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??

#Algebra #DeMoivre #HjalmarPhillippe

Easy Math Editor

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Comments

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.

Curtis Clement - 5 years, 10 months ago

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

Curtis Clement - 5 years, 10 months ago

Yes

Hjalmar Orellana Soto - 5 years, 10 months ago

See here.

Micah Wood - 5 years, 10 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.

Hjalmar Orellana Soto - 5 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$