# MandysNotes

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:

$x^{2} + y^{2} = r^{2} .$