complex numbers roots help (1 Viewer)

Bisu · Jan 20, 2010

i) Solve the equation z^6=0 using De Moivre's theorem and express the roots in the form a + ib

ii) Find the fourth roots of -8-8(rt3)i and express your answers in the form a+ib.

For ii) .. could you please use the mod-arg form, then using De Moivre's theorem method?

Thank you

Trebla · Jan 20, 2010

Bisu said:
i) Solve the equation z^6=0 using De Moivre's theorem and express the roots in the form a + ib

ii) Find the fourth roots of -8-8(rt3)i and express your answers in the form a+ib.

For ii) .. could you please use the mod-arg form, then using De Moivre's theorem method?

Thank you

$i) $Assuming you meant $z^6=1\\\\z = 1, cis \frac{\pi}{3}, cis \frac{2\pi}{3}, cis \pi, cis \frac{4\pi}{3}, cis \frac{5\pi}{3}\\\\= 1, \dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}, - \dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}, - 1, - \dfrac{1}{2}- i\dfrac{\sqrt{3}}{2}, \dfrac{1}{2}-i\dfrac{\sqrt{3}}{2}$

$ii) z^4= 8(- 1 - \sqrt{3}i)\\\\$Let $ z = rcis \theta\quad (r, \theta $ are real$) \\\\r^4cis 4\theta = 16 cis \left (-\frac{2\pi}{3}+2k\pi \right )\:$ for integer $k\\\\\Rightarrow r = 2, \theta = - \dfrac{\pi}{6}+\dfrac{k\pi}{2}\\\\k = -1, z_1 = 2cis \left (-\frac{2\pi}{3} \right )\\\\k = 0, z_2 = 2cis \left (-\frac{\pi}{6} \right )\\\\k = 1, z_3 = 2cis \left (\frac{\pi}{3} \right )\\\\ k = 2, z_4 = 2cis \left (\frac{5\pi}{6} \right )$

Affinity · Jan 25, 2010

Bisu said:
i) Solve the equation z^6=0 using De Moivre's theorem and express the roots in the form a + ib

ii) Find the fourth roots of -8-8(rt3)i and express your answers in the form a+ib.

For ii) .. could you please use the mod-arg form, then using De Moivre's theorem method?

Thank you

wouldn't i) be 0,0,0,0,0,0

complex numbers roots help (1 Viewer)

Bisu

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Trebla

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Affinity

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