 2 i))^6 + 1$

$\displaystyle {zz}({k}){=}\sqrt[6]{|{z6}|} e^{i \arg \left(\frac{2 \pi k}{6}+\frac{\text{z6}}{6}\right)}$

$\displaystyle k = {0, 1, 2, 3, 4, 5}$

z1, z2, z3, z4, z5, z6=
$\displaystyle -\sqrt[3]{\frac{3}{5}} \sqrt[12]{149250101163100633121} e^{i \tan ^{-1}\left(\frac{7146825580544}{2687695088383}\right) }$
$\displaystyle -\sqrt[3]{\frac{3}{5}} \sqrt[12]{149250101163100633121} e^{i \tan ^{-1}\left(\frac{64321430224896}{24189255795447-31250 \pi }\right)}$
$\displaystyle -\sqrt[3]{\frac{3}{5}} \sqrt[12]{149250101163100633121} e^{i \tan ^{-1}\left(\frac{64321430224896}{24189255795447-62500 \pi }\right)}$
$\displaystyle -\sqrt[3]{\frac{3}{5}} \sqrt[12]{149250101163100633121} e^{i \tan ^{-1}\left(\frac{21440476741632}{8063085265149-31250 \pi }\right)}$
$\displaystyle -\sqrt[3]{\frac{3}{5}} \sqrt[12]{149250101163100633121} e^{i \tan ^{-1}\left(\frac{64321430224896}{24189255795447-125000 \pi }\right)}$
$\displaystyle -\sqrt[3]{\frac{3}{5}} \sqrt[12]{149250101163100633121} e^{i \tan ^{-1}\left(\frac{64321430224896}{24189255795447-156250 \pi }\right)}$
• Sep 16th 2012, 06:36 PM
timeforchg
Re: 6 distinct roots of complex number
Quote:

Originally Posted by MaxJasper
$\displaystyle {z6}=1+\left(\frac{64 i (1-i)}{1+2 i}\right)^6$

$\displaystyle z6 = ((64 i (1 - i))/(1 + 2 i))^6 + 1$

$\displaystyle {zz}({k}){=}\sqrt[6]{|{z6}|} e^{i \arg \left(\frac{2 \pi k}{6}+\frac{\text{z6}}{6}\right)}$

$\displaystyle k = {0, 1, 2, 3, 4, 5}$

z1, z2, z3, z4, z5, z6=
$\displaystyle -\sqrt[3]{\frac{3}{5}} \sqrt[12]{149250101163100633121} e^{i \tan ^{-1}\left(\frac{7146825580544}{2687695088383}\right) }$
$\displaystyle -\sqrt[3]{\frac{3}{5}} \sqrt[12]{149250101163100633121} e^{i \tan ^{-1}\left(\frac{64321430224896}{24189255795447-31250 \pi }\right)}$
$\displaystyle -\sqrt[3]{\frac{3}{5}} \sqrt[12]{149250101163100633121} e^{i \tan ^{-1}\left(\frac{64321430224896}{24189255795447-62500 \pi }\right)}$
$\displaystyle -\sqrt[3]{\frac{3}{5}} \sqrt[12]{149250101163100633121} e^{i \tan ^{-1}\left(\frac{21440476741632}{8063085265149-31250 \pi }\right)}$
$\displaystyle -\sqrt[3]{\frac{3}{5}} \sqrt[12]{149250101163100633121} e^{i \tan ^{-1}\left(\frac{64321430224896}{24189255795447-125000 \pi }\right)}$
$\displaystyle -\sqrt[3]{\frac{3}{5}} \sqrt[12]{149250101163100633121} e^{i \tan ^{-1}\left(\frac{64321430224896}{24189255795447-156250 \pi }\right)}$

Hi Max,

This makes me more confused. Am suffering from brain damage! haha