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• Users: saint_n
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1. ### Need help with a complex inequality?

for p = 1 |\frac{1}{2}(a+b)|\leq(\frac{1}{2}|(a+b)|)-ab isnt this false because you subtracting a ab on the RHS?
2. ### Need help with a complex inequality?

so you saying that |\frac{1}{2}(a+b)|^{p}\leq(\sqrt[p]{\frac{1}{2}}|(a+b)|)^{p}-pab \leq \frac{1}{2}(|a|^p+|b|^p)
3. ### Need help with a complex inequality?

Need help with a complex inequality?? hey! i been trying to do this inequality for a 2 hrs now and can't seem to prove it |\frac{1}{2}(a+b)|^p \leq \frac{1}{2}(|a|^p+|b|^p) where a,b are complex numbers Can anyone suggest a way?? thanks
4. ### Need help with a proof

ok,,thanks i also found the Fundamental Theorem of Algebra which I am reading up now. Thank you everyone! Cheers
5. ### Need help with a proof

I figured that we have to use complex numbers but we weren't told a method how to factorise complex polynomials with say n degree.Can you tell me a method of how to factorize a complex polynomial?? You mentioned that the quote above is the theorema egregium of Algebra which I am having...
6. ### Need help with a proof

the highest we ever gone up to is probably degree 5 that i know of..i'll try look up the theorem and see what you people are talking about
7. ### Need help with a proof

Do you know the name of it because of never heard of it.I wouldn't know where to start.If i have the name i can probably do some research on it and continue from there maybe
8. ### Need help with a proof

I don't see where you going.The roots you can get from the product..do want me to replace z with \cos\theta+i\sin\theta hmmm...with the LHS = (\cos\theta+i\sin\theta)^{n+1}-1=(\cos(n+1)\theta+i\sin(n+1)\theta)-1 and RHS =...
9. ### Need help with a proof

After learning a little latex maybe this looks better z^{n+1} - 1 =\Pi (z - e^\frac{2 \pi i(j+1)}{n+1}) where the product goes from 0<= j <= n
10. ### Need help with a proof

sorry!yes i was missing something in the original equation.So it suppose to look like this (z^(n+1))-1 = Product[(z-exp[2*Pi*i*(j+1)/(n+1)]) , 0<=j<=n] where Product[(z-exp[2*Pi*i*(j+1)/(n+1)]) , 0<=j<=n] = (z-exp[2*Pi*i*(0+1)/(n+1)]*(z-exp[2*Pi*i*(1+1)/(n+1)]*...*(z-exp[2*Pi*i*(n+1)/(n+1)]...
11. ### Need help with a proof

Hey!I have a tut question and I am having problems proving (z^(n+1))-1 = (z-exp[2*Pi*i*(j+1)/(n+1)]) where 0<=j<=n I tried doing it by induction which is easy for the 1st case with n = 1,I assume the case n-1 but then i get stuck with the last case. How will i know that all the z's with...