What theorem does this remind you of?

Algebra Level 5

True or false :

$\sqrt{r(\cos\theta + i\sin\theta)}=\sqrt{r}\left(\cos\left(\dfrac{\theta}{2}\right)+i\sin\left(\dfrac{\theta}{2}\right)\right)$

where $r>0$ .

False True

2 solutions

Jon Haussmann
Mar 10, 2017

For $r = 1$ and $\theta = 2 \pi$ , $\sqrt{r (\cos \theta + i \sin \theta)} = \sqrt{1} = 1,$ but $\sqrt{r} \left( \cos \frac{\theta}{2} + i \sin \frac{\theta}{2} \right) = -1.$

Aditya Narayan Sharma
Apr 27, 2016

Relevant wiki: De Moivre's Theorem

LHS = $\sqrt{r}e^{i\frac{\theta}{2}}$ = RHS .

As De-Moivre's theorem states : $cos\theta+isin\theta=e^{i\theta}$

Moderator note:

With complex exponentiation, we have to be careful with the values that we're picking. The equality refers to "one of these set of values is equal to one of these set of values". If we picked particular values, then they might not be equal.

That isn't De Moivre's theorem, its Euler's Formula . Also you cannot apply De Moivre's theorem here since its applicable only if the index is an integer :3

Nihar Mahajan - 5 years, 1 month ago

I've got a doubt about De Moivre's Theorem. Why can't you apply it for non integer numbers? I tried to prove this theorem by taking Euler's Formula this way: $e^{ix}=cis(x)$ $(e^{ix})^k=(cis(x))^k$ $e^{i(kx)}=(cis(x))^k$ $cis(kx)=(cis(x))^k$ For any real $k$ .....

Hjalmar Orellana Soto - 5 years, 1 month ago

Hjalmar, my guess is that non-integer exponentiation of iota would be a messy affair. Also compare your proof with the proof here: https://goo.gl/OXMZUq. You'll find that the exponentiation law used is for integer powers. Your question demands a better answer, and I too am looking forward to one.

Arjun Bahuguna - 5 years ago

@Arjun Bahuguna – The exponention law must be on integers when we are talking about real numbers, with complex it doesn't matter, realize that that's the only fact that may make a mistake for non integer numbers... that's what I think

Hjalmar Orellana Soto - 5 years ago

Yeah Mistake. euler's formulae. If the index is rational i.e. not an integer, then one of the values is this.

Aditya Narayan Sharma - 5 years, 1 month ago

Euler's Formula, not Euler's formulae ;}

Aditya Sky - 5 years, 1 month ago

What theorem does this remind you of?

2 solutions

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