ROOTS OF UNITY
When you take the square root of 1, you will get two answers: 1 and –1. But what happens when you take the cube root of 1? All these while, you studied that there’s only 1 cube root, which is 1 itself. But the actual fact is that it has 3 roots, –1/2 + √3/2 i, –1/2 - √3/2 i, and 1.
From here, you start guessing that probably the 4th root of 1 will yield 4 roots, 5th root of 1 will yield 5 roots and so on! You are correct. So in this section, I will teach you how to find the roots of unity. Here unity means ‘one’, not ‘bersatu padu’. You will see this term very often in higher level physics.
Let us try finding the cube root of 1. If you use 3√1, you won’t get anywhere. We need to use de Moivre’s theorem. So expressing 1 in polar form, we get |z| = 1, arg z = 2π (It is actually 0, but I recommend you using this instead) and we have 1 = cos 2π + i sin 2π. Let us continue from here:
Does it look familiar? Now you might want to try to use de Moivre’s theorem to solve it. But wait! de Moivre’s theorem is true for integers only, and that if n is not an integer, it is only one of the possible solutions. Which means, there are other solutions to be found. Let’s use de Moivre’s theorem first:
Now I will teach you how to find the remaining roots: you need to add 2nπ to the angle to get another root. You do this all the way until you reach the angle 2π (or, not exceeding 2π). n here denotes the exponent that you use, here it is 1/3, so you need to add 2π/3 to the angles. This means that we have
So we have found the 3 roots of the cube root of unity. By the way, it could have been faster to write if we used the exponential form, so 3√1 = e(2/3)πi, e(-2/3)πi and e2πi (which is actually 1). Did you notice that cos (4/3)π + i sin (4/3)π is actually cos (2/3)π – i sin (2/3)π, and it is actually the conjugate of cos (2/3)π + i sin (2/3)π? So it seems there is a faster way of finding the roots. Putting in mind that 1 is always a root of itself (and –1 is also a root ONLY if n is even, for example the square root, or the 4th root), you just need to add 2π/3 to the angles up to and not exceeding π, and you already know the rest of the roots, because every root’s conjugate is also a root!
You can try to solve the roots of unity for the 4th root, 5th root all the way up to the 9th root or beyond. Let me tell you some applications of this:
1. If you were to plot the roots of unity on the Argand diagram, they form a circle with |z| = 1, and they are equally spaced from each other! Here is the Argand diagram for the cube roots of unity:
This actually applies to the roots of any number, just that |z| might equal to something else.
2. Using this information for roots of unity, you can find all the real and complex roots of any other number. For example, 4√16 = 4√1 × 2 or 6√-64 = 6√1 × 2i . The terms 2 and 2i refer to the absolute root of 16 and –64 respectively. Using de Moivre’s theorem to find out the 4th root and 6th root of 1, you can just multiply the roots of unity with 2 and 2i respectively to get the required roots for the answer. If you remembered that the 4th roots of unity are 1, –1, i and –i respectively, you get 4√16 = 2, –2, 2i, –2i. Try solving the other one on your own. There are many questions of this kind, for example, solving for z as in z5 = –8 + 8i, or (z + 2i)2 = 4. This needs some practice.
3. Now you will probably face polynomial questions with complex coefficients, like
2ix2 +3x + 3 + 4i = 0. So be ready to know how to solve.
4. Try adding up all the roots of unity. You will be surprised that for every nth root of unity, all the roots add up to 0! Try answering the following question using this piece of information:
By considering the ninth roots of unity, show that
Hint: Find the all the ninth roots of unity, sum them up (equals to 0), and you’ll prove the equation. Also, Remember what you learned in high school Maths, the 2 complex roots in a quadratic expressions are conjugates of one another. This means that for whatever nth root of unity, other than 1 (or –1), the rest of the complex roots are paired up such that one root is sure to be the conjugate of another.
