In the pre-2017 syllabus, the only exposure students had to vectors was in the application of complex numbers to solve geometric problems.
In the 2017 syllabus, and continuing into the newly released 2024 syllabus, the introduction of Euclidean vectors and their operations such as the dot product means that there is now a second framework one can use to approach such geometric problems.
The advantage of a complex numbers approach lies in the fact that multiplying a complex number that represents a vector by \(e^{i\theta}\) (oh whoops this is now removed in the upcoming 2024 syllabus so I really mean \(\cos(\theta) + i \sin (\theta)\)) results in a rotation of the vector by \(\theta\). The equivalent notion of rotation with purely Euclidean vector methods requires university first year linear algebra with linear transformations and their matrix representations – well beyond the scope of the syllabus, of course.
So some tools, i.e. rotation, that are easy to model with a complex number approach does not have an in-syllabus equivalent when using a Euclidean vectors approach.
Conversely, however, using the dot product with Euclidean vectors does have an in-syllabus equivalent with complex numbers – albeit not mentioned explicitly in the syllabus. In this article we aim to make explicit this connection – after all, we should be making links and connections between topics in our study of Mathematics.
Recovering The Dot Product
The vector \(\left(\begin{matrix} a \\ b \end{matrix}\right)\) where \(a,b\in \mathbb{R}\) is in one-to-one correspondence with the complex number \(a + ib\), where \(i^2 = -1\).
Recall the vector dot product: \( \left(\begin{matrix} a \\ b \end{matrix}\right)\cdot\left(\begin{matrix} c \\ d \end{matrix}\right)=ac + bd\).
So we wish to have an operation with complex numbers \( z = a + ib \) and \(w = c + id \) that will also give \( ac + bd \), thus recovering the dot product for their corresponding vectors. Now observe that:
\[\begin{align*}\overline{(a + ib)}(c+id) &= (a-ib)(c+id)\\& = ac + adi – bci + bd.\end{align*}\]
Now taking the real part of this expression recovers \(ac + bd\), thus giving us back the dot product of the vectors represented by \(v\) and \(w\).
I like to use angle bracket notation to denote this:
\[ \langle z, w \rangle = \operatorname{Re}(\bar{z}w).\]
Now we can use this to solve vector geometry problems phrased in the language of complex numbers!
It’s not too difficult to see the following properties.
- If \(\operatorname{Re}(\bar{z}w) = 0\) then \(z\) and \(w\) are perpendicular (as vectors).
- \(\langle z,z \rangle = \operatorname{Re}(\bar{z}z) = \bar{z}z = |z|^2\).
- \(\langle z,w\rangle = \langle w,z\rangle\).
- \(\cos\theta = \frac{\langle z,w\rangle}{|z||w|}\), where \(\theta\) is the angle between the vectors represented by \(z\) and \(w\).
We’ve effectively defined an inner product structure on \(\mathbb{C}\), by transferring the one we already know, the dot product in \(\mathbb{R}^2\) over. More reading can be done here: https://en.wikipedia.org/wiki/Inner_product_space
Example Problem
When I was taking a substitute class, a student had brought along their tutoring homework and asked me how to do a problem that he was stuck on. The question was allegedly created by a staff member at the school but was copied into the tutoring work sheet.
To avoid theft of intellectual property, the following question is an adaptation of that problem to highlight the same technique used to solve it.
Let \(z_1, z_2, z_3, z_4\in \mathbb{C}\) represent points in the complex plane \(A,B,C,D\) respectively.
The points \(O,B,D\) are collinear, where \(O\) is the origin, and \(A,B,C\) are also collinear. If \(|AB|:|AC| = 1:4\) and \(|OB|:|OD| = 1:k\), find the value of \(k\) in terms of \(z_1\) and \(z_2\) and \(\theta\) where \(\theta = \angle AOB\) if \(OC\) is perpendicular to \(CD\).
Solution
The assumptions given can be summarised as follows:
- \(z_4 = kz_2\)
- \(z_3 – z_1 = 4(z_2 – z_1)\), so \(z_3 = 4z_2 – 3 z_1\).
- \( z_3 \perp (z_4 – z_3)\).
The third assumption can be rephrased in the language of the dot product that we have recovered above:
\[ \langle z_3, z_4 – z_3\rangle = 0.\]
Using the definition \(\langle z,w\rangle\):
\[ \operatorname{Re}(\bar{z_3}(z_4 – z_3) = 0.\]
Substituting the expressions for \(z_3\) and \(z_4\):
\[ \operatorname{Re}(\overline{(4z_2 – 3z_1)}((k – 4)z_2 + 3z_1)) = 0.\]
Expanding this will give:
\[\operatorname{Re}(4(k-4)\bar{z_2}z_2 + 12\bar{z_2}z_1 – 3(k-4)\bar{z_1}z_2 – 9\bar{z_1}z_1)=0.\]
Since taking the real part of a complex number is linear and \(\langle z,w\rangle = \langle w,z\rangle\):
\[4(k-4)|z_2|^2 + (24-3k) \langle z_1, z_2\rangle – 9 |z_1|^2 = 0.\]
Expanding:
\[4k|z_2|^2 – 16|z_2|^2 + 24\langle z_1, z_2\rangle – 3k\langle z_1, z_2\rangle – 9|z_1|^2 = 0\]
Rearranging for \(k\):
\[k = \frac{16|z_2|^2 – 24 \langle z_1, z_2 \rangle + 9|z_1|^2}{4|z_2|^2 – 3\langle z_1, z_2\rangle}.\]
Now since \(\langle z_1, z_2\rangle = |z_1||z_2|\cos\theta\):
\[k = \frac{16|z_2|^2 – 24|z_1||z_2|\cos\theta+ 9|z_1|^2}{4|z_2|^2 – 3|z_1||z_2|\cos\theta}.\]
Imaginary Part?
If you take the imaginary part of \(\bar{z}w\), we get \(ad – bc\). It was recently assessed in the 2023 Extension 1 HSC that the area of the triangle spanned by the vectors \(a\mathbf{i} + b\mathbf{j}\) and \(c\mathbf{i} + b\mathbf{j}\) is equal to \(\frac{1}{2}|ad – bc|\).
So it turns out that \(\operatorname{Im}(\bar{z}w)\) is equal to the (signed) area of the parallelogram spanned by the vectors represented by \(z\) and \(w\).
Now if this will be used in a question in the future? Who knows.