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??

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

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 · 1 year, 11 months ago

Log in to reply

See here. – Micah Wood · 1 year, 11 months ago

Log in to reply

– Hjalmar Orellana Soto · 1 year, 11 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.Log in to reply

Do you mean: \[\ Cis( \phi ) = Cos( \phi ) + iSin( \phi ) \ ? \ ... (i = \sqrt{-1} ) \] – Curtis Clement · 1 year, 11 months ago

Log in to reply

– Hjalmar Orellana Soto · 1 year, 11 months ago

YesLog in to reply