Going Round in Circles

Or : Multiplying by \(i\).

The beginning of complex numbers is often attributed to Cardano (1501 – 1576).  Complex numbers provide a solution to the equation \(x^2 = -1\).

They are an extension of the number system from \(\Bbb R\) (the real numbers) to \(\Bbb C\) (the complex numbers).

Complex numbers take the form :

\(z = a + bi\),

where \(a,b\) are real numbers and \( i = \sqrt -1\).

The normal rules of arithmetic and algebra apply with the additional rule that :

\(i^2 = -1\)

When two complex numbers are multiplied together the real numbers, \(a\) and \(b\), act as scalars and the imaginary number, \(i\), acts as a rotation about the origin, \(0\), (counter-clockwise from the positive real axis). Multiplication by \(i\) on its own leads to a rotation of \(\pi/2\) radians in the complex plane.

\(\begin{align}1\cdot i & =  i & = i\\
i\cdot i & = i^2 & = -1\\
-1\cdot i & = i^3 & = -i\\
-i\cdot i & =  -i^2 & = 1 \end{align}\)

OldTrout, Going Postal

Starting at any point on the real axis, (Re), and repeatedly mutilplying by \(i\) traces larger or smaller circles.

The multiplication of two complex numbers gives another complex number.

Let :

\(z = a + bi\) and \(w = c + di\)

Then :

\(\begin{align} zw & = (a + bi)(c + di)\\
& = ac + bdi^2 + adi + bci\\
& = ac-bd + (ad + bc)i \end{align}\)

Taking our starting point as one again, this time we multiply by \(1 + i\) repeatedly.

\(\begin{align}1(1 + i) & = 1 + i\\
(1 + i)(1 + i) & = 2i\\
2i(1 + i) & = -2 + 2i\\
(-2 + 2i)(1 + i) & = -4\\
-4(1 + i) & = -4-4i\\
(-4 – 4i)(1 + i) & = -8i\\
(-8i)(1 + i) & = 8-8i\\
(8 – 8i)(1 + i) & = 16 \end{align}\)

OldTrout, Going Postal

We still have a rotational locus but now in the form of a spiral.

Each multiplication by \(1 + i\) leads to a rotation by \(\pi/4\) and an increase in distance from the origin by \(\sqrt 2\).

This is because the point \(1 + i\) is at a distance \(\sqrt 2\) from the origin, \(0\), (by Pythagorus) and the angle that this point makes to the positive real axis is arctan \((1/1) = \pi/4\), where arctan is the inverse of the tangent function.

We look at an example.

Let :

\(z = 3 + 4i\) and \(w = 5 + 2i\)

Then :

\(zw = 7 + 26i\),

by the above formula.

We can break it down to see what is happening.

\((3 + 4i)5 = 15 + 20i\)

OldTrout, Going Postal

The point has been scaled up by a factor of \(5\) and the angle is left unchanged.

\((3 + 4i)2i = -8 + 6i\)

OldTrout, Going Postal

The point has been scaled up by a factor of \(2\) and rotated by an angle of \(\pi/2\).

The addition of these two parts gives us the point \(7 + 26i\)

OldTrout, Going Postal

In terms of distances and angles (the polar representation) :

\(z = (r_1, \theta), w = (r_2, \phi)\) and \(zw = (r_1r_2, \theta + \phi)\).

\(z = 3 + 4i = (5, \theta)\), where \(\theta = \)arctan\((4/3) \approx 0.9273\) radians \(\approx 53.13^\circ\)

\(w = 5 + 2i = (\sqrt29, \phi)\), where \(\phi = \)arctan\((2/5) \approx 0.3805\) radians \(\approx 21.80^\circ\)

\(zw = 7 +26i = (5\sqrt29, \theta + \phi)\), where\(\theta + \phi =\)arctan\((26/7) \approx 1.3078\) radians \(\approx 74.93^\circ\)

 

Next time we will rotate in three different directions using the quaternions.
 

© OldTrout \(2019\)
 

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