User Rating:     / 0
PoorBest
Written by Bryce Ringwood

Please review the section on complex numbers, or you might get a bit lost.  Here's Eulers equation:

\color \orange {e^{ix} = cos(x) + i sin(x)}

Its easy to prove, we simply assume that $e^{ix}$ is a complex number, and we express it in polar coordinates:

e^{ix}={r(cos(\theta) + i sin(\theta))}

Now we differentiate both sides with respect to x:

\displaystyle {ie^{ix}=(cos(\theta ) + i sin(\theta )){dr\over dx} +r (-sin(\theta ) + i cos(\theta)){d{\theta }\over dx}}
so:

\displaystyle {ir({(cos(\theta) + i sin(\theta))} = (cos(\theta)+ i sin(\theta )){dr\over dx} +r (-sin(\theta ) + i cos(\theta)){d{\theta }\over dx}}

(We use  the product rule to arrive at the above result - see previous article)

If we equate real and imaginary parts:

\displaystyle{ircos(\theta)=isin(\theta){dr\over dx }+ircos(\theta){d\theta \over dx}}

\displaystyle{-rsin(\theta )= cos(\theta ){dr\over dx}-rsin(\theta){d\theta \over dx}}
These equations are satisfied if:

\displaystyle{{dr\over dx}=0}
and
\displaystyle {{{d\theta}\over dx} = 1}

This tells us the radius is a constant and that $\theta = x$ - which we couldn't just assume. When $x=0$, then $cos(0)=1$, so $r=1$, since $e^{0i}=1$.

This shows the theorem to be true.

The important thing is that we can express a complex number as $re^{i\theta}$, which we need to do for impedance calculations.

Aknowlegements to Wikpedia for bringing this method to my attention. I was going to do it another way, but I prefer this.

##### Fallout

Here are some interesting relationships that result from fiddling about with the above equation. They are all quite easy to derive:

$\displaystyle{{e^{i\pi }}= -1}$

$\displaystyle{sin(\theta ) ={{e^{i\theta} - e^{-i\theta }}\over 2i}}$

$\displaystyle{cos(\theta) ={{e^{i\theta } + e^{-i\theta }}\over 2}}$

Joomla template by a4joomla