Euler’s formula is a fundamental equation in complex analysis that establishes a deep relationship between trigonometric functions and the exponential function. It is expressed as:
\[ e^{i\theta} = \cos(\theta) + i\sin(\theta) \]
where:
– \( e \) is the base of the natural logarithm,
– \( i \) is the imaginary unit, with \( i^2 = -1 \),
– \( \theta \) is a real number, representing an angle in radians.
This formula is particularly significant because it connects the exponential function with the sine and cosine functions, providing a powerful tool for analyzing and understanding complex numbers and their properties. Euler’s formula is also the basis for many other important results in mathematics, including Euler’s identity, which is a special case of Euler’s formula when \( \theta = \pi \):
\[ e^{i\pi} + 1 = 0 \]
This identity elegantly links five of the most important constants in mathematics: \( e \), \( i \), \( \pi \), 1, and
Euler’s formula can be proven in several ways, including using power series expansions, differential equations, or integration. One of the most straightforward proofs uses the power series expansions for the exponential, sine, and cosine functions:
Power Series Expansions
The exponential function \( e^z \) for a complex number \( z \) can be expanded as a power series:
\[ e^z = \sum_{n=0}^{\infty} \frac{z^n}{n!} \]
For \( z = i\theta \), we have:
\[ e^{i\theta} = \sum_{n=0}^{\infty} \frac{(i\theta)^n}{n!} \]
Separate into Real and Imaginary Parts
Now, we separate the series into real and imaginary parts. Note that \( i^2 = -1 \), \( i^3 = -i \), \( i^4 = 1 \), and so on. This periodic behavior of powers of \( i \) allows us to group terms involving \( i \) in a systematic way:
\[ e^{i\theta} = \sum_{n=0}^{\infty} \frac{(i\theta)^n}{n!} = \sum_{n=0}^{\infty} \frac{i^n \theta^n}{n!} \]
Grouping Terms
We can group terms by separating the even and odd powers of \( n \):
\[ e^{i\theta} = \sum_{k=0}^{\infty} \frac{(i\theta)^{2k}}{(2k)!} + \sum_{k=0}^{\infty} \frac{(i\theta)^{2k+1}}{(2k+1)!} \]
Even Powers (Real Part)
For the even powers, \( (i\theta)^{2k} = (i^2)^k \theta^{2k} = (-1)^k \theta^{2k} \):
\[ \sum_{k=0}^{\infty} \frac{(i\theta)^{2k}}{(2k)!} = \sum_{k=0}^{\infty} \frac{(-1)^k \theta^{2k}}{(2k)!} = \cos(\theta) \]
Odd Powers (Imaginary Part)
For the odd powers, \( (i\theta)^{2k+1} = i (i^2)^k \theta^{2k+1} = i (-1)^k \theta^{2k+1} \):
\[ \sum_{k=0}^{\infty} \frac{(i\theta)^{2k+1}}{(2k+1)!} = i \sum_{k=0}^{\infty} \frac{(-1)^k \theta^{2k+1}}{(2k+1)!} = i \sin(\theta) \]
Combining Both Parts
Now, combining the real and imaginary parts, we get:
\[ e^{i\theta} = \cos(\theta) + i\sin(\theta) \]
This completes the proof of Euler’s formula using the power series expansions.
Proof of Euler’s Formula Using the Laplace Transform
Euler’s formula states that for any real number \(\theta\),
\[ e^{i\theta} = \cos(\theta) + i\sin(\theta). \]
This proof utilizes the Laplace transform to establish the formula.
The Laplace Transform
The Laplace transform of a function \( f(t) \) is defined as:
\[ \mathcal{L}\{f(t)\} = \int_0^\infty e^{-st} f(t) \, dt, \]
where \( s \) is a complex number.
Laplace Transform of \( e^{i\theta t} \)
Consider the function \( f(t) = e^{i\theta t} \). The Laplace transform of \( f(t) \) is:
\[ \mathcal{L}\{e^{i\theta t}\} = \int_0^\infty e^{-st} e^{i\theta t} \, dt. \]
Combining the exponents, we get:
\[ \mathcal{L}\{e^{i\theta t}\} = \int_0^\infty e^{-(s – i\theta)t} \, dt. \]
Evaluate the Integral
To evaluate this integral, note that it converges when the real part of \( s \) is greater than zero:
\[ \int_0^\infty e^{-(s – i\theta)t} \, dt = \left[ \frac{e^{-(s – i\theta)t}}{-(s – i\theta)} \right]_0^\infty. \]
Evaluating the limits, we find:
\[ \left[ \frac{e^{-(s – i\theta)t}}{-(s – i\theta)} \right]_0^\infty = \left. \frac{1}{s – i\theta} \right|_0^\infty = \frac{1}{s – i\theta}. \]
Thus,
\[ \mathcal{L}\{e^{i\theta t}\} = \frac{1}{s – i\theta}. \]
Separate into Real and Imaginary Parts
Next, we express \( \frac{1}{s – i\theta} \) in terms of its real and imaginary components. Multiply the numerator and denominator by the complex conjugate of the denominator:
\[ \frac{1}{s – i\theta} = \frac{1}{s – i\theta} \cdot \frac{s + i\theta}{s + i\theta} = \frac{s + i\theta}{s^2 + \theta^2}. \]
This can be written as:
\[ \frac{1}{s – i\theta} = \frac{s}{s^2 + \theta^2} + i \frac{\theta}{s^2 + \theta^2}. \]
Recognize the Laplace Transforms
We recognize that these fractions correspond to the Laplace transforms of \( \cos(\theta t) \) and \( \sin(\theta t) \):
\[ \mathcal{L}\{\cos(\theta t)\} = \frac{s}{s^2 + \theta^2}, \]
\[ \mathcal{L}\{\sin(\theta t)\} = \frac{\theta}{s^2 + \theta^2}. \]
Thus,
\[ \mathcal{L}\{e^{i\theta t}\} = \mathcal{L}\{\cos(\theta t)\} + i \mathcal{L}\{\sin(\theta t)\}. \]
Inverse Laplace Transform
Taking the inverse Laplace transform of both sides, we get:
\[ e^{i\theta t} = \cos(\theta t) + i\sin(\theta t). \]
Conclusion
By setting \( t = 1 \), we obtain Euler’s formula:
\[ e^{i\theta} = \cos(\theta) + i\sin(\theta). \]
This completes the proof using the Laplace transform.