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 π=16arctan154arctan1239. \pi = 16 \arctan \frac{1}{5} - 4 \arctan \frac{1}{239}. Use the above identity to calculate π\pi to 100 decimal places.

Note by Calvin Lin
6 years, 1 month ago

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Although I wasn't able to find the value of π\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 44. arctan15=θ \arctan \frac {1} {5} = \theta or tanθ=15 \tan \theta = \frac {1} {5} . Evaluating tan4θ=120/119tan 4\theta = 120/119 using the identity tan2θ=2tanθ1tanθtan 2\theta = \frac{2 \tan \theta } {1 - \tan \theta}. Now, converting the equation into polar form, 119+120i239+i=(1+i) \frac {119 + 120i}{239 + i} = (1 + i) which is true. I don't know how to find the value of π\pi. I think I am close. Please help.

Lokesh Sharma - 6 years, 1 month ago

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

Mark Hennings - 6 years, 1 month ago

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OH yes, thanks.

Lokesh Sharma - 6 years, 1 month ago

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

Lokesh Sharma - 6 years, 1 month ago

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

Mark Hennings - 6 years, 1 month ago

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@Mark Hennings π4\frac{\pi}{4}

Lokesh Sharma - 6 years, 1 month ago

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@Lokesh Sharma There's your π\pi

Mark Hennings - 6 years, 1 month ago

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

Lokesh Sharma - 6 years, 1 month ago

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@Lokesh Sharma Taking the argument of a complex number is finding the θ\theta value of the polar form. Thus we are using the polar form when we take (5+i)4=2(1+i)(239+i)(5+i)^4 = 2(1+i)(239+i) and obtain 4Arg(5+i)=Arg(1+i)+Arg(239+i)4tan115=14π+tan11239 \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.

Mark Hennings - 6 years, 1 month ago

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@Mark Hennings Ah... Sorry, I didn't get it. Instead of doing this why don't we just use that π=arctan11 \pi = \arctan \frac {1}{1} which we get from this equation (1+i)=(5+i)4(239+i) (1 + i) = \frac {(5 + i)^4}{(239 + i)} . I was expecting an answer which can be calculated without involving arctan\arctan at all.

Lokesh Sharma - 6 years, 1 month ago

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Use the Taylor Series for arctan(x)\arctan(x) with a sufficient number of terms. Use the Lagrange Error Bound to show that the error is less than 1010010^{-100}.

Ricky Escobar - 6 years, 1 month ago

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But the question says to find it using polar form!

Lokesh Sharma - 6 years, 1 month ago

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

Ricky Escobar - 6 years, 1 month ago

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@Ricky Escobar 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 tan115\tan^{-1}\tfrac15 accurately, since tan115  =  j=0(1)j2j+1(15)2j+1  =  j=0(1)j22j+1(2j+1)102j+1 \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 1010 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, tan11239  =  j=0(1)j2j+1(1239)2j+1 \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 239239 - but you don't need nearly as many terms, since 1239\tfrac{1}{239} is small. You need 66 terms of the tan115\tan^{-1}\tfrac15 series, but only 22 terms of the tan11239\tan^{-1}\tfrac{1}{239} series, for example, to get π\pi to 88 decimal places. You would only need 2121 terms of the second series, and about 7171 terms of the first, to get π\pi accurate to 100100 decimal places.

Mark Hennings - 6 years, 1 month ago

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

Taehyung Kim - 6 years, 1 month ago

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180.17

Narasimha Rao B L - 6 years, 1 month ago

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