10. Complex numbers#
10.1. Introduction#
Consider the equation
where \(a\neq0\). Previously you probably learned that Equation (10.1.1) only has solutions when \(D=b^2-4ac\) was non-negative and in that case you would have the two solutions
Now consider the case that you really need solutions of Equation (10.1.1) even if \(D\) is negative. This might sound strange, but you will encounter this case in many engineering applications, for example in solving second-order linear differential equations. It turns out that the solution in Equation (10.1.2) is still valid when \(D<0\), which means that you need to take the square root of a negative number.
Luckily, mathematicians have found a way to handle this, namely complex numbers. Before we start treating complex numbers, we first need to introduce the special number \(i\). After that, we will introduce the complex numbers and some other terminologies.
10.2. Complex numbers#
Definitions
We start with considering an easier form of Equation (10.1.1), namely
From this equation we find that \(D=-1\) and if we just straight forward apply Equation (10.1.2), we get
The square root of \(-1\) obviously is a problem. Therefore we introduce the special number \(i\):
Definition 10.2.1
The imaginary unit \(i\) is a number defined by the equation
Because \(i\) is defined to be a number (mind you, it is not a real number), we also assume \(i\) behaves just like any normal number.
Using the imaginary unit \(i\) we can prove the following theorem:
Theorem 10.2.1
Let \(a\) be a positive real number. Then the two numbers \(x_-=-ai\) and \(x_+=ai\) are solutions to the equation \(x^2=-a^2\).
Proof of Theorem 10.2.1
First we consider \(x_-=-ai\) and take its square:
This shows that \(x_-=-ai\) is indeed a solution to the equation \(x^2=-a^2\).
We repeat the same for \(x_+=ai\):
We also find that \(x_+=ai\) is a solution to the equation \(x^2=-a^2\).
This means we can rewrite our two solutions as
Doesn’t this already look simpler?
Now we have defined \(i\), we can revisit the general case given in Equation (10.1.1). Let us start with an example:
Example 10.2.1
Consider the second degree polynomial \(p(z) =z^2+2z+5\) and we want to solve the equation \(p(z)=0\).
We are going to do this by first rewriting \(p\) to the form \(p(z) = (z+p)^2+q\) for some numbers \(p\) and \(q\).
Expanding gives that we want \(z^2+2z+5=z^2+2pz + (p^2+q)\), thus \(2=2p\) and \(5=p^2+q\). The first equation gives us \(p=1\). Plugging this into the second equation, we obtain \(5=1+q\), so \(q=4\). Therefore, \(z^2+2z+5=(z+1)^2+4\).
To solve \(z^2+2z+5=0\), we can now write
Note that we used Theorem 10.2.1.
Going from \(z^2+2z+5\) to \((z+1)^2+4\) is called completing the square. You can also immediately see that the minimal value of the parabola \(y=z^2+2z+5\) for real values of \(z\) equals 4 (as \((z+1)^2\geq 0\) for all real \(z\)), and the minimum is obtained when \(z=-1\).
In general, you can write any polynomial \(az^2+bz+c\) in the form \(a ((z+p)^2+q)\) by first factoring out the \(a\), subsequently choosing the \(p\) such that the linear term (the term involving \(z\)) is correct, and letting \(q\) be the remainder. Using this form, you can then determine the zeros of the polynomial.
As you can see in Example 10.2.1, we now found two numbers that are of the form \(a+bi\), where \(a\) and \(b\) are real numbers (for short \(a\in\mathbb{R}\) and \(b\in\mathbb{R}\)). A number like this is called a complex number:
Definition 10.2.2
A complex number is a number of the form
where \(a\in\mathbb{R}\) and \(b\in\mathbb{R}\).
The set of all complex numbers is denoted by the symbol \(\mathbb{C}\), and is called the complex plane.
Such complex numbers we usually denote with the letter \(z\) (if we only have one). A complex number also has some special parts, which we define next:
Definition 10.2.3
If \(z=a+bi\) is a complex number (with \(a\in\mathbb{R}\) and \(b\in\mathbb{R}\)), the real part \(\Re{z}\) is defined as
and the imaginary part \(\Im{z}\) is defined as
To make everything concise, we also define the relation between the set of real numbersr \(\mathbb{R}\) and the set of complex numbers \(\mathbb{C}\):
Definition 10.2.4
Assume \(x\in\mathbb{R}\). Then we define that \(x\in\mathbb{C}\) with \(\Re{x}=x\) and \(\Im{x}=0\).
Assume \(z\in\mathbb{C}\) and \(\Im{z}=0\). Then we define that \(z\in\mathbb{R}\).
Just like we can visualise real numbers on a number line, we can visualise complex numbers. For this we need two axes, one indicating the value of the real part of a complex number and one indicating the imaginary part of the same complex number. In Figure 10.2.1 you can see this visualisation. The horizontal axis always indicates the real part, while the vertical axis always represents the imaginary part.
Fig. 10.2.1 Visualisation of the complex plane \(\mathbb{C}\).#
Operations with complex numbers
With complex numbers we can do the same operations as with real numbers:
Theorem 10.2.2
If \(z=a+bi\) and \(w=c+di\) are complex numbers (with \(a,b,c,d\in\mathbb{R}\)), then the following numbers are again complex numbers:
Proof of Theorem 10.2.2
We prove the four results by working each out separately. We start with the addition:
Then we treat subtraction the same:
Now we turn to multiplication:
Finally we look into the division:
Besides these four standard operations we have one more:
Definition 10.2.5
If \(z=a+bi\) is a complex number (with \(a,b\in\mathbb{R}\)), the complex conjugate \(\overline{z}\) is defined as
which is also a complex number.
