Author Topic: interesting complex number question (Read 709 times) Tweet Share

Martoman · « **on:** February 24, 2010, 06:43:29 pm »

A good mate of mine passed on a question which he deems quite nice, which it is, but I can't seem to get an "elegant" solution to it. Using hacky maths sure, but am wondering if the community can go one better than me :S

Question is from the essentials book, and asks for $Arg(1 + cos(\theta) + isin(\theta))$ in terms of $\theta$

GerrySly · « **Reply #1 on:** February 24, 2010, 07:21:47 pm »

Just getting the mind moving, some random thoughts

$\begin{align*} \text{Let } z&=1+\cos{(\theta)}+i\sin{(\theta)}\\ \therefore \text{Arg}(z)&=\tan^{-1}{(\frac{\sin{(\theta)}}{1+\cos{(\theta)}})}\\ &=\tan^{-1}{(\tan{(\frac{\theta}{2})})}\\ &=\frac{\theta}{2} \end{align*}$

Not sure if there are restrictions or what not, that's where I would go

Martoman · « **Reply #2 on:** February 24, 2010, 07:31:16 pm »

Yes that is shorter than mine. Buuutttt how did you get $tan(\frac{\theta}{2})$ to be equal to $\frac{sin(\theta)}{1+cos(\theta)}$

Martoman · « **Reply #3 on:** February 24, 2010, 07:50:56 pm »

Ah ok. For reference, http://www.mathalino.com/reviewer/derivation-of-formulas/derivation-of-the-half-angle-formulas.

I just didn't know this. You can do this question without knowing this, BUT it takes a good page and a half of working.

QuantumJG · « **Reply #4 on:** February 24, 2010, 08:15:11 pm »

I got it pretty easy:

$\dfrac{ sin(\theta) }{1+ cos(\theta) } = tan( \dfrac{\theta}{2} )$

$sin(\theta) = 2sin( \dfrac{\theta}{2} )cos( \dfrac{\theta}{2} )$

$cos(\theta) = 2cos ^{2} ( \dfrac{\theta}{2} ) - 1$

$\dfrac{ sin(\theta) }{1+ cos(\theta) } = \dfrac{ 2sin( \dfrac{\theta}{2} )cos( \dfrac{\theta}{2} ) }{ 1 + 2cos ^{2} ( \dfrac{\theta}{2} ) - 1 }$

$= \dfrac{ sin( \dfrac{\theta}{2} ) }{ cos( \dfrac{\theta}{2} )}$

$= tan( \dfrac{\theta}{2} )$

TrueTears · « **Reply #5 on:** February 24, 2010, 08:42:33 pm »

are there any restrictions? i hate doing tan^-1 because it is not general at all

QuantumJG · « **Reply #6 on:** February 24, 2010, 10:58:56 pm »

Wouldn't θ be restricted to (-π,π)? I'm just saying because arctan(x) is restricted to $( - \dfrac{\pi}{2},\dfrac{\pi}{2} )$ .

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