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# Do you know De Moivre's formula ?

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??

Note by Hjalmar Orellana Soto
2 years, 7 months ago

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Is 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.

- 2 years, 7 months ago

See here.

- 2 years, 7 months 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.

- 2 years, 7 months ago

Do you mean: $\ Cis( \phi ) = Cos( \phi ) + iSin( \phi ) \ ? \ ... (i = \sqrt{-1} )$

- 2 years, 7 months ago

Yes

- 2 years, 7 months ago