We can combine the complex conjugate with the first four operations, which gives the following theorem:
Theorem 10.2.3
If \(z\) and \(w\) are a complex numbers, then the following identities hold:
Proof of Theorem 10.2.3
We show each of the identities, one after the other, where we assume \(z=a+bi\) and \(w=c+di\), \(a,b,c,d\in\mathbb{R}\):
Double conjugation:
Addition and conjugation:
Substraction and conjugation:
Multiplication and conjugation:
Division and conjugation:
Even better, we can relate the complex conjugate with the real and imaginary part of a complex number:
Theorem 10.2.4
If \(z\) is a complex number, then the following identities hold:
Proof of Theorem 10.2.4
We show each of the identities, one after the other, where we assume \(z=a+bi\), \(a,b\in\mathbb{R}\):
Conjugation and real part:
Conjugation and imaginary part:
Conjugation and product:
From the second identity above we can even deduce the next theorem:
Theorem 10.2.5
Assume \(z\in\mathbb{C}\). \(z\in\mathbb{R}\) if and only if \(z=\overline{z}\).
Proof of Theorem 10.2.5
Assume \(z\in\mathbb{C}\).
If \(z\in\mathbb{R}\), then \(\Im{z}=0\). The second identity of Theorem 10.2.4 then gives \(\frac{z-\overline{z}}{2i}=0\), which gives in turn \(z=\overline{z}\).
If \(z\notin\mathbb{R}\), then \(\Im{z}\neq0\). The second identity of Theorem 10.2.4 then gives \(\frac{z-\overline{z}}{2i}\neq0\), which gives in turn \(z\neq\overline{z}\).
Geometric interpretation of the complex conjugate
First, we look at the complex conjugation. This is a relatively straightforward case, as it involves only a single number. Recall that the complex conjugate \(\bar z\) changes the sign of the imaginary part of the number \(z\). That is \(\overline{a+bi} = a-bi\). As the imaginary part of a complex number corresponds to the second coordinate of its representation in the complex plane, this implies that the number is reflected in the real axis (the horizontal axis). See Figure 10.2.2.
Fig. 10.2.2 Complex conjugation reflects a number in the real axis.#
Geometric interpretation of addition
The geometric interpretation of adding complex numbers should look familiar to you. Indeed if we add \(z=a+bi\) and \(w=c+di\) the new number is \(z+w=(a+c)+(b+d)i\), so we add the real and imaginary parts. This means we add the coordinates of the corresponding points, just as if we were adding vectors. Thus, geometrically we can add two complex numbers by following the parallelogram rule. That is, the lines from the origin to the two complex numbers form two sides of a parallelogram with vertices \(0\), \(z\), \(z+w\), and \(w\). See Figure 10.2.3.
Fig. 10.2.3 Adding complex numbers follows the parallelogram rule.#
If we want to interpet the other operations such as multiplication easily, we first need another way of writing complex numbers. But before we do that we will turn our attention first to solving equations.
10.3. Solving equations#
The reason for introducing complex numbers is to ensure that more equations have solutions, for example \(z^2+1=0\). In this section, we consider equations involving complex numbers. This means that the solutions may be complex, but also that the coefficients in the equations can be complex.
We already solved quadratic equations using a new technique called completing the square and in this section you will learn more ways to solve equations.
We introduced complex numbers to give the equation \(x^2+1 = 0\) a solution. It appears that something much stronger holds, namely, that every polynomial equation with coefficients in \(\mathbb{C}\), for instance \((1+i)x^4 - 2x^2 + x = 10i\), has solutions in \(\mathbb{C}\). This is the content of the following theorem.
Theorem 10.3.1 (Fundamental Theorem of algebra)
Consider a polynomial \(p(z)\) of degree \(n\),
where the coefficients \(a_n, a_{n-1}, \ldots, a_0\) are complex numbers and \(a_n\neq 0\). Then you can factor the polynomial in linear terms, that is
for some complex numbers \(b_1, b_2, \ldots, b_n\).
Observe that this factorization means that \(b_1, b_2, \ldots, b_n\) are the zeros (= roots) of the polynomial \(p(z)\). It might happen that the \(b\)’s are complex numbers, even if \(a_1, a_2, \ldots, a_n\) are real.
We will not discuss the proof of this theorem, as that requires much more mathematics. However, we will illustrate the theorem using some examples.
Example 10.3.1
Consider \(p(z)=z^2+3z+2\). Then we know that we can factor the polynomial as \(p(z) = (z+2)(z+1)\) and thus find the zeros as \(-2\) and \(-1\).
Example 10.3.2
Consider the equation \(z^2=-1\). We can always rewrite an equation to an equation where one side is equal to zero by moving everything to one side. Thus this equation corresponds to \(z^2+1=0\). We can now factor \(z^2+1\) to \((z+i)(z-i)\) and thus find that \(i\) and \(-i\) are the two solutions to this equation.
Example 10.3.3
While a polynomial of degree \(n\) can be factored in \(n\) linear terms, and we have \(n\) values \(b_i\), this does not mean that there are \(n\) distinct zeros. For example \(p(z) = z^2+2z+4\) can be factored as \(p(z)=(z+2)^2\) and thus only has \(z=-2\) as a solution. However, the term \((z+2)\) occurs twice in the factorization. We therefore say that the multiplicity of the zero \(-2\) is equal to two.
