Calculating pi - part 3

Up a level : Euler's number e --- and a bit of pi
Previous page : Calculating pi - part 2
Next page : Trig from the compound formula

Adding angles

Remember that when we multiply complex numbers, 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 reaches –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 successive powers of 1+ib/n , for various values of b, starting with the number we found on the previous page.

Exploring (1+ib/n)ⁿ

We can adjust n and also b, starting with b = the value we previously 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 complex numbers, we multiply their modulus, so

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

Then you may try to verify that

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

One way to do this is furher down the page.

So, we could rewrite the above as

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

But

$\displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,{{\left( {1+\frac{x}{n}} \right)}^{n}}={{e}^{x}}$

so,

$\displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,\sqrt{{{{{\left( {1+\frac{{{{b}^{2}}/n}}{n}} \right)}}^{n}}}}=\sqrt{{\underset{{n\to \infty }}{\mathop{{\lim }}}\,{{e}^{{{{b}^{2}}/n}}}}}={{e}^{0}}=1$

So, the distance to the origin will basically stay at 1.

Now, let us look at the half circle. We get that by adding up the smaller sides of n small triangles rotated about the origin.

If n is large, the triangle’s height approaches the short side length, and in our case, that height is b/n, since our triangle represents the number 1+ib/n. The sum of those lengths is b, and that number will approach the arclength of a half a unit circle as n approaches infinity. The arclength of a half a unit circle is π, so b must indeed be π, and we have thus found a way of calculating π using the compound formula that was used to define e.

The values of e and π are indeed deeply connected. They are basically two sides of the same (circular) coin.

Radians

calculating that value, we added n small length b/n, but on the other hand, we have also added n small angles. It would then be quite natural to identify those lengths with the angles. That means that the value π could not just be seen as the arc length of half a unit circle, but also the angle 180°. This will be the angle unit radians where 180°=π.

One could argue that 180° is simply another way to write π.

Up a level : Euler's number e --- and a bit of pi
Previous page : Calculating pi - part 2
Next page : Trig from the compound formula

Last modified: May 9, 2026 @ 20:40