## FANDOM

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Euler's formula is a famous formula discovered by Leonhard Euler. It provides a link between complex algebra and trigonometry. It states that, for any angle θ,

$e^{\theta i} = \cos{\theta} + i \sin{\theta}$

Famously, this gives rise to Euler's identity:

$e^{\pi i} + 1 = 0 \therefore e^{\pi i} = -1$

Multiplying a point in the complex plane by Euler's formula results in rotation of that point by an angle θ. This is the rationale for the theta function.

## TheoremEdit

Euler's formula works because of its link to trigonometry and its application to the complex plane.

Consider a right triangle, as shown here:

The trigonometric ratios tell us:

$\cos{\theta} = \frac{adjacent}{hypotenuse} = \frac{a}{c}$
$\sin{\theta} = \frac{opposite}{hypotenuse} = \frac{b}{c}$

If you allow point A to lie at the origin of the complex plane (0, 0), then point B lies on the real axis and point C lies on the imaginary axis. Thus, point C becomes the sum of the real part and the imaginary part; i.e. C = a+bi.

Since the trig ratios allow us to define the angle θ, we can express this as cos θ + i sin θ. This implies that side a = cos θ and side b = i sin θ, which fits nicely with the parameters we've described here.

In 1740, Euler collaborated with other mathematicians at the time to obtain his formula eθi = cos θ + i sin θ, after John Bernoulli noted that

$\frac{1}{(1-x)^{2}} = \frac{1}{2} (\frac{1}{1-xi} + \frac{1}{1+xi}$and
$\int{\frac{dx}{1+ax}} = \frac{1}{a} \ln{(1+ax)} + C$

and Roger Cotes noted that

$xi = \ln{(\cos{x} + i \sin{x})}$

50 years later, Caspar Wessel described complex numbers as residing in the complex plane, allowing subsequent mathematicians to apply Euler's formula to the complex plane.

## Euler's Formula applied to θEdit

Below is a list of formulas for θ radians.

$e^{0} = 1$
$e^{\frac{\pi}{6}i} = (\frac{\sqrt{3}}{2} + \frac{1}{2}i)$
$e^{\frac{\pi}{4}i} = (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i)$
$e^{\frac{\pi}{3}i} = (\frac{1}{2} + \frac{\sqrt{3}}{2}i)$
$e^{\frac{\pi}{2}i} = i$
$e^{\frac{2 \pi}{3}i} = (-\frac{1}{2} + \frac{\sqrt{3}}{2}i)$
$e^{\frac{3 \pi}{4}i} = (-\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i)$
$e^{\frac{5 \pi}{6}i} = (-\frac{\sqrt{3}}{2} + \frac{1}{2}i)$
$e^{\pi i} = -1$
$e^{\frac{7 \pi}{6}i} = (-\frac{\sqrt{3}}{2} - \frac{1}{2}i)$
$e^{\frac{5 \pi}{4}i} = (-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}i)$
$e^{\frac{4 \pi}{3}i} = (-\frac{1}{2} - \frac{\sqrt{3}}{2}i)$
$e^{\frac{3 \pi}{2}i} = -i$
$e^{\frac{5 \pi}{3}i} = (\frac{1}{2} - \frac{\sqrt{3}}{2}i)$
$e^{\frac{7 \pi}{4}i} = (\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}i)$
$e^{\frac{11 \pi}{6}i} = (\frac{\sqrt{3}}{2} - \frac{1}{2}i)$
$e^{2 \pi i} = e^{0} = 1$
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