Your History

Menu

Eulers Formula

Prerequisites

Product Rule | \((u \cdot v)' = u' \cdot v + v' \cdot u\)
Derivative of exp(x) | \((e^{x})' = e^{x}\)
Derivative of Sine | \((\sin(x))' = \cos(x)\)
Derivative of Cosine | \((\cos(x))' = -\sin(x)\)

Description

This is Euler's Formula, which shows an equivalent form of a complex number, to it's polar form: Polar Form of a Complex Number.

\[\htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000129}{j} \cdot \htmlClass{sdt-0000000024}{\theta}} = \htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000024}{\theta}) + \htmlClass{sdt-0000000129}{j} \htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000024}{\theta})\]

Symbols Used:

This is a commonly used symbol to represent an angle in mathematics and physics.
\( e \)

This symbol represents Euler's constant. It is approximately \(2.718\).
\( \cos \)

This is the symbol for cosine, a trigonometric function that calculates the ratio of the adjacent side to the hypotenuse of a right-angled triangle.
\( \sin \)

This is the symbol for sine, is a trigonometric function that represents the ratio of the opposite side to the hypotenuse in a right-angled triangle.
\( j \)

This symbol represents the imaginary unit, which is defined as the square root of \(-1\). \( j = \sqrt{-1}\). It is the most fundamental unit in the field of complex numbers, allowing for the expression of numbers that cannot be represented on the real number line.

Derivation

  1. Consider a function: \[\htmlClass{sdt-0000000096}{f}(\htmlClass{sdt-0000000024}{\theta}) = \frac{\htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000024}{\theta}) + \htmlClass{sdt-0000000129}{j} \cdot \htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000024}{\theta})}{\htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000129}{j} \cdot \htmlClass{sdt-0000000024}{\theta}}} \equiv \htmlClass{sdt-0000000035}{e}^{-\htmlClass{sdt-0000000129}{j} \cdot \htmlClass{sdt-0000000024}{\theta}} (\htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000024}{\theta}) + \htmlClass{sdt-0000000129}{j} \cdot \htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000024}{\theta}))\]
  2. From here, we can take the derivative of both sides using the Product Rule, which is defined as:
    \[(\htmlClass{sdt-0000000008}{u} \cdot \htmlClass{sdt-0000000019}{v})\htmlClass{sdt-0000000025}{'} = \htmlClass{sdt-0000000008}{u}\htmlClass{sdt-0000000025}{'} \cdot \htmlClass{sdt-0000000019}{v} + \htmlClass{sdt-0000000019}{v}\htmlClass{sdt-0000000025}{'} \cdot \htmlClass{sdt-0000000008}{u}\]
    to get...
    \[ \htmlClass{sdt-0000000096}{f}\htmlClass{sdt-0000000025}{'}(\htmlClass{sdt-0000000024}{\theta}) = (\htmlClass{sdt-0000000035}{e}^{-\htmlClass{sdt-0000000129}{j} \cdot \htmlClass{sdt-0000000024}{\theta}})\htmlClass{sdt-0000000025}{'} \cdot (\htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000024}{\theta}) + \htmlClass{sdt-0000000129}{j} \cdot \htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000024}{\theta})) + ( \htmlClass{sdt-0000000035}{e}^{-\htmlClass{sdt-0000000129}{j} \cdot \htmlClass{sdt-0000000024}{\theta}})(\htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000024}{\theta}) + \htmlClass{sdt-0000000129}{j} \cdot \htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000024}{\theta}))\htmlClass{sdt-0000000025}{'}\]
  3. From here we can use the derivative of \(\htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000003}{x}}\), \(\htmlClass{sdt-0000000127}{\sin}\) and \(\htmlClass{sdt-0000000124}{\cos}\):

    \[(\htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000003}{x}})\htmlClass{sdt-0000000025}{'} = \htmlClass{sdt-0000000035}{e}^{x}\]
    \[(\htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000003}{x}))\htmlClass{sdt-0000000025}{'} = \htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000003}{x})\]
    \[(\htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000003}{x}))\htmlClass{sdt-0000000025}{'} = -\htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000003}{x})\]

  4. to get...
    \[\htmlClass{sdt-0000000096}{f}\htmlClass{sdt-0000000025}{'}(\htmlClass{sdt-0000000024}{\theta}) = \htmlClass{sdt-0000000035}{e}^{-\htmlClass{sdt-0000000129}{j} \cdot \htmlClass{sdt-0000000024}{\theta}}(\htmlClass{sdt-0000000129}{j} \htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000024}{\theta}) - \htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000024}{\theta})) - \htmlClass{sdt-0000000129}{j} \cdot \htmlClass{sdt-0000000035}{e}^{-\htmlClass{sdt-0000000129}{j} \cdot \htmlClass{sdt-0000000024}{\theta}} \cdot (\htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000024}{\theta}) + \htmlClass{sdt-0000000129}{j} \htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000024}{\theta})) = 0\]
  5. From here it it follows that \(\htmlClass{sdt-0000000096}{f}(\htmlClass{sdt-0000000024}{\theta})\) is a constant, as the derivative is \(0\).
  6. If we plug in \( (\htmlClass{sdt-0000000024}{\theta} = 0)\) into our original function, we get the value of 1. Because \(\htmlClass{sdt-0000000096}{f}(\htmlClass{sdt-0000000024}{\theta})\) is constant (based on point (5)), we can say that...
    \[\htmlClass{sdt-0000000096}{f}(\htmlClass{sdt-0000000024}{\theta}) = \frac{\htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000024}{\theta}) + \htmlClass{sdt-0000000129}{j} \cdot \htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000024}{\theta})}{\htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000129}{j} \cdot \htmlClass{sdt-0000000024}{\theta}}} = 1\]
  7. Finally, we can multiply both sides by \(\htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000129}{j} \cdot \htmlClass{sdt-0000000024}{\theta}}\) to get...
    \[\htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000129}{j} \cdot \htmlClass{sdt-0000000024}{\theta}} = \htmlClass{sdt-0000000124}{\cos}(\htmlClass{sdt-0000000024}{\theta}) + \htmlClass{sdt-0000000129}{j} \htmlClass{sdt-0000000127}{\sin}(\htmlClass{sdt-0000000024}{\theta})\]
    as required.