Calculating pi – part 3

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Adding angles

Remember that when we multiply comlex nubers we add their angles. So when multiplying 1+ib/n over and over again we add a small angle over and over again. When this reach –1 we have rotated half a turn.  For our particular value 3.14… that we just found,  and for large n this will happen when we have multiplied our number by itself n times.

In the following page you can explore what happens with sucsessive powers of 1+ib/n , for varius values of b, starting with the number we found on the previous page.

We can adjust n and also b, starting with b = the value we previous found.  Please explore this a bit.

Now set n to the maximum value and see what happens when you change the value of b.  If b= 3,14…. you basically get a half circle. For half that value, a quarter of a circle, for twice the value a complete circle and so on.  How much you get of a circle is proportional to b.  Also, the radius will be one. We have that

\displaystyle \left| {1+\frac{{ib}}{n}} \right|=\sqrt{{1+\frac{{{{b}^{2}}}}{{{{n}^{2}}}}}}

And, when multiplying comlex numbers we multiply their mudulus, so

\displaystyle \left| {{{{\left( {1+\frac{{ib}}{n}} \right)}}^{n}}} \right|={{\left( {\sqrt{{1+\frac{{{{b}^{2}}}}{{{{n}^{2}}}}}}} \right)}^{n}}

Then you may verify for yourself that

\displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,{{\left( {\sqrt{{1+\frac{{{{b}^{2}}}}{{{{n}^{2}}}}}}} \right)}^{n}}=1

 

If we use radianas as our unit of angles then half a turn equals π radians, and for small angles  tan–1(b/n) apraches b/n. That means that the sum of those angles must be  π , and thus that the number we found on the previous page must be π . We have thus found that we can find the value of π using  the same formula as we used to find e.

Up a level : Euler's number e --- and a bit of pi
Previous page : Calculating pi using the compound formula
Next page : Calculating pi - part 2Last modified: Apr 28, 2026 @ 20:38