Lecture 1: Complex Number System

Definition of $\mathbb C$

Definition 1 (Complex Number). A complex number is an ordered pair of real numbers of the form $(x,\, y)$ . The number $x$ is called the real part of $z$ and the number $y$ is called the imaginary part of $z$ . Two complex numbers $(x_1,\, y_1)$ and $(x_2,\, y_2)$ are equal if and only if $x_1=x_2$ and $y_1=y_2$ .

Notation 1. The set of all complex numbers is denoted by $\mathbb C$ .

Notation 2. For a given complex number $z=(x,\, y)$ , we use $\operatorname{Re}(z)$ and $\operatorname{Im}(z)$ to denote the real and imaginary components respectively. That is $\operatorname{Re}z=x$ and $\operatorname{Im}z=y$ .

We are interested in translating in some way the usual addition and multiplication operations on $\mathbb R$ into $\mathbb C$ .

Definition 2 (Addition and Multiplication). Let $z_1=(x_1,\, y_1)$ and $z_2=(x_2,\, y_2)$ be two complex numbers. We define the addition of $z_1$ and $z_2$ , denoted by $z_1+z_2$ , and the multiplication of them, denoted by $z_1 z_2$ , respectively as the complex numbers $z_1+z_2:= (x_1+x_2,\, y_1+y_2),$ and $z_1 z_2\equiv (x_1,\, y_1)(x_2,\, y_2):=(x_1 x_2-y_1y_2, x_1y_2+y_1 x_2).$

The following properties can be easily verified.

Exercise 3. Let $z_1$ , $z_1$ and $z_3$ be complex numbers. Show that: $\begin{array}{rcl} z_1+z_2 = z_2+z_1, && z_1z_2 = z_2z_1\\ z_1+z_2+z_3= z_1+(z_2+z_3), & & z_1z_2z_3=z_1(z_2z_3),\\ z_1(z_2+z_3)=z_1z_2+z_1z_3 & & \\ \end{array}$

For each complex number, we define its inverse element under addition, which we call the negation, and its inverse element under multiplication, which we call the inverse, as follows.

Definition 4 (Negation and Inverse). Let $z=(x,\, y)$ be a complex number. The negation of $z$ , denoted by $-z$ , is the complex number defined as $-z := (-x,\, -y).$ The inverse of $z$ , denoted by $\dfrac{1}{z}$ or $z^{-1}$ , is the complex number defined as $\frac{1}{z}:=(\frac{x}{x^2+y^2},\, -\frac{y}{x^2+y^2})$ provided that $z\neq (0,\, 0)$ .

Example 5. Let $z=(1,\, 2)$ , then $-z=(-1,\, -2)$ and $z^{-1}=(\dfrac{1}{5},\, -\dfrac{2}{5})$ .

Example 6. Let $z=(3,\, -4)$ , then $-z=(-3, 4)$ and $z^{-1}=(\dfrac{3}{25},\, \dfrac{4}{25})$ .

Exercise 7. For any $z\in\mathbb C$ , verify that (i) $z+(-z)=(0,\, 0)$ , and (ii) if $z\neq (0,\, 0)$ , then $z z^{-1}= z^{-1}z=(1,\, 0)$ .

Exercise 8. Let $z_1\in\mathbb C$ and $z_2\in\mathbb C$ be non-zero complex numbers. Prove that $\dfrac{1}{z_1}\dfrac{1}{z_2}=\dfrac{1}{z_1z_2}.$

Once we define the addition and multiplication operations, we could use them to define subtraction and division as follows.

Definition 9 (Subtraction and Division). Let $z_1=(x_1,\, y_1)$ and $z_2=(x_2,\, y_2)$ be two complex numbers. We define the subtraction of $z_1$ by $z_2$ , denoted by $z_1-z_2$ , and the division of $z_1$ by $z_2$ (provided that $z_2\neq (0,\, 0)$ ), denoted by $\dfrac{z_1}{z_2}$ or $z_1/z_2$ , respectively as the complex numbers $z_1-z_2:=(x_1-x_2,\, y_1-y_2)\,,$ and $\frac{z_1}{z_2}:= z_1\dfrac{1}{z_2} \equiv (\dfrac{x_1x_2+y_1y_2}{x_2^2+y_2^2},\, \frac{x_2y_1-x_1y_2}{x_2^2+y_2^2})\,.$

Example 10. Let $z_1=(1,\, 1)$ , and $z_2=(1,\, -2)$ . Then $z_1/z_2=(-\dfrac{1}{5},\, \dfrac{3}{5})$ .