In particular, we see that any polynomial of degree \(n\) has \(n\) complex zeros counting multiplicity:
Theorem 10.3.2
If \(\{z_1,z_2,\ldots,z_k\}\) is the set of unique zeros of a polynomial \(p\) of degree \(n\) with \(p(z)=a_n z^n + a_{n-1} z^{n-1} + \cdots + a_1 z + a_0\), the polynomial \(p\) can be written as
where \(\alpha_j,j=1,2,\ldots,k\) are positive integers and
Proof of Theorem 10.3.2
Assume \(\{z_1,z_2,\ldots,z_k\}\) is the set of unique zeros of a polynomial \(p\) of degree \(n\). Then following Theorem 10.3.1, we can write
for some complex numbers \(b_1, b_2, \ldots, b_n\).
Because \(p(z_1)=0\) for \(j\in\{1,\ldots,k\}\), we must have that \(\alpha_1\in\{1,\ldots,n\}\) numbers out of the set \(\{b_1, b_2, \ldots, b_n\}\) must equal \(z_1\). Without loss of generality we may assume those are \(b_1,\ldots,b_{\alpha_1}\). This is also means that Equation (10.3.1) transforms to
We can repeat the above argument for \(z_2\): we must have that \(\alpha_2\in\{1,\ldots,n-\alpha_1\}\) numbers out of the set \(\{b_{\alpha_1+1}, \ldots, b_n\}\) must equal \(z_2\). Without loss of generality we may assume those are \(b_{\alpha_1+1},\ldots,b_{\alpha_1+\alpha_2}\). This is also means that Equation (10.3.2) transforms to
Repeating this argument for \(z_3,\ldots,z_k\) leads to desired formulae.
Definition 10.3.1
If \(z_j\) is a zero of a polynomial \(p\) of degree \(n\) with \(p(z)=a_n(z-z_1)^{\alpha_1}(z-z_2)^{\alpha_2}\cdots(z-z_k)^{\alpha_k}\), the (algebraic) multiplicity of \(z_j\) is equal to \(\alpha_j\).
Example 10.3.4
The third degree polynomial \(p(z) = z^3-4z^2\) can be factored as \(p(z) = z^2(z-4) = (z-0)^2(z-4)\). Therefore, it has zeros 4 and 0, where the multiplicity of 4 is equal to one and the multiplicity of 0 is equal to two. The degree of the polynomial is 3, which is equal to the sum of the multiplcities of its zeros (\(1+2=3\)).
Example 10.3.2 showed that both \(z=i\) and its complex conjugate \(\overline{z}=-i\) where roots of the polynomial \(p(z)=z^2+1\). One might wonder whether it is always the case that both \(z\) and its complex conjugate \(\overline{z}\) are both roots of a given polynomial. It can be shown that this is the case if all coefficients are real valued.
Theorem 10.3.3
Let \(p\) be a polynomial with real coefficients. If \(p(z)=0\), then \(p(\overline{z})=0\) as well, and the algebraic multiplicities of \(z\) and \(\overline{z}\) are the same.
Proof of Theorem 10.3.3
Consider a polynomial \(p\) of degree \(n\), \(\sum_{j=0}^n a_j z^j\), where the coefficients \(a_n, a_{n-1}, \ldots, a_0\) are real valued numbers and \(a_n\neq 0\).
First we show that \(p(\overline{z})=0\) by considering \(\overline{p(z)}\) twice:
but also
Combining these two results gives the desired \(p(\overline{z})=0\).
Now we focus on the algebraic multiplicity. Theorem 10.3.3 shows that we can write \(p(z)\) as
where \(\alpha_j,j=1,2,\ldots,k\) are positive integers,
and \(z_1,z_2,\ldots,z_k\) are the unique zeros of \(p\).
If \(p(z)=0\), then \(z=z_j\) for some \(j\in\{1,2,\ldots,k\}\) and \(z\neq z_i\) for \(i\in\{1,2,\ldots,k\}\setminus\{j\}\). Without loss of generality we can assume \(j=1\).
This means that also \((z-z_1)^{\alpha_1}=0\) and \((z- z_i)^{\alpha_i}\neq0\) for \(i\in\{2,\ldots,k\}\), and even more that \(\overline{(z-z_1)}^{\alpha_1}=0\) and \(\overline{(z- z_i)}^{\alpha_i}\neq0\) for \(i\in\{2,\ldots,k\}\).
From the last it even follows that \((\overline{z}-\overline{z_1})^{\alpha_1}=0\) and \((\overline{z}- \overline{z_i})^{\alpha_i}\neq0\) for \(i\in\{2,\ldots,k\}\).
Using the previous results, we look at \(p(\overline{z})\):
As all terms except the first term \((\overline{z}-\overline{z_1})^{\alpha_1}\) are nonzero, and the first term is zero, we find that \(\overline{z}=\overline{z_1}\), \(\alpha_1\) times. In other words, the algebraic multiplicity of \(\overline{z}\) is \(\alpha_1\).
Polynomial Division
Next, we consider a method you can use whenever you know one root of a polynomial. The fundamental theorem of Algebra says that if \(p(z)\) is a polynomial such that \(p(b)=0\) for some \(b\), then \(p(z) = a_n(z-b)(z-b_2)\cdots (z-b_n) = (z-b) q(z)\) for another polynomial \(q(z)=a_n(z-b_2)\cdots (z-b_n)\). Thus, we divide the polynomial \(p(z)\) by \((z-b)\) in this case and obtain a new polynomial. To find the zeros of \(p\) we now just have to find the zeros of the quotient \(\nicefrac{p(z)}{z-b}\) and add \(b\) to this list.
