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Cubic root from (1+i)
Cubic root from (1+i)Let's calculate cubic root from (1+i), or in another style of writing: ∛(1+i). This is very easy to do in Euler or the exponential form of complex numbers. How to do it? \[ \begin{align*} z &= 1 + i \\ z &= \sqrt{1^2 + 1^2} = \sqrt{2} \\ \theta &= \tan^{1}\frac{1}{1} = \frac{\pi}{4} \\ z &= \sqrt{2},e^{i\pi/4} \\ z^{1/3} &= \sqrt[3]{\sqrt{2}}e^{i(\pi/12 + 2k\pi/3)},; k=0,1,2 \\ &= \sqrt[6]{2},(\cos(\frac{\pi}{12} + \frac{2k\pi}{3}) + i\sin(\frac{\pi}{12} + \frac{2k\pi}{3})),; k=0,1,2 \end{align*}\]For more details and some comments please check this video:

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