How would i go about solving this?
The equation is as in the title:
$z^3+i+1=0$
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hint: rewrite as $z^{3} = -i -1$. do you know how to take complex roots? – pwerth Aug 25 '17 at 14:51
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2Solving $z^n=a$ in the complexes can be done using the polar representations of complex numbers. – Angina Seng Aug 25 '17 at 14:53
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1$z = (-1 - i)^{1/3}$. Of course, that's not the best answer you can give. What theorems of complex analysis do you know? – Kaynex Aug 25 '17 at 14:53
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Please use MathJax. – José Carlos Santos Aug 25 '17 at 14:54
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Do you know about polar form of writing complex numbers? – ajotatxe Aug 25 '17 at 14:57
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Related question: https://math.stackexchange.com/questions/2203770/how-to-find-all-values-of-z-such-that-z3-8i?noredirect=1&lq=1. – Dietrich Burde Aug 25 '17 at 15:05
2 Answers
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HINT: the first solution is given by $$z_1=1/2\,{2}^{2/3}-i/2{2}^{2/3}$$ then you can divide your equation by $$z-z_1$$
Dr. Sonnhard Graubner
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Hint:
$z^3 = -1-i = \sqrt2\left(-\dfrac1{\sqrt2}-i\dfrac1{\sqrt2}\right) = \sqrt2\left[\cos\left(\dfrac{5\pi}4+2n\pi\right)+i\sin\left(\dfrac{5\pi}4+2n\pi\right)\right]$
Can you use De Moivre's theorem and proceed?
George Law
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