Euler's formula

Complex numbers: The complex plane

Euler's formula

We can define the following function of real numbers to complex numbers on the unit circle: $f(\varphi)=\cos\varphi + \mathrm{i}\sin\varphi$ We now pretend that we can differentiate this function in the usual way: $\begin{aligned}f'(\varphi)&=\lim_{h\to 0}\frac{f(\varphi+h)-f(\varphi)}{h}\\ \\ &= \lim_{h\to 0}\frac{\cos(\varphi+h)-\cos(\varphi)}{h}+ \mathrm{i}\cdot \lim_{h\to 0}\frac{\sin(\varphi+h)-\sin(\varphi)}{h}\\ \\ &= \cos'(\varphi)+ \mathrm{i}\sin'(\varphi)\\ \\ & -\sin\varphi+\mathrm{i}\cos\varphi\\ \\ &= \mathrm{i}\cdot(\cos\varphi + \mathrm{i}\sin\varphi) \\ \\ &=\mathrm{i}\cdot f(\varphi)\end{aligned}$ We see that the derivative of the function $f$ is equal to a constant times the function self. For real functions this leads to the definition of the exponential function. We screw up our courage and define the imaginary power of $e$ $e^{\,\mathrm{i}\,\varphi}$ as $\cos\varphi + \mathrm{i}\sin\varphi$ . This is the famous formula of Leonhard Euler (1707-1783), who introduced the power of $e$ in a different way and proved that this fits well into the general mathematical structure of complex functions.

$e^{\,\mathrm{i}\,\varphi}=\cos\varphi + \mathrm{i}\sin\varphi$ In particular: $e^{\pi\,\mathrm{i}}=-1$

Write $e^{\,-\tfrac{3}{4}\!\pi\,\mathrm{i}}$ in standard form.

$\begin{aligned}e^{\,-\tfrac{3}{4}\!\pi\,\mathrm{i}}&=\cos(-\tfrac{3}{4}\!\pi)+\mathrm{i}\,\sin(-\tfrac{3}{4}\!\pi) & \color{blue}{\text{Euler's formula}}\\ &= -\tfrac{1}{2}\!\sqrt{2}-\tfrac{1}{2}\!\sqrt{2}\,\mathrm{i}& \color{blue}{\text{simplification}}\end{aligned}$

New example

In the new format of imaginary power of $e$ , the multiplication and division of complex numbers on the unit circle look a lot neater because apparently the usual calculation rules of powers of $e$ may be used.

$\begin{aligned} e^{\,\mathrm{i}\,\varphi_1}\cdot e^{\,\mathrm{i}\,\varphi_2}&=e^{\,\mathrm{i}\,(\varphi_1+\varphi_2)}\\ \\ \frac{e^{\,\mathrm{i}\,\varphi_1}}{e^{\,\mathrm{i}\,\varphi_2}}&=\e^{\,\mathrm{i}\,(\varphi_1-\varphi_2)}\end{aligned}$

Write $e^{\,\tfrac{1}{2}\!\pi\,\mathrm{i}}\cdot e^{\,-\tfrac{1}{6}\!\pi\,\mathrm{i}}$ in standard form.

$\begin{aligned}e^{\,\tfrac{1}{2}\!\pi\,\mathrm{i}}\cdot e^{\,-\tfrac{1}{6}\!\pi\,\mathrm{i}} &=e^{\,(\tfrac{1}{2}\!\pi\,-\tfrac{1}{6}\!\pi)\,\mathrm{i}} & \color{blue}{\text{calculation rules for imaginary }e\text{-powers}}\\ &= e^{\,{{\pi}\over{3}}\,\mathrm{i}} & \color{blue}{\text{simplification}}\\ &= \cos({{\pi}\over{3}})+\mathrm{i}\sin({{\pi}\over{3}})& \color{blue}{\text{Euler's formula}}\\ &= {{1}\over{2}}+{{\sqrt{3}\, \ii}\over{2}}& \color{blue}{\text{simplification}}\end{aligned}$

New example

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The trigonometric functions can also be written as expressions in imaginary powers of $e$ ; these are two other famous formulas of Euler.

$\cos\varphi = \frac{e^{\,\mathrm{i}\,\varphi}+e^{-\mathrm{i}\,\varphi}}{2}\qquad\text{and}\qquad \sin\varphi = \frac{e^{\,\mathrm{i}\,\varphi}-e^{-\mathrm{i}\,\varphi}}{2\,\mathrm{i}}$