Calculating pi using the compound formula

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We have previously found that

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

How about if we explore what this will do if x is an imaginary number? We will look at this, pretending we don’t know how to calculate e^z for complex numbers, and that we don’t know the value of π. But we do know how to multiply complex numbers in Cartesian form a+bi.

So, we will explore

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

for various values of b. We can calculate this quite easily by squaring a number of times, as discussed on a previous page.

To start with, if b=0 we get

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

I.e., just as expected. How about other values? We can test this on this page:

Calculating e^ib using the compound formula

We, quite obviously, get the real answer 1 for b=0.

We will now try to find what other value(s), of b, if any, that will also give a pure real answer. In particular what value could possibly give us the answer –1.

For b=2 we get about –0.416 + 0.909i, and b=4 gives ≈– 0.654 – 0.757i .
One might thus guess that a value somewhere between 2 and 4 would give us an imaginary part of 0. We may test with b=3, and then we get – 0.990 + 0.141i. This looks rather promising. The real part is close to –1 and the imaginary part somewhat close to 0.

We will examine this further in the following page.

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

Last modified: May 8, 2026 @ 21:03