Recent content by silontes

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    Undergrad Solving z^6 = 1 using De Moivre's Theorem |z|^6(cos 6θ + isin6θ) = 1

    I don't understand why Z^6 = |Z|^6(\cos(6\theta) + i\sin(6\theta) ) = 1 equals to 1( 1+0i)
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    Undergrad Solving z^6 = 1 using De Moivre's Theorem |z|^6(cos 6θ + isin6θ) = 1

    Sorry for being unclear, but I think it's the step in your explanation that turning the De Moivre theorem to polar form. How do you do that?
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    Undergrad Solving z^6 = 1 using De Moivre's Theorem |z|^6(cos 6θ + isin6θ) = 1

    For that question, here is the answer the prof wrote. Find all the solutions of Z^6 = 1. |z|^6(cos6θ + isin6θ) |z|^6 = 1 |z| = 1 Therefore: 1(1 +0i) <-- don't get how the prof got to here. cos6θ = 1 sin6θ = 0 6θ = 0+n(2π) = n(π/3) so 0, π/3, 2π/3, π...
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    Undergrad Solving z^6 = 1 using De Moivre's Theorem |z|^6(cos 6θ + isin6θ) = 1

    Hi guys, just a simple question. I need to find all solutions of z^6 = 1. I need to change that to |z|^6(cos 6θ + isin6θ) = 1 and the prof wrote → 1(1+0i). How did the prof do this? The rest of the question I understand. It's just that one step still stumps me. Thanks