As we know \(\sqrt{1}=1\) But we know too that \(cis(2π)=1\), so we can take the square root and find that \(\sqrt{cis(2π)}=\sqrt{1}\). By De Moivre's formula \(cis(π)=1\) or \(cis(2π)=1\) \(=> -1=1\) or \(1=1\). What's wrong with it??

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TopNewestIs it something to do with the fact that you have to take the positive root when dealing with complex numbers? I know \(\sqrt{ab} = ( \sqrt{a}) \times\ ( \sqrt{b}) \) isn't applicable to complex numbers, but I'm not sure if that proves useful here. – Curtis Clement · 2 years, 1 month ago

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See here. – Micah Wood · 2 years, 1 month ago

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– Hjalmar Orellana Soto · 2 years, 1 month ago

So, what happens with this? \(e^{ix}=cis(x)\) \(=>(e^{ix})^{k}=(cis(x))^{k}\) \(=>e^{i(kx)}=cis(kx)\) \(=>(cis(x))^{k}=cis(kx)\) For every real \(k\) this should work.Log in to reply

Do you mean: \[\ Cis( \phi ) = Cos( \phi ) + iSin( \phi ) \ ? \ ... (i = \sqrt{-1} ) \] – Curtis Clement · 2 years, 1 month ago

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– Hjalmar Orellana Soto · 2 years, 1 month ago

YesLog in to reply