To divide a polynomial by another polynomial you can use a long division. Let us recall how this worked for ordinary fractions.
Example 10.3.5
Let us calculate \(\frac{97813}{382}\). In Figure 10.1 on the left, you see the American notation for Long division, on the right the corresponding Dutch notation. Everything in red is usually not written down, but included here to clarify what happens. (Note that the calculations in the middle are true regardless of whether the red parts are included or not.)
Fig. 10.1 Example of long division for ordinary fractions.#
Here, the following calculation has been written down concisely:
First, subtract as many multiples of \(100\cdot 382\) from 97813 as possible (or multiples of \(100\) from \(\frac{97813}{382}\)). Next, you do the same with the remainder, and one digit lower, so you subtract as many multiples of \(10\cdot 382\) from the remainder \(21413\). You continue until you can’t subtract the numerator even once from the remainder (or until you have as many digits as you want).
We can do the same thing for polynomials.
Example 10.3.6
Consider \(p(z) = z^3+3z^2+z-5\). You may notice that \(z=1\) is a root; \(p(1)=0\). Thus \(z-1\) must be a factor. If we calculate this division we obtain (on the left again US notation, on the right Dutch notation):
Fig. 10.2 Example of long division for polynomials.#
Note that the remainder is zero, as we should have expected.
Here, we essentially calculate
As a consequence, we see that \(z^3+3z^2+z-5=(z-1)(z^2+4z+5)\), so it equals \(0\) if either \(z-1=0\) or \(z^2+4z+5=0\). Completing the square gives \(z^2+4z+5=(z+2)^2+1\), so the zeros are \(z=-2\pm i\). Thus, the roots of \(z^3+3z^2+z-5\) are \(z=1\) and \(z=-2\pm i\).
10.4. The polar form of complex numbers#
Modulus and argument
To consider the multiplication of complex numbers, it is best to first consider the polar coordinates of a complex number. Polar coordinates is a concept that works for points in a plane. The idea is that instead of looking at the \(x\) and \(y\) coordinates of a point, we describe the point by the distance to the origin and the direction from the origin.
The distance from zero to the point in the complex plane, we call the modulus \(|z|\). By using Pythagoras theorem and with \(z=a+bi\), it holds that \(|z|=\sqrt{a^2+b^2}\) and equivalently \(|z|=\sqrt{z\overline{z}}\). We often denote the modulus by the symbol \(r\), so \(r=|z|\). The direction is designated by the angle measured from the positive real axis in a counterclockwise direction towards the ray from zero through the point. This angle, we call the argument or \(\arg(z)\) as seen in Figure 10.4.1. We often denote the argument by the symbol \(\theta\), so \(\theta=\arg(z)\). The argument uses the convention similar to the unit circle: the direction straight the right corresponds to 0 radians, up corresponds to \(\frac{1}{2}\pi\) radians, to the left to \(\pi\) radians and down to \(\frac{3}{2}\pi\) radians.
Fig. 10.4.1 The polar coordinates of a point in the complex plane \(\C\) are the distance \(r=|z|\) from zero to the point and the angle \(\theta=\arg(z)\) measured from the positive real axis in a counterclockwise direction towards the ray from zero through the point.#
Notice that the argument is not uniquely defined, as you can always go a full circle extra and add \(2\pi\) radians to the angle. For example, the number \(1\) has argument 0 (as it is on the positive real axis), but also \(2\pi\), \(4\pi\), and \(-2\pi\) (etc.). In order to make a uniform choice, we sometimes work with the principal value of the argument, which is by definition the unique value of the argument between \(-\pi\) and \(\pi\). We write the principal value using a capital A. Thus we have \(-\pi < \Arg{z} \leq \pi\).
Example 10.4.1
Suppose \(z=3+3i\). We find by using Pythagoras that the modulus (the distance to the origin) equals \(|z|=\sqrt{3^2+3^2}=3\sqrt{2}\). The argument, the corresponding angle, equals \(\frac14\pi\) as you can see in Figure 10.3.
Fig. 10.3 The complex number \(3+3i\).#
Example 10.4.2
Suppose \(w=2+3i\). We can still use Pythagoras for the modulus and obtain \(|w|=\sqrt{2^2+3^2} = \sqrt{13}\). The argument can’t be deduced immediately from a picture, see Figure 10.4, but we do see that
Therefore \(\arg(w) = \arctan\left(\frac32\right) \approx 0.982794\).
Fig. 10.4 The complex number \(2+3i\).#
Example 10.4.3
As a final example we consider \(v=-1+2i\). Using Pythagoras theorem once again, we find that \(|v|=\sqrt{(-1)^2+2^2} = \sqrt{5}\). For the argument, we obtain that, just as in the previous example, \(\tan(\arg(v)) = \frac{2}{-1}\), so we would expect that \(\arg(v) = \arctan( -2) \approx -1.10715\). But this answer is negative, while we can see in Figure 10.5 that the true argument is something between \(\frac12\pi\) and \(\pi\). Thus, this argument cannot be correct. If we multiply both the real and imaginary parts of a complex number by \(-1\), then the quotient stays the same. Thus in this case the arctangent gives the argument of \(1-2i\) instead. Fortunately, we can easily find the correct argument as it is exactly \(\pi\) higher. We find \(\arg(v) = \arctan(-2) + \pi \approx 2.03444\).
Fig. 10.5 The complex number \(-1+2i\).#
You always have to check whether the value you find with the arctangent gives the correct angle. As the range of the arctangent is \(\left(-\frac12\pi,\frac12\pi\right)\) you can only find the correct argument if the complex number is to the right of the imaginary axis, that is, if the real part is positive.
