[Analytic Number Theory] Deriving Machin's Formula using Arguments of Complex Numbers

We can derive Machin's formula by inspecting the complex number relation (5+i)4=2(1+i)(239+i)(5+i)^4 = 2 (1+i) (239 + i) (1).

Essentially, what we are doing is relating each quantity of Machin's formula to the argument of a complex number.

First, we recall that the sum of arguments of complex numbers is the argument of their product, and the argument of a complex number raised to a power is simply that number times the argument. That is, the relations Arg(z,w)=Arg(z)+Arg(w)Arg(z,w) = Arg(z) + Arg(w) , and Arg(zk)=k(Arg(z))Arg(z^k) = k( Arg(z)) hold for all zCz \in \mathbb{C}.

Taking the argument of both sides of (1), we get Arg((5+i)4)=Arg(2(1+i)(239+i))Arg((5+i)^4) = Arg( 2(1+i)(239+i)) 4(Arg(5+i))=Arg(2(1+i))+Arg(239+i)4(Arg(5+i)) = Arg (2(1+i)) + Arg(239+i) 4(Arg(5+i))Arg(239+i)=Arg(2(1+i))4(Arg(5+i)) -Arg(239+i)= Arg (2(1+i)) Since Arg(a+bi)=arctan(ba)Arg(a+bi) = \arctan (\frac{b}{a}), RHS =arctan(22)=arctan(1)=π4= \arctan(\frac{2}{2}) = \arctan(1) = \frac{\pi}{4}.

So we have the following relation: 4arctan(15)arctan(1239)=π44 \arctan(\frac{1}{5}) - \arctan(\frac{1}{239}) = \frac{\pi}{4} as required.

It is also possible to derive other Machin-like formulas via the same method using complex numbers. You can try deriving the following formulas as an exercise:

Euler's π4=arctan12+arctan13.\frac{\pi}{4} = \arctan \frac{1}{2} + \arctan \frac{1}{3}. Hermann's π4=2arctan12arctan17.\frac{\pi}{4} =2 \arctan \frac{1}{2} - \arctan \frac{1}{7}. Hutton's π4=2arctan13+arctan17.\frac{\pi}{4} = 2\arctan \frac{1}{3} + \arctan \frac{1}{7}.

Note by Bright Glow
3 years, 2 months ago

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