This week, we learn about Polar Form.

You may first choose to read the post Complex Numbers if you have not already done so.

How would you use polar form to solve the following?

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[John Machin, 1706] Show that \[ \pi = 16 \arctan \frac{1}{5} - 4 \arctan \frac{1}{239}. \] Use the above identity to calculate \(\pi \) to 100 decimal places.

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## Comments

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TopNewestAlthough I wasn't able to find the value of \(\pi\). I did the following thinking. Dividing both sides by 4. In order to reduce the right hand side to complex number, the coefficient of first term has to be reduced to \(4\). \( \arctan \frac {1} {5} = \theta\) or \( \tan \theta = \frac {1} {5} \). Evaluating \(tan 4\theta = 120/119 \) using the identity \(tan 2\theta = \frac{2 \tan \theta } {1 - \tan \theta}\). Now, converting the equation into polar form, \( \frac {119 + 120i}{239 + i} = (1 + i) \) which is true. I don't know how to find the value of \(\pi\). I think I am close. Please help.

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Use the Taylor Series for \(\arctan(x)\) with a sufficient number of terms. Use the Lagrange Error Bound to show that the error is less than \(10^{-100}\).

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But the question says to find it using polar form!

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Think about \((5+i)^4\) as well...

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OH yes, thanks.

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The equation reduces to \( (1 + i) = \frac {(5 + i)^4}{(239 + i)} \). Still no sign of \( \pi \) yet.

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Multiply both sides by \(i\) and use exp function on both sides. QED (I think)

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180.17

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