$(2Z+1)^6 = -1$
I have tried:
\begin{align}W^6 &= -1\\ &= e^{(pai/2+2pai*k)/6}\\ K &= {0..5}\end{align}
And I got 6 answers. What to do now?
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The editing process has been lost here. Could someone please edit the question appropriately. I can edit, but I'm unsure what the OP is looking for, especially with regards to "2*_+1" - i can guess but this is insufficient when editing questions, which I believe Lioness99a has done. – May 15 '17 at 11:09
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$$(2z+1)^6=-1=e^{(2m+1)i\pi}$$ where $m$ is any integer
$$2z+1=e^{(2m+1)i\pi/6}$$ where $m\equiv0,1,2,3,4,5\pmod6$
Now use $e^{2iy}-1=\cos2y+i\sin2y-1=-2\sin^2y+2i\sin y\cos y=2i\sin y(\cos y+i\sin y)$
lab bhattacharjee
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Can you explain again the Now use $e^{2iy}-1=\cos2y+i\sin2y-1=-2\sin^2y+2i\sin y\cos y=2i\sin y(\cos y+i\sin y)$ – Fayubla May 15 '17 at 11:23
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@Fayubla, See https://math.stackexchange.com/questions/3510/how-to-prove-eulers-formula-eit-cos-t-i-sin-t – lab bhattacharjee May 15 '17 at 11:25