Introductory university-level calculus, linear algebra, abstract algebra, probability, statistics, and stochastic processes.
Geometric interpretations of complex arithmetic
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The addition and multiplication of complex numbers have interesting geometric interpretations in terms of translation, rotation, and scaling on the complex plane.
Addition is translation
Let \(z_1 = a_1 + ib_1\) and \(z_2 = a_2 + ib_2\) where \(a_i, b_i \in \mathbb{R}\) for \(i = 1, 2\). Then, \(z_1 + z_2 = (a_1 + a_2) + i(b_1+b_2)\). Therefore, on the complex plane, adding \(z_1\) to \(z_2\) to obtain \(z_1 + z_2\) corresponds to translating the point \((a_1, b_1)\) by the vector \((a_2, b_2)\) to obtain the point \((a_1+a_2, b_1+b_2)\). Thus, the four points, \(0, z_1, z_1+z_2\), and \(z_2\) comprise a parallelogram (Figure 2).
Figure 2. Adding two complex numbers.
Multiplication by a real number is scaling
Let \(c \in \mathbb{R}\) and \(z = a + ib\in \mathbb{C}\) with \(a, b\in\mathbb{R}\). Then, \(cz = ca + i(cb)\), which corresponds to the point \((ca, cb)\) on the complex plane. Meanwhile, we have \(|cz| = |c|\cdot|z|\), so the modulus is scaled by \(|c|\). If \(c > 0\), \(cz\) and \(z\) are in the same direction from the origin; if \(c < 0\), then they are in the opposite directions from the origin (Figure 3).
Figure 3. Multiplying by a real number is scaling.
Multiplication by a complex number \(u\) with \(|u|=1\) is rotation around the origin by \(\arg u\).
Figure 4. Multiplication is rotation.
Theorem
Let \(z, u\in\mathbb{C}\) with \(|u| = 1\). If we multiply \(z\) by \(u\), then \(z\) is rotated around the origin by an angle of \(\text{Arg}(u)\) on the complex plane (Figure 4).
Proof.
(Step 1) Note \(1 \in \mathbb{C}\) and \(|1| = 1\). If we multiply 1 by \(u\), we have \(u\cdot 1 = u\) so \(\text{Arg}(u\cdot 1) = \text{Arg}(u)\). Thus, 1 is rotated around the origin by \(\text{Arg}(u)\).
(Step 2) Now consider \(i \in \mathbb{C}\). \(|i| = 1\). Suppose \(z = a + ib\). If we multiply \(z\) by \(i\), we have \(z\cdot i = (a + ib)i = -b + ia\). Note that \(|z|^2 = a^2 + b^2\) and \(|z\cdot i|^2 = |z|^2|i|^2 = a^2 + b^2\). On the complex plane, the squared distance between \(z\) and \(zi\) is given by \(|z - zi|^2 = |(a + b) + i(b - a)|^2 = (a + b)^2 + (b-a)^2 = 2a^2 + 2b^2\). Therefore we have
By the Pythagorean theorem, the angle made by \(z\), 0 (the origin), and \(zi\) is the right angle, \(\pi/2\). Furthermore, this rotation is counter-clockwise. Hence multiplying by \(i\) rotates \(z\) around the origin by \(\pi/2\). However, \(\text{Arg}(i) = \pi/2\). In other words, \(\text{Arg}(zi) = \text{Arg}(z) + \text{Arg}(i)\).
(Step 3) Let \(z,w\in\mathbb{C}\). \(|u(z-w)| = |u|\cdot|z-w| = 1\cdot |z-w| = |z-w|\). Thus multiplying by \(u\) preserves distance.
(Step 4) Let \(z\in \mathbb{C}\). Consider the triple of distances \((|z - 1|, |z-0|, |z-i|)\). These are the three distances of \(z\) from \(1, 0\) and \(i\), respectively.
It is noted that this \(z\) is the only point on the complex plane which has this particular triple of distances from \(1, 0\) and \(i\). Now, \((|uz - u|, |uz - 0|, |uz - ui|) = (|z - 1|, |z-0|, |z-i|)\) since multiplying by \(u\) preserves distances. Note that the triangle with vertices \(u, 0, ui\) is obtained by rotating the triangle with \(1, 0, i\) by an angle \(\text{Arg}(u)\). If a point has a distance from \(u, 0, ui\) equal to \(|z-1|, |z|, |z-i|\), then that point must also be obtained by rotating \(z\) by an angle \(\text{Arg}(u)\).
Thus, we conclude that multiplication by \(u\) has the effect of rotating points in the complex plane by \(\text{Arg}(u)\). ■
Next, consider multiplying \(z\) by an arbitrary non-zero complex number \(\alpha\). We can decompose \(\alpha\) as
If we define \(u = \frac{\alpha}{|\alpha|}\), then \(|u| = 1\). Of course, \(|\alpha|\) is a positive real number. Thus, multiplication by \(\alpha\) can be decomposed into two steps: (1) rotating by \(\arg u\), then (2) scaling by \(|\alpha|\):
\[z \stackrel{\text{rotate}}{\longmapsto} uz \stackrel{\text{scale}}{\longmapsto} |\alpha|(uz) = \alpha z.\]
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