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Sep
11
comment Useful device in complex analysis (Perron's formula)
My original reference was Havil (2003), pp. 200-201. I'm not at home at the moment, but I think this was his book "Gamma"
Sep
11
revised Useful device in complex analysis (Perron's formula)
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Sep
11
comment Useful device in complex analysis (Perron's formula)
I read this in a book last week and made quick note of it due to its obvious usefulness. It's also mentioned in a video on Pretentious Analytic Number Theory by Granville - start the video at 10 mins, 30 second: youtube.com/watch?v=RpuhtlFqFLw
Sep
11
revised Useful device in complex analysis (Perron's formula)
edited title
Sep
10
asked Useful device in complex analysis (Perron's formula)
Sep
10
revised Proving $\arg(zw)=\arg(z)+\arg(w)$
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Sep
10
comment Proving $\arg(zw)=\arg(z)+\arg(w)$
I think that last edit should get across what I actually mean.
Sep
9
revised Proving $\arg(zw)=\arg(z)+\arg(w)$
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Sep
9
comment Proving $\arg(zw)=\arg(z)+\arg(w)$
Ah, yes I was always aware of that problem but I was only ever considering principal arguments. I think I just limited the scope of the question which wouldn't do as you quite rightly said.
Sep
9
revised Proving $\arg(zw)=\arg(z)+\arg(w)$
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Sep
9
comment Proving $\arg(zw)=\arg(z)+\arg(w)$
The principal argument is a well known concept. What do you mean by the usual sum?
Sep
9
revised Proving $\arg(zw)=\arg(z)+\arg(w)$
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Sep
9
comment Proving $\arg(zw)=\arg(z)+\arg(w)$
In that case, $\text{Arg}(-i)=3\pi/2$, and so $\text{Arg}(-i)+\text{Arg}(-i)=3\pi\equiv \pi$. But $\text{Arg}(-i\times-i)=\text{Arg}(-1)=\pi$. This is correct given arguments in $[0,2\pi)$.
Sep
9
comment Proving $\arg(zw)=\arg(z)+\arg(w)$
I think that fixes it, but I'm sure you will throw another spanner my way @ Did.
Sep
9
revised Proving $\arg(zw)=\arg(z)+\arg(w)$
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Sep
9
revised Proving $\arg(zw)=\arg(z)+\arg(w)$
[Edit removed during grace period]
Sep
9
answered Proving $\arg(zw)=\arg(z)+\arg(w)$
Sep
9
comment Find the generating function of the sequence $a_n = \sum\limits_{k=0}^n k(k-1)$
For reference, the answer is $$\frac{2 x^2}{(x-1)^4}.$$
Sep
8
answered Inequalities with logarithms and limits
Sep
8
revised Is anything known about $2\pi$ integer multiple arguments of the cosine integral?
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