If the real part is negative, the argument is between \(\frac12\pi\) and \(\frac32\pi\) (or between \(-\frac12\pi\) and \(-\frac32\pi\) depending on which direction you want to consider) and outside the range of the arctangent. To get the correct value for the argument in these cases, you have to add or subtract \(\pi\) from the arctangent.
If we write \(z=a+bi\) and \(r=|z|\) and \(\theta=\arg(z)\), then we have the following formulas for calculating \(a\) and \(b\) from \(r\) and \(\theta\):
For calculating \(r\) and \(\theta\) from \(a\) and \(b\) we can use
where it should be noted you cannot say \(\theta= \arctan(b/a)\) in general. For \(a\neq 0\), we can use
In particular, we see that the complex number with modulus \(r\) and argument \(\theta\) equals \(r\bigl(\cos(\theta) + i\sin(\theta)\bigr)\) and vice versa. The form \(r\cos(\theta) + ir\sin(\theta)\) is called the polar form of the complex number:
Definition 10.4.1
The polar form of a complex number \(z=a+bi\) is defined as
where \(r=|z|\) is the modulus of \(z\) and \(\theta=\arg(z)\) is an argument of \(z\).
Geometric interpretation of conjugation, multiplication and division
First we consider conjugation. Suppose \(z\) is the complex number with modulus \(|z|=r\) and argument \(\arg(z)=\theta\). Hence, \(z=r\cos(\theta) + i r\sin(\theta)\) and \(\overline{z}=r\cos(\theta) - i r\sin(\theta)\). This least equation we can rewrite:
From the last line we see that conjugating a complex number means negating the argument of that complex number.
We can now see what happens to the product of two complex numbers. Suppose we have the complex number \(z\) with modulus \(|z|=r\) and argument \(\arg(z)=\theta\). Hence, \(z=r\cos(\theta) + i r\sin(\theta)\). The second complex number we consider is \(w\) with modulus \(|w|=s\) and argument \(\arg(w) = \phi\); thus \(w=s\cos(\phi) + i s \sin(\phi)\). We can then calculate the product using the addition formulas for cosine and sine.
We recognize this product as the number with modulus \(|zw|=rs\) and argument \(\arg(zw) = \theta+\phi\). In particular we find:
Theorem 10.4.1
If you take the complex conjugate of a complex number \(z\), the modulus remains the same and the argument is negated:
-
\(|\overline{z}| = |z|\),
-
\(\arg(\overline{z}) = -\arg(z)\).
If you multiply two complex numbers \(z\) and \(w\), you multiply the moduli and add the arguments:
-
\(|zw| = |z| \cdot |w|\),
-
\(\arg(zw) = \arg(z) + \arg(w)\).
If you divide the complex number \(z\) by the complex number \(w\neq0\) you divide the modulus of \(z\) by the modulus of \(w\) and subtract the argument of \(w\) from the argument of \(z\):
-
\(\left|\frac{z}{w}\right| = \frac{|z|}{|w|}\),
-
\(\arg\left(\frac{z}{w}\right) = \arg(z) - \arg(w)\).
Proof of Theorem 10.4.1
Proof for conjugation
With \(|z|=r\) and \(\arg{z}=\theta\), Equation (10.4.1) shows that \(|\overline{z}|=|z|\) and \(\arg(\overline{z})=-\arg(z)\).
Proof for multiplication
With \(|z|=r\), \(|w|=s\), \(\arg{z}=\theta\) and \(\arg{w}=\phi\), Equation (10.4.2) shows that \(|zw| = |z| \cdot |w|\) and \(\arg(zw) = \arg(z) + \arg(w)\).
Proof for division
For division (assuming \(w\neq0\)) we do the following:
where the last line shows that \(\left|\frac{z}{w}\right| = \frac{|z|}{|w|}\) and \(\arg\left(\frac{z}{w}\right) = \arg(z) - \arg(w)\).
You can see each of these operations illustrated in Figures 10.4.2, 10.4.3 and 10.4.4.
Fig. 10.4.2 Conjugating a complex number means negating the argument and keeping the modulus the same.#
Fig. 10.4.3 Multiplying complex numbers means adding the arguments and multiplying the moduli.#
Fig. 10.4.4 Dividig complex numbers means substracting the arguments and dividing the moduli.#
10.5. Euler’s formula#
Euler’s formula for the polar form of a complex number
Given that the polar coordinates of a complex number are so convenient and that the polar form \(r\left(\cos(\theta) + i\sin(\theta)\right)\) is such a long expression to write down, we would like to have a simple way of representing the complex number with given modulus \(r\) and argument \(\theta\).
Therefore we introduce the following identity:
Definition 10.5.1
Let \(\theta\in\mathbb{R}\). Then
which is called Euler’s formula.
Using this definition we even can prove the following theorem:
Theorem 10.5.1
Let \(z=a+bi\) with \(a,b\in\mathbb{R}\) and define \(r=|z|\) and \(\theta=\arg{z}\). Then
Proof of Theorem 10.5.1
The proof is relatively straight forward:
We can even show that the following property of the derivative still is true:
Theorem 10.5.2
Let \(\theta\in\mathbb{R}\). Then
Proof of Theorem 10.5.2
The proof is again straight forward:
You might think it is strange to use the number \(e\) and superscripts which would suggest some sort of power of \(e\). But it turns out this is very convenient as this expression satisfies the rules of calculation for exponentials. Indeed the calculation in Equation (10.4.2) shows that
This corresponds precisely with the rules for multiplying exponentials.
