Complex
Complex numbers¶
\(f\) is the definition of \(sin(\theta)\)
\(g\) is the definition of \(cos(\theta)\)
\(\theta\) is the angle from the right of the \(x\) axis to the line \(h\)
now, I will define \(f(x)\) as a function that takes in a real number and outputs a complex number that is \(x\) radians around the unit circle, so like the geogebra file above, but in the complex plane. using the file, we can say that \(f(x) = cos(x) + isin(x)\), and any complex number \(z\) has an angle \(\theta\) measured in radians and a distance \(r\) from the origin, so \(z\) equals \(rf(\theta)\), eccept pepole don't use \(f(\theta)\) because there are other complex functions \(f(z)\), and \(rcos(x) + irsin(x)\) dosen't reach the high bar for perfection set by mathematicians
so now, the question is: solve for \(f(x)\)
i'll start by multaplying \(f(x) \cdot f(y)\)
and using angle addition derived in the trigonomitry section...
hmmm \(f(x)f(y) = f(x + y)\) sounds faniliar... Oh right! This is an exponential, but what's the base?
while, a common proof that I once used that if the derivitave of \(g(x)\) is \(g(x)\), then \(g(x) = ce^x\), it has a method of:
\(\frac{g(x + dx) - g(x)}{dx} = g(x)\), \(g(x + dx) = g(x)(1 + dx)\), \(g(x + 2dx) = g(x)(1 + dx)^2\), \(g(x + ndx) = g(x)(1 + dx)^n\), \(g(0 + \frac{x}{dx} dx) = g(x) = g(0)(1 + dx)^{\frac{x}{dx}}\)
and then, using facts from Calculus part 2, \(g(x) = g(0)e^x = ce^x\).
So I will use a proof like that, the proof will go like this:
proof¶
\(cos(x)\) reaches a peak at \(0\), so \(cos'(0) = 0\)
while as you zoom into the right of the unit circle, the hight is the distance travled and the sine of a small angle is that angle, so \(sin'(0) = 1\)
and if you saw Calculus part 2, you know that...
so suprisingly, the base of the exponential is \(e^i\), eccept many of you would of guessed that because of eueler's identity, which I was trying to derive, anyways there is one more step (more like three but you get the point)
thus:
proof complete!
now what?¶
well everything! Now complex multiplication is mutch easier, for example \(z \cdot w = r e^{i \theta} \rho e^{i \psi} = r \rho e^{\theta + \psi}\)
and if multiplying by a real number is scaling the distance from the origin, then complex multiplication has the mutch more intuitave interpritation that most pepole teach: multiplying the magnitudes and adding the angles.
another interpritation of complex multiplication is that if \(w(z) = wz\) then \(w(1) = w\) and scaling the input scales the output and then rotating the input rotates the output, in other words complex multiplication looks like scaling and rotating
Eccept I think that just saying things like "multiplying by a number on the unit circle rotates a point counterclockwise" which by the way is true, needs rigoris proof. And I wanted to add more cases, but this page only needs one more fact before saying "now everything" so that the right of the page makes sense, anywas the fact is that the only reason why mathematicians use \(e^{i \theta}\) is either to show that \(e^{i \theta} e^{i \psi} = e^{i(\theta + \psi)}\), to show that \(\frac{d}{dx} e^{ix} = ie^{ix}\) which makes sense if you think about it, or to confuse complex number beginners possibly making them not want to learn about complex numbers in the first place
now everything¶
quaternions!
jk
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ok, if you want quaternions, than...
you have to find it!
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