Euler's Formula

Ching-Kuang Shene (冼鏡光), Professor Emeritus
Department of Computer Science
Michigan Technological University
Houghton, MI 49931
USA
Created October 5, 2025

Few people who are interested in mathemtics do not know Euler's Identity, one of the most beautiful equations in human's history, where \( i = \sqrt{-1} \) \[ e^{\pi i} + 1 = 0 \] The Euler Identity has all the basic units in nature into a single formula. This is a direct consequence of Euler's formula, where \( z \) is a complex number: \[ e^{z i} = \cos(z) + i\sin(z) \] If \( z = \pi \), we have Euler's Identity. Now if we replace \( z \) by \( -z \), we have \[ e^{-z i} = \cos(z) - s\sin(z) \] Then, multiplying \( e^{zi} \) and \( e^{-zi} \) together yields \( e^{-zi}e^{-zi}= e^{iz+(-iz)} = e^0 = 1 = \cos^2(z) + \sin^2(z) \).

This is a very6 simple proof of the Pythagorean Identity. However, we have a few simple questions before fully "fully" convinced:

This page proves that Euler's formula is indeed independent of the Pythagorean Identity. There are multiple ways of definining \( e^x \), \( \sin() \) and \( \cos(x) \). Here, we choose to use the power-series approach, which should be available in many complex analysis textbooks. We define \( e^x \), \( \sin() \) and \( \cos(x) \) as follows, where \( z \) is a complex number: $$ \begin{eqnarray*} e^z &=& \sum_{n=0}^{\infty } \frac{z^n}{n!} \\ \sin(z) &=& \sum_{n=0}^{\infty} \frac{(-1)^n z^{2n+1}}{(2n+1)!} \\ \cos(z) &=& \sum_{n=0}^{\infty} \frac{(-1)^n z^{2n}}{(2n)!} \end{eqnarray*} $$ These series are well-define on the complex plane. Note that they agree with their real counterparts. From the definition of \( e^z \) we have the following $$ \begin{eqnarray*} e^{iz} &=& \sum_{n=0}^{\infty} \frac{iz^n}{n!} \\ &=& 1 + \frac{iz}{1!} + \frac{(iz)^2}{2!} + \frac{(iz)^3}{3!} + \frac{(iz)^4}{4!} + \frac{(iz)^5}{5!} + \frac{(iz)^6}{6!} + \frac{(iz)^7}{7!} + \frac{(iz)^8}{8!} + \cdots \\ &=& 1 + iz - \frac{z^2}{2!} - i\frac{z^3}{3!}+ \frac{z^4}{4!} + i\frac{z^5}{5!} - \frac{z^6}{6!} - i\frac{z^7}{7!} + \frac{z^8}{8!} + \cdots \\ &=& \left( 1-\frac{z^2}{2!} + \frac{z^4}{4!} - \frac{z^6}{6!} + \frac{z^8}{8!} + \cdots \right) +i \left( z-\frac{z^3}{3!} + \frac{z^5}{5!} - \frac{z^7}{7!} + \frac{z^9}{9!} + \cdots \right) \\ &=& \cos(z) + i\sin(z) \end{eqnarray*} $$ Thus, the relationship among \( e^z \), \( \sin(z) \) and \( \cos(z) \) aew established independent of the Pythagorean Idcentity. Next, we shall prove \( e^{z_1}e^{z_2} = e^{z_1+z_2} \). To this end, we need to find the product of two power series. Suppose \( \sum_{n=0}^{\infty} c_nx^n \) and \( \sum_{n=0}^{\infty} d_n x^n \) are two power series. Their product is computed as follows: \[ \left( \sum_{n=0}^{\infty} c_nc^2 \right)\cdot \left( \sum_{n=0}^{\infty} d_nc^2 \right) = \sum_{n=0}^{\infty} \left( \sum_{i=0}^n c_i\cdot d_{n-i}\right) x^n \] Then, we have $$ \begin{eqnarray*} e^{z_1}\cdot e^{z_2} &=& \left( \sum_{n=0}^{\infty} \frac{z_1^n}{n!} \right) \cdot \left( \sum_{n=0}^{\infty} \frac{z_2^n}{n!} \right) = \sum_{n=0}^{\infty} \left( \sum_{i=0}^n \frac{z_1^i}{i!} \cdot \frac{z_2^{n-i}}{(n-i)!} \right) &=& \sum_{i=0}^{\infty}\left( \frac{1}{i!(n-i)!} z_1^i z_2^{n-i} \right) \\ &=& \sum_{i=0}^{\infty}\frac{1}{n!}\left( \frac{n!}{i!(n-i)!} z_1^i z_2^{n-i} \right) \\ &=& \sum_{n=0}^{\infty} \frac{(z_1+z_2)^n}{n!} = e^{z_1+z_2} \end{eqnarray*} $$ Note that \( n! \) is mulltiplied to each term inside the parentheses and hence my be cancelled by dividing \( n! \) outside of the parentheses. Additionally, we used the Binomial Theorem to obtain the final result. This is certainly independent of the Pythagorean Identity.

Finally, \( e^{iz}\cdot e^{-iz} = e^{(iz + (-iz))} = e^0 = 1 \) holds and we have Euler's formula. Consequently, the proof using Euler's formula is independent of the Pythagorean Identity.

In summary, before citing Euler's formula, we have to know that it is independent of the Pythagorean Identity. Otherwise, we have a circular reasoning.

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