Thus, you can calculate with this strange notation \(re^{i\theta}\) for complex numbers just as you would if you were indeed taking imaginary powers of \(e\) (\(=2.71828\ldots\)). Some deep mathematics show that the definition given here is the only reasonable way to define taking imaginary exponents.
In practice, this polar notation of complex numbers is convenient to use when you take products or powers, whereas the \(a+bi\) notation is more convenient when you have to add complex numbers.
Example 10.5.1
Let us calculate \((1+i)^6\). As this is a high power of a complex number, we use Euler’s formula for the polar form. First, we write \(1+i=\sqrt{2} e^{i \frac{\pi}{4}}\). Thus, we have
Example 10.5.2
Let us calculate \(\dfrac{(1-\sqrt{3}i)^3}{(2+2i)^6}\). We have \(1-\sqrt{3}i=2e^{-i\frac{\pi}{3}}\) and \(2+2i=2\sqrt{2} e^{i\frac{\pi}{4}}\). Thus, we obtain
De Moivre and other trigonometric identities
The notation invented by Euler of \(e^{i\theta} = \cos(\theta) + i\sin(\theta)\) allows us to quickly derive trigonometric identities. The most famous one is De Moivre’s identity \(e^{in\theta} = (e^{i\theta})^n\), which seems obvious now, but was discovered by De Moivre decades before the exponential notation was introduced and is a lot more impressive in the form
Example 10.5.3
De Moivre’s identity allows us to find an expression for \(\cos(3\theta)\) in terms of \(\cos(\theta)\) and \(\sin(\theta)\). Indeed, expanding the right hand side of the identity we have
Comparing the real and imaginary parts on both sides of this equation we find
You can also easily derive other formulas.
Example 10.5.4
Formulas for \(\cos(\theta+\phi)\) and \(\sin(\theta+\phi)\) are often used in calculus courses. These formulas can be derived using De Moivre’s identity.
Thus \(\cos(\theta+\phi) = \cos(\theta) \cos(\phi) - \sin(\theta) \sin(\phi)\) and \(\sin(\theta +\phi)=\sin(\theta) \cos(\phi) +\cos(\theta) \sin(\phi)\).
Solving \(z^n=w\)
The most basic equations we want to solve are of the form \(z^n=w\) for a given complex number \(w\), where \(z\) is the variable we want to solve for. Let’s consider an example:
Example 10.5.5
Consider the equation \(z^3=-16+16i\). We know it has 3 complex solutions, as it is a third degree equation. If we write \(z=a+bi\) and expand (to find \(a\) and \(b\)), we get a very large expression which is not easy to work with.
If we write \(z=re^{i\phi}\) in polar coordinates instead, we can easily express \(z^3=r^3e^{3i\phi}\). We also have to express the right hand side in polar coordinates: \(-16+16i=16\sqrt{2} e^{\frac34 \pi i}\). Comparing the modulus and argument of these two expressions, we find
Taking a cube root, we find \(r=2\sqrt{2}\). Note that \(r>0\) is real, so here we need to consider only the single positive real solution, we don’t want to find the complex solutions.
Moreover, we have \(3\phi = \frac34\pi\), so \(\phi = \frac14\pi\). This gives the solution \(z=2\sqrt{2} e^{\frac14 \pi i}\). But this is just one solution and there ought to be two more by the fundamental theorem of algebra. So what are the remaining two?
As you know, the argument is only defined up to a multiple of \(2\pi\). Thus, when we get the equation \(3\phi = \frac34\pi\), we should actually write \(3\phi = \frac34\pi + 2\pi k\) for some integer \(k\). Dividing this by \(3\) gives \(\phi = \frac14\pi + \frac23 \pi k\). We see that different values of \(k\) give different values of \(\phi\). For \(k=0\), we obtain \(\phi=\frac14\pi\) as before. For \(k=1\), we obtain \(\phi = \frac14\pi + \frac23\pi = \frac{11}{12}\pi\). For \(k=2\), we have \(\phi = \frac14\pi + \frac43\pi = \frac{19}{12}\pi\). For \(k=3\), we obtain \(\phi = \frac14\pi + 2\pi\). This gives the same complex number as \(\phi=\frac14\pi\), as the argument is shifted by one full period. Indeed, if we add a multiple of \(3\) to \(k\), the argument of \(\phi\) is shifted by a multiple of \(2\pi\) and thus the corresponding solution \(z\) does not change. Therefore, only the cases \(k=0\), \(1\), and \(2\) suffice to obtain all solutions.
The three solutions, \(z_0\), \(z_1\) and \(z_2\), to the equation \(z^3=-16+16i\) thus are
You can find a visualisation of these three solutions in Figure 10.6.
Fig. 10.6 The three solutions from Example 10.5.5.#
We can generalize the method for solving \(z^n=w\) from the example above:
Algorithm 10.5.1 (Solving \(z^n=w\))
-
Write \(z=re^{i\phi}\) (for unknown \(r\) and \(\phi\)) and express the right hand side \(w\) in polar coordinates.
-
Obtain equations for the modulus \(r\) and argument \(\phi\) by equating the modulus and argument on both sides of this equation.
-
Solve for \(r\) (you only need the single positive real solution).
-
Solve for \(\phi\), remembering to add \(+2\pi k\) first to the right hand side.
-
You obtain all solutions to the equation by taking \(n\) (the degree of the equation) subsequent values of \(k\) in your expression of \(\phi\).
