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??

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.

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.

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## 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.

Log in to reply

See here.

Log in to reply

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: \[\ Cis( \phi ) = Cos( \phi ) + iSin( \phi ) \ ? \ ... (i = \sqrt{-1} ) \]

Log in to reply

Yes

Log in to reply