Example 11. Let $z_1=(2,\, 3)$ , and $z_2=(3,\, -2)$ . Then $z_2^{-1}=(\dfrac{3}{3^3+(-2)^2},\, \dfrac{2}{3^3+(-2)^2})$ . Therefore $z_1/z_2=z_1z_2^{-1} =(2,\, 3)(\dfrac{3}{13},\, \dfrac{2}{13})=(0,\, 1)$ .

We can easily prove the following property of the division operation.

Theorem 12. Let $z_1$ , $z_2 (\neq (0,\, 0))$ and $z_3 (\neq (0,\, 0))$ be arbitrary complex numbers. Then $\frac{z_1}{z_2}=\frac{z_1 z_3}{z_2 z_3}\,.$

Proof. This follows from the following calculations. $\frac{z_1}{z_2}=z_1\frac{1}{z_2}=z_1\frac{1}{z_2}(z_3\frac{1}{z_3})=(z_1z_3)(\frac{1}{z_2}\frac{1}{z_3})=(z_1z_3)\dfrac{1}{z_2z_3}=\dfrac{z_1z_3}{z_2z_3}\,.$ ◻

The Real Numbers as a Subset of the Complex Plane

One important observation about the complex number system is that the subset of complex numbers whose imaginary part is $0$ has exactly the same arithmetics as the set of real numbers. This can be easily seen from the following theorem.

Theorem 13. For any $x$ , $y\in\mathbb R$ , we have $(i)\ (x,\, 0)+(y,\, 0)=(x+y,\, 0), \qquad (ii)\ (x,\, 0)-(y,\, 0)=(x-y,\, 0),$ $(iii)\ (x,\, 0)(y,\, 0)=(xy,\, 0), \qquad (iv)\ \dfrac{(x,\, 0)}{(y,\, 0)}=(\dfrac{x}{y},\, 0),\ \ \mbox{if}\ \ y\neq 0.$

Remark 14. The above theorem says that for any given real number $x\in\mathbb R$ , we can identify the complex number $z=(x,\, 0)$ with the real number $x$ . Therefore, real numbers are simply complex numbers with zero imaginary parts. This way, we make $\mathbb R$ a subfield of $\mathbb C$ .

Moreover, the above theorem allows us to have the following representation of complex numbers.

Theorem 15. Let $z=(x,\, y)$ be a complex number, and $i=(0,\, 1)$ . Then $z=x+iy\,.$

Proof. This follows from the following calculation: $x+iy=(x,\, 0)+(0,\, 1)(y,\, 0) =(x,\, 0)+(0,\, y)=(x,\, y),$ where we have used the fact that $x\equiv(x,\, 0)$ for any $x\in\mathbb R$ . ◻

Exercise 16. Prove that $i^2=-1.$

Complex conjugate

An important concept in complex analysis is complex conjugate.

Definition 17 (Complex Conjugate). Let $z=x+iy$ be a complex number. The complex conjugate of $z$ , denoted by $\bar z$ , is the complex number defined as $\bar{z}: = x+i(-y) \equiv x-iy.$

Example 18. (i) $z=2-2i$ , then $\bar z=2+2i$ ; (ii) $z=5+i$ , then $\bar z=5-i$ .

Example 19. Let $z=x+iy$ , then $z\bar z=(x+iy)(x-iy)=x^2-y(-y)=x^2+y^2$ .

Remark 20. If we take $z_1=1$ , $z_2=z$ and $z_3=\bar z$ in Theorem Theorem 12, then we have $\dfrac{1}{z}=\dfrac{\bar z}{z\bar z},\ \ \mbox{if}\ \ z\neq 0.$ This is a very useful relation.