-
Combine the solution for \(r\) and the \(n\) values for \(\phi\) to obtain the \(n\) solutions.
Adding trigonometric functions
Quite often, you come across expressions where a cosine and a sine of identical frequency are added. If you plot a function of the form \(f(t)=b\cos(\omega t) + c\sin(\omega t)\), you notice that it becomes a new single wave. You can use complex numbers in a smart way to rewrite \(f(t)\) to the form \(A \cos(\omega t -\phi)\) as a single cosine with shifted argument. The variable \(A\) gives the amplitude of the combined wave and the variable \(\phi\) gives the phase-shift.
Example 10.5.6
Take \(f(t) = \cos(2t) + \sqrt{3} \sin(2t)\). If you plot the graph of this function (see Figure 10.7), you notice it is a single wave.
Indeed, we have
We first wrote both the cosine as the sine as real parts of complex exponentials. For the sine, we use that \(-ie^{i\theta} = -i\cos(\theta) + \sin(\theta)\), so \(\Re{-ie^{i\theta}} = \sin(\theta)\). Subsequently, we can take out the common factor \(e^{2it}\); it is a common factor as the periods of both the cosine and sine are identical. Next, we rewrite \(1-i\sqrt{3}\) in polar coordinates and work out what the result is.
Fig. 10.7 The graph of the sum of a cosine and a sine of identical period is a sinusoid as well.#
Example 10.5.7
In the same way you can add two cosines (or sines) with shifted arguments.
10.6. Derivations of Euler’s formula#
Using a scalar initial value problem
To find Euler’s formula, consider the initial value problem
Because we assumed that \(i\) behaves like any other number, we can solve this initial value problem, which leads to the solution
Now we have a solution, we put it aside and focus on another function \(q\), which we define as
Let us calculate first the value of \(q\) in \(\theta=0\):
This indicates that \(q\) satisfies the same initial condition from Equation (10.6.1) as the function \(y\) from Equation (10.6.2). Could it also be that \(q\) is a solution to the differential equation from Equation (10.6.1)? Let us investigate by looking at the first derivative of \(q\):
So we found that our function \(q\) from Equation (10.6.3) is also a solution to initial value problem from Equation (10.6.1).
But because this initial value problem can only have one unique solution, the function \(q\) from Equation (10.6.3) must be the same function as the first solution \(y\) in Equation (10.6.2). This means that we found Euler’s formula:
Using series
Euler’s formula can also be derived using series. You may already be familiar with the following three series
We can use these series to derive Euler’s formula.
First consider the following 12 powers of the complex number \(\theta i\) with \(\theta\in\mathbb{R}\):
Do you notice the pattern that even powers give real numbers and odd powers give complex numbers with zero real part? And also that for the list of even powers the sign flips each time? And the same for the odd powers?
Now let us consider \(e^{\theta i}\) and expand the series of the exponential function using these patterns:
As you can see we have arrived at Euler’s formula.
10.7. Exercises#
Exercise 10.7.1
Evaluate the given expression and write your answer in the form \(a+bi\).
-
\((3-2i)+(4+3i)\)
-
\((4-2i)-(7+4i)\)
-
\(\overline{2-6i}\)
-
\((1-3i)(2+4i)\)
-
\((\overline{5-4i})(2+i)\)
-
\(\displaystyle\frac{8+4i}{2-2i}\)
-
\(\displaystyle\frac{6}{3+3i}\)
-
\(i^3\)
-
\(i^{50}\)
-
\((2 + i)^{2}\)
Exercise 10.7.2
Find the roots of the quadratic polynomial by completing the square.
-
\(2z^2-4z+10\)
-
\(z^2+(1-i)z-i\)
-
\(z^2+4z-12\)
Exercise 10.7.3
Find \(h(z)=\frac{p(z)}{q(z)}\) if it is a polynomial using long division.
-
\(p(z)=2z^5-3z^4-5z^3+2z^2-z-1\) and \(q(z)=2z^3+z^2-z+1\)
-
\(p(z)=z^4+(3+i)z^3 + (- 2+4i)z^2 + (-3 + 2i)z+1-i\) and \(q(z)=z^2+iz-1+i\)
-
\(p(z)=z^4+z^3-3z^2+5z-2\) and \(q(z)=z^2-z+1\)
Exercise 10.7.4
Find the roots of \(p(z)\) and their multiplicities. Certain roots may already be given.
(Hint: use long division and completing squares.)
-
\(p(z)=z^3-2z^2-5z+6\) with \(p(1)=0\)
-
\(p(z)=z^4-2z^3+2z^2-2z+1\) with \(p(i)=0\)
-
\(p(z)=z^4-5z^3+2z^2+22z-20\) with \(p(3+i)=0 \)
Exercise 10.7.5
Find the argument and modulus of the complex number \(z\).
-
\(z=1-i\)
-
\(z=3+\sqrt{3}i\)
Exercise 10.7.6
Write the complex number in polar form \(r\cos(\theta) +i r\sin (\theta)\) with \(-\pi \leq \theta \leq \pi\).
-
\(-1+i\)
-
\(3\sqrt{3}-3i\)
Exercise 10.7.7
Write the complex number in the form \(a+bi\).
-
\(2 \cos(\frac{\pi}{6})+ i 2 \sin(\frac{\pi}{6})\)
-
\(4 \cos(\frac{-3\pi}{4})+ i 4 \sin(\frac{-3\pi}{4})\)
-
\(e^{\frac12\pi i}\)
-
\(e^{\frac14\pi i}\)
-
\(2e^{-\frac13\pi i}\)
Exercise 10.7.8
Find the polar forms of \(z\) and \(w\) and calculate \(zw\), \(\frac{z}{w}\) and \(\frac{1}{z}\).
