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Saturday, 14 May 2011 00:00

The Pythagorean Theorem from Euler’s Formula

By  Gideon
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Consider any complex number \[ z, \ \]with modulus

 

\[ \sqrt{z\bar{z} } = |z| = r \ ,\]

where \[ r \in \mathbb{R^{+}}\ . \]

Note that \[ \frac{z}{r} \ \]would be a complex number with modulus one:

\[ | \frac{z}{r} | = | \frac{z}{|z|} | = 1. \]

Therefore we can write \[ \frac{z}{r} \ \ \]

as: \[ \frac{z}{r} = e^{i\theta} \ ,\]

for some \[ \theta \ \in \mathbb{R}\ , \]

\[ 0 \leq \theta \leq 2\pi . \]

Now, \[ \frac{z}{r} = \cos{\theta} + i \sin{\theta} . \]

So that:

\[ z = r \cos{\theta}+ ir \sin{\theta} \]

 

\[ = r e^{i\theta} = e^{\ln{r} + i\theta} . \]

One immediate consequence of this is the Pythagorean Theorem:

Consider \[ z \ \]as a vector in the complex plane with \[|z| = r. \]

The real part of z is the projection of z onto the real axis:

\[ \text{Re}(z) = z\cos{\theta} = x . \]

The imaginary part of z is the projection of z onto the imaginary axis:

\[ \text{Im}(z) = z\sin{\theta} = y . \]

Now x, y, and z, make a right triangle, where z is the hypotenuse, and z satisfies:

\[ z\bar{z} = r^{2} = re^{i\theta}re^{-i\theta} \]

\[ = ( r\cos{\theta} + ir \sin{\theta})( r\cos{\theta} + ir \sin{\theta}) \]

\[ = (x + iy)(x – iy) = x^{2} + y^{2}. \]

Or:

\[ x^{2} + y^{2} = r^{2} . \]

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