Complex Variables Proof

josephtw · #1 $Old$ July 1st, 2009, 08:42 PM

So to prove that |sin(nTheta/2)/sin(Theta/2)|<=n for theta!={0,+/-2pi,+/-4pi,...} and n a positive integer I'm told that I can assign z=e^iTheta and replace the left side of the inequality with |(1-z^n)/(1-z)|. After that I just divide to get |1+z+z^2+...z^n-1| which is less than the sum of the magnitudes by the triangle inequality and thusly less than or equal to n. My problem is that I don't see why |sin(nTheta/2)/sin(Theta/2)|=|(1-z^n)/(1-z)| is true.

I don't see how 1-z could equal sin(Theta/2) because 1-z could clearly be complex while sin(Theta/2) is real, so I guess the relationship must be strictly between the magnitudes. (Since the |a/b|=|a|/|b|, right?) And so all I need is to figure out why |1-z|=|sin(Theta/2)| (I'm hoping the numerator follows from that)

Any advice or corrections would be appreciated,

Thanks

simplependulum · #2 $Old$ July 2nd, 2009, 12:26 AM

can we prove it by indution ?

Replace $\frac{\theta}{2} ~$ by $\phi$

Assume that $| \sin{ k \phi } | \leq k | \sin{ \phi } |$

When $n = k+1$

$| \sin{ (k+1) \phi } |$

$= | \sin{ k \phi + \phi } |$

$= | \sin{k \phi} \cos{\phi } + \cos{ k \phi} \sin{ \phi} |$

$\leq | \sin{k \phi} \cos{\phi } | + | \cos{ k \phi} \sin{ \phi} |$

$\leq (k | \sin{ \phi } |) |\cos{ \phi }| + | \cos{ k \phi} \sin{ \phi} |$

$\leq k | \sin{ \phi } | + | \sin{ \phi} |$

$= (k+1) | \sin{\phi} |$

The equality holds when $n= 2m+1 , \theta = (2m+1) \pi , zero$

Oh , I forgot that $\theta ~~$ is a complex number , i'll try to correct my answer

josephtw · #3 $Old$ July 2nd, 2009, 06:35 AM

Yeah, that induction approach makes sense to me (thanks), but then I wouldn't be using any of what was introduced in the text. I feel that there must be some property about complex numbers I should use to determine that |1-z|=|sin(theta/2)|. I also thought, like you, that the the 1/2 coefficient is virtually meaningless given that it's just in front of a single variable.

Now what you said about theta being complex...I'm not sure that's the case. To be fair it's not stated explicitly either way in the question, but it's just that within this text theta has thus far only been used to represent the real argument (principle or otherwise). Of course if theta is real then sin(theta) is real, which led me to think the relationship must be between the magnitudes, since 1-z could be complex.

Anyway thanks.

malaygoel · #4 $Old$ July 2nd, 2009, 06:59 AM

$z=e^{i\theta}$
........which means that z is a complex number of magnitude 1 and arguement $\theta$

$\left |1-z\right |$

= $\left |1-e^{i\theta}\right |$

= $\left |1-(cos\theta+isin\theta)\right |$

= $\left |(1-cos\theta)-isin\theta\right |$

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

= $\sqrt{2-2cos\theta}$

= $\left |2sin\frac{\theta}{2}\right |$

josephtw · #5 $Old$ July 2nd, 2009, 08:04 AM

Thanks, that's exactly what I needed. From there |1-z^n|=|2sin(nTheta/2)| is clear and you can pull out and cancel the twos. Thanks.

simplependulum · #6 $Old$ July 2nd, 2009, 09:31 PM

I just want to share a math technique :

$| 1 - e^{ i \theta} |$

$= | ( e^{ i \frac{ \theta}{2} }) ( e^{ - i \frac{\theta}{2}}- e^{ i \frac{\theta}{2}})|$

$= |e^{ i \frac{ \theta}{2} }| | -2 i \sin{ \frac{\theta}{2}} |$

$= |2 \sin{ \frac{\theta}{2}} |$