We can derive Machin's formula by inspecting the complex number relation \((5+i)^4 = 2 (1+i) (239 + i)\) (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 \(Arg(z,w) = Arg(z) + Arg(w) \), and \(Arg(z^k) = k( Arg(z)) \) hold for all \(z \in \mathbb{C}\).

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

So we have the following relation: \[4 \arctan(\frac{1}{5}) - \arctan(\frac{1}{239}) = \frac{\pi}{4}\] 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 \[\frac{\pi}{4} = \arctan \frac{1}{2} + \arctan \frac{1}{3}.\] Hermann's \[\frac{\pi}{4} =2 \arctan \frac{1}{2} - \arctan \frac{1}{7}.\] Hutton's \[\frac{\pi}{4} = 2\arctan \frac{1}{3} + \arctan \frac{1}{7}.\]

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