# Polar Form

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?

> >

[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. Note by Calvin Lin
6 years, 8 months ago

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Although 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.

- 6 years, 8 months ago

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

- 6 years, 8 months ago

OH yes, thanks.

- 6 years, 8 months ago

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

- 6 years, 8 months ago

More accurately, $(5+i)^4 \; =\; 476 + 480i \; =\; 2(238 + 240i) \; = \; 2(1+i)(239 + i)$ so, taking arguments, $4\mathrm{Arg}(5+i) \; = \; \mathrm{Arg}(1+i) + \mathrm{Arg}(239+i)$ (all arguments are small enough that this is true). What is $\mathrm{Arg}(1+i)$?

- 6 years, 8 months ago

$\frac{\pi}{4}$

- 6 years, 8 months ago

There's your $\pi$

- 6 years, 8 months ago

But how can I eat it to find its taste. Haha, I mean how can I find its value using polar form.

- 6 years, 8 months ago

Taking the argument of a complex number is finding the $\theta$ value of the polar form. Thus we are using the polar form when we take $(5+i)^4 = 2(1+i)(239+i)$ and obtain $\begin{array}{rcl} 4\mathrm{Arg}(5+i) & = & \mathrm{Arg}(1+i) + \mathrm{Arg}(239+i) \\ 4\tan^{-1}\tfrac15 & = & \tfrac14\pi + \tan^{-1}\tfrac{1}{239} \end{array}$ which is what we want.

- 6 years, 8 months ago

Ah... Sorry, I didn't get it. Instead of doing this why don't we just use that $\pi = \arctan \frac {1}{1}$ which we get from this equation $(1 + i) = \frac {(5 + i)^4}{(239 + i)}$. I was expecting an answer which can be calculated without involving $\arctan$ at all.

- 6 years, 8 months ago

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 $10^{-100}$.

- 6 years, 8 months ago

But the question says to find it using polar form!

- 6 years, 8 months ago

I think the point is that you prove the identity using the polar form, and then use the Taylor Series to approximate $\pi$ to 100 digits. Is there any other way to get such an approximation besides using a series?

- 6 years, 8 months ago

Absolutely. The historical importance of this formula is that it was easy to use in pre-calculator days (like the early 1700s!). It is easy to calculate $\tan^{-1}\tfrac15$ accurately, since $\tan^{-1}\tfrac15 \; = \; \sum_{j=0}^\infty \frac{(-1)^j}{2j+1}\big(\tfrac15\big)^{2j+1} \; = \; \sum_{j=0}^\infty \frac{(-1)^j 2^{2j+1}}{(2j+1)10^{2j+1}}$ and dividing by $10$ is easy. Even if you have to use a large number of terms, and do all the calculations by hand, they are not too hard. On the other hand, $\tan^{-1}\tfrac1{239} \; = \; \sum_{j=0}^\infty \frac{(-1)^j}{2j+1}\big(\tfrac1{239}\big)^{2j+1}$ has terms that are quite hard work to calculate - there is no easy way to divide by $239$ - but you don't need nearly as many terms, since $\tfrac{1}{239}$ is small. You need $6$ terms of the $\tan^{-1}\tfrac15$ series, but only $2$ terms of the $\tan^{-1}\tfrac{1}{239}$ series, for example, to get $\pi$ to $8$ decimal places. You would only need $21$ terms of the second series, and about $71$ terms of the first, to get $\pi$ accurate to $100$ decimal places.

- 6 years, 8 months ago

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

- 6 years, 8 months ago

180.17

- 6 years, 8 months ago