The following result is obvious from the definition but is very useful.

Lemma 21. For any $z\in\mathbb C$ , we have $\operatorname{Re}z=\frac{z+\bar z}{2}, \qquad \operatorname{Im}z=\frac{z-\bar z}{2i}, \qquad \bar{\bar z}=z\,.$

Theorem 22 (Complex Conjugate Commutes with Arithmetic Operations). For any $z_1\in\mathbb C$ and $z_2\in\mathbb C$ , the following equalities hold: $\begin{aligned} \overline{z_1\pm z_2}=\bar z_1 \pm \bar z_2\\ \overline{z_1z_2}=\bar z_1 \bar z_2\\ \overline{z_1/z_2}=\bar z_1/\bar z_2 \end{aligned}$

Proof. We only prove the last equality. Let $z=x+iy\neq 0$ . By definition of inverse and complex conjugate, we verify that $\overline{\dfrac{1}{z}}=\overline{\dfrac{x-iy}{x^2+y^2}} = \dfrac{x+iy}{x^2+y^2} = \dfrac{x-i(-y)}{x^2+(-y)^2}= \dfrac{1}{\bar z}\,.$ The rest follows from simple calculations here: $\overline{z_1/z_2}=\overline{z_1 \frac{1}{z_2}} = \bar z_1 \overline{(\frac{1}{z_2})} = \bar z_1 \frac{1}{\bar z_2} = \bar z_1 /\bar z_2\,.$ This completes the proof. ◻

Exercise 23. Prove the properties of complex numbers stated in Theorem Theorem 22.

Polar representation

From our presentation, it is clear that we can uniquely identify a complex number with a point in $\mathbb R^2$ (an example of which is the xy-plane in our 3-dimensional Euclidean space). It is also very convenient to think of $z=x+iy$ as the vector that goes from the origin $(0,0)$ in the $xy$ plane to the point $(x,y)$ which we denote by $\overrightarrow{OZ}$ .

It is well-know that any point $(x,y)$ in the $xy$ plane can be represented using the polar system $(r,\theta)$ where $r$ is the distance of the point to the origin while $\theta$ is the angle formed between the vector $\vec{x}$ and $\overrightarrow{OZ}$ (when we go from $\vec{x}$ and $\overrightarrow{OZ}$ anti-clockwise). The representation reads $x=r\cos\theta, \qquad y=r\sin\theta,$ which naturally gives us $\label{EQ:Polar Representation} z=r(\cos\theta+i\sin\theta).$

It is easy to verify that under this representation, we have $x^2+y^2=r^2(\cos^2\theta+\sin^2\theta)=r^2,$ which then leads to $r=\sqrt{x^2+y^2}.$

Definition 24 (Modulus and Argument). Let $z=x+iy\in\mathbb C$ be a given complex number. The modulus of $z$ , denoted by $|z|$ , is the nonnegative number $|z|=\sqrt{x^2+y^2}.$ The argument of $z$ , denoted by $\operatorname{arg}z$ , is the angle $\theta$ in the representation [EQ:Polar Representation].

It is extremely important to understand that the argument of $z$ , $\operatorname{arg}z$ (i.e. the $\theta$ in the polar representation) is $2\pi-$ periodic. That is, if we add $2k\pi$ ( $k=0, \pm 1, \pm 2, ...$ ) to $\theta$ , the $x$ and $y$ values won’t change (we remain at the same point). This is to say that $\operatorname{arg}z$ is multi-valued. We set the following convention: $\operatorname{arg}z$ always takes the form $\theta=\theta_0+2k\pi, \qquad k=0, \pm 1, \pm 2, ...$ where $-\pi<\theta_0 \le \pi$ ¹ is called the principle value of the argument, often denoted with $\operatorname{Arg} z$ (that is $\operatorname{Arg} z=\theta_0$ ).