-
\(z=1+\sqrt{3} i\) and \(w=\sqrt{3}+i\)
-
\(z=2\sqrt{3}-2i\) and \(w=2-2i\)
Exercise 10.7.9
Solve the equation by finding the relevant roots of the complex number in the form \(a+bi\).
-
\(z^4=1\)
-
\(z^3=-i\)
Exercise 10.7.10
Use De Moivre’s identity to derive a quadruple angle formula.
In other words to express \(\cos(4\theta)\) and \(\sin(4\theta)\) in terms of \(\cos(\theta)\) and \(\sin(\theta)\).
Exercise 10.7.11
Solve the equation by finding the relevant roots of the complex number in the Euler form.
-
\(z^4=-16i\)
-
\(z^5=16\sqrt{2}+i16\sqrt{2}\)
-
\(z^3=4\cos(\frac{\pi}{6})+4i\sin(\frac{\pi}{6})\)
10.8. Solutions#
Solution to Exercise 10.7.1
-
\(7+i\)
-
\(-3-6i\)
-
\(2+6i\)
-
\(14-2i\)
-
\(6+13i\)
-
\(1+3i\)
-
\(1-i\)
-
\(-i\)
-
\(-1\)
-
\(3 + 4i\)
Solution to Exercise 10.7.2
-
\(1+2i, 1-2i\)
-
\(-1, i\)
-
\(2, -6\)
Solution to Exercise 10.7.3
-
\(h(z)=z^2-2z-1\)
-
\(h(z)=z^2+3z-1\)
-
\(h(z)\) is not a polynomial.
Solution to Exercise 10.7.4
-
The roots of \(p(z)\) are \(1, -2, 3\), all with multiplicity 1.
-
The roots of \(p(z)\) are \(i, -i, 1\), where \(i\) an \(-i\) have multiplicity 1 and \(1\) has multiplicity 2.
-
The roots of \(p(z)\) are \(3+i, 3-i, 1, -2\), all with multiplicity 1.
Solution to Exercise 10.7.5
-
\(|z|=\sqrt{2}\) and \(\Arg{z}=-\frac{\pi}{4}\) or \(\arg(z)=-\frac{\pi}{4}+2k\pi\) for some integer \(k\);
-
\(|z|=2\sqrt{3}\) and \(\Arg{z}=\frac{\pi}{6}\) or \(\arg(z)=\frac{\pi}{6}+2k\pi\) for some integer \(k\).
Solution to Exercise 10.7.6
-
\(\sqrt{2}\cos(\frac{3\pi}{4})+ i\sqrt{2}\sin(\frac{3\pi}{4})\)
-
\(6 \cos(\frac{-\pi}{6})+ i 6 \sin(\frac{-\pi}{6})\)
Solution to Exercise 10.7.7
-
\(\sqrt{3}+i\)
-
\(-2\sqrt{2}-2\sqrt{2}i\)
-
\(i\)
-
\(\frac{1}{\sqrt{2}} +\frac{1}{\sqrt{2}} i\)
-
\(1-\sqrt{3} i\)
Solution to Exercise 10.7.8
-
\(z=2e^{\frac13\pi i}\) and \(w=2e^{\frac16\pi i}\), thus \(zw=4i\), \(\frac{z}{w}=\frac{\sqrt{3}}{2}+\frac{1}{2} i\) and \(\frac{1}{z}=\frac{1}{4}-\frac{\sqrt{3}}{4} i\).
-
\(z=4e^{-\frac16\pi i}\) and \(w=2\sqrt{2}e^{-\frac14\pi i}\), thus \(zw=8 \sqrt{2} \cos\left(-\frac{5\pi}{12} \right) + 8\sqrt{2}\sin\left(-\frac{5\pi}{12} \right) i\), \(\frac{z}{w}=\sqrt{2} \cos\left(\frac{\pi}{12} \right) + \sqrt{2} \sin\left(\frac{\pi}{12} \right) i\) and \(\frac{1}{z}=\frac{\sqrt{3}}{8}+\frac{1}{8} i\).
Solution to Exercise 10.7.9
-
\(1, -1, i, -i\)
-
\(i, -\frac{1}{2}\sqrt{3}-i\frac{1}{2}, \frac{1}{2}\sqrt{3}-i\frac{1}{2}\)
Solution to Exercise 10.7.10
\(\cos(4\theta)=\cos^4(\theta)+\sin^4(\theta)-6\cos^2(\theta)\sin^2(\theta)\)
and
\(\sin(4\theta)=4\cos^3(\theta)\sin(\theta)-4\cos(\theta)\sin^3(\theta)\).
Solution to Exercise 10.7.11
-
\( 2e^{i\frac{3\pi}{8}}, 2e^{i\frac{7\pi}{8}}, 2e^{i\frac{11\pi}{8}}, 2e^{i\frac{15\pi}{8}}\)
-
\(2e^{i\frac{\pi}{20}}, 2e^{i\frac{9\pi}{20}}, 2e^{i\frac{17\pi}{20}}, 2e^{i\frac{5\pi}{4}}, 2e^{i\frac{33\pi}{20}}\)
-
\(4^{\frac{1}{3}}e^{i\frac{\pi}{18}}, 4^{\frac{1}{3}}e^{i\frac{13\pi}{18}}, 4^{\frac{1}{3}}e^{i\frac{25\pi}{18}}\)