5. The square of one root is actually the conjugate of itself, which is just another root! So if w is a cube root of 1, then 1 + w + w2 might have 2 answers. If w is a real root, then it will be 3, but it is either of the complex roots, the answer will be 0. Try verifying this.
COMPLEX INTEGRATIONI hope you have learned the chapter on Integration in Maths T already. You solved the integral
by using integration by parts, and handing over from left to right and etc. But you could actually solve such an equation by complex integration. Let me show you how:
Do you think this is easier compared to the integration by parts? I thought so, just don’t be careless.
LOCI IN THE COMPLEX PLANE
Still remember learning loci in form 3 Mathematics? It’s read as ‘lo-sai’ by the way, not ‘lo-kee’ or ‘lo-chee’. Loci is actually just a representation of an equation in a graph or diagram. In a cartesian plane, you should be familiar with the equations for circles, straight lines, or ellipse (please study Coordinate Geometry in Maths T before starting this section). Here, the representation of complex loci in the Argand diagram is similar to the one in polar coordinates (which you do not learn in STPM anyhow). Anyway, I’ll start by introducing you 6 types of common loci. Throughout this section, z is a variable, while z1 or z2 will be complex numbers, and k, r, c or θ are just constants.
1. |z - z1| = r
This basically represents a circle, with the centre z1, while r is the radus of the circle. For example, take |z + 2 + 2i| = 2√2, we have
2. arg (z – z1) = θ
This is a half line, starting from the point z1, making an angle of θ with the positive real axis, going all the way to infinity. Taking arg (z + 1 + i) = π/4, we have
3. |z – z1| = |z – z2|
This is a perpendicular bisector, in between the two complex numbers z1 and z2. To find the distance between z1 and z2, just find |z1 – z2|, and to find the mid-point, use (z1 – z2)/2. Taking |z – 1 – i| = |z + 2 + 2i|, we have
4. |z – z1| = k |z – z2|
Adding a constant ‘k’ really changes this loci compared to the one before. This loci turned into a circle instead! It is the ratio between the distances between 2 complex numbers. Probably this is the hardest one, because you need to find out the centre of the circle before you can draw it out. Taking |z – 2| = 3 |z + 2i|, we can’t easily draw it out straight away. Using the fact that z = x + i y and using the definition of modulus, you can change the equation into
and you can further solve the equation into an equation of a circle, which is
and now we plot the loci onto the Argand diagram, which is
5. arg [(z – z1) / (z – z2)] = θ
This one is hard to explain in terms of Cartesian coordinates. It is actually a circle drew out by 2 lines connecting 2 complex number points at a fixed angle. Using the equation
The 2 points are (0, 2) and (0, 0) respectively. We draw 2 lines out of these 2 points, such that they intersect at some point which nicely makes the angle π/4. So this is shown below:
The loci is not the blue line. It is the arc of a circle traced out by the tip of the arrow shown below.
6. |z – z1| + |z – z2| = k
This one plots out an ellipse, where z1 and z2 are the foci of the ellipse, and k is the sum of the distance between z1 and z2. To plot this one, you also need to change the equation into cartesian coordinates as in the previous example. I leave this to you as an exercise. A general ellipse looks like this:
Now that you know all the possible types of loci, you need to practice plotting them when given a loci equation. Then, you should also know how to convert them into equations of cartesian coordinates. Besides, you should also know how to transform loci. I give you a few general rules:
1. w = z2 transforms a line into another line, and a circle into another circle.
2. w = (z1 + z) / (z2 – z) transforms a circle into a bisector.
3. w = 1/(z1 – z) transforms a circle into a line.
4. w = kz – c/z transforms a circle into an ellipse, where k and c are constants.
5. w = z̅ also transforms a circle into a circle.
I don’t think this is in the syllabus, but I will give you an example.
For the transformation w2 = z, find the locus w for |z| = 5.
z = 5 (cos θ + i sin θ)
w = √z = √5 [ cos (θ/2) + i sin (θ/2) ]
.·. w is a circle, with radius √5 and centre at 0.