Therefore, in the polar representation of $z$ , the radius $r$ is simply the modulus of $z$ , that is, $|z|$ . The polar representation of a complex number $z$ thus takes the following final form $z=|z|(\cos\theta + i\sin\theta)\,.$

For any real-valued $y$ , we define the exponential function $e^{iy}$ as $\label{EQ:Euler's Formula} e^{iy} := \cos y+ i\sin y.$ The relation [EQ:Euler's Formula] is called Euler’s formula. It is an extremely useful tool in the study of trigonometric functions as we will see a little later. For the moment, let us simply note that Euler’s formula allows us to have a more compact polar representation for a complex number $z=|z|(\cos\theta+ i\sin\theta)$ called the polar exponential representation: $z=|z|e^{i\theta}.$ This representation will be used often later.

Example 25. Let $z=1+i$ , then we can write $|z|=\sqrt{1^2+1^2}=\sqrt{2}$ . Therefore $z=\sqrt{2}(\dfrac{1}{\sqrt{2}}+i\dfrac{1}{\sqrt{2}})$ . This gives us that $\cos\theta=\frac{1}{\sqrt{2}}$ and $\sin\theta=\frac{1}{\sqrt{2}}$ . Therefore $\theta=\dfrac{\pi}{4}+2k\pi$ , $k=0, \pm 1, \pm 2, \cdots$ . This gives $1+i=\sqrt{2}e^{i(\frac{\pi}{4}+2k\pi)},\ k\in\mathbb Z\,.$

Example 26. Let $z=\sqrt{3}+i$ . Then $|z|=\sqrt{3+1}=2$ . Therefore $z=2(\dfrac{\sqrt{3}}{2}+\dfrac{1}{2} i)$ . This gives $\cos\theta=\frac{\sqrt{3}}{2}$ and $\sin\theta=\dfrac{1}{2}$ . Therefore $\theta=\pi/6+2k\pi$ , $k\in\mathbb Z$ . This gives $\sqrt{3}+i=2e^{i(\pi/6+2k\pi)},\ \ k\in\mathbb Z\,.$

The Triangle Inequality

The concept of modulus allows us to compare sizes of complex numbers. The following result is straightforward to prove.

Theorem 27 (Properties of Modulus). Let $z$ , $z_1$ and $z_2$ be arbitrary complex numbers. The following equalities hold: $\begin{aligned} |\operatorname{Re}(z)|\le |z|,\quad |\operatorname{Im}(z)|\le |z|\\ |z|=|\bar z|=|-z|,\\ |z|^2=z\bar z,\\ |z_1z_2|=|z_1||z_2|\\ |z^{-1}|=|z|^{-1},\ \ \ \mbox{if}\ z\neq 0. \end{aligned}$

Proof. The first two identities are trivial. The third one can be verified easily with the definition. The last equality can be deduced easily from the fourth. Assuming that the four identity is true. Then $|zz^{-1}|=|z||z^{-1}|$ . This means $|z||z^{-1}|=1$ . Therefore $|z^{-1}|=|z|^{-1}$ . Proof of the fourth identitie is left as an exercise. ◻

Exercise 28. Prove the fourth identity in Theorem Theorem 27.

One of the most useful results in complex analysis is the following Triangle Inequality. It allows us to bound size of summation of two complex numbers by the summation of their sizes.

Theorem 29 (Triangle Inequality). For any $z_1\in\mathbb C$ and $z_2\in \mathbb C$ , we have $\label{EQ:Triangle Ineq} ||z_1|-|z_2||\le |z_1+z_2|\le|z_1|+|z_2|.$

Proof. (i) The second inequality is a result of the following calculation. $\begin{gathered} |z_1+z_2|^2=(z_1+z_2)(\bar z_1+\bar z_2)=z_1\bar z_1+z_2\bar z_2 + z_1\bar z_2+\bar z_1 z_2=|z_1|^2+|z_2|^2+2\operatorname{Re}(z_1\bar z_2)] \\ \le |z_1|^2+|z_2^2|+2|z_1||z_2|=(|z_1|+|z_2|)^2. \end{gathered}$ (ii) To prove the first inequality, we observe, using the result from part (i), that $|z_1|=|z_1+z_2-z_2|\le |z_1+z_2|+|z_2| \implies |z_1|-|z_2|\le |z_1+z_2|,$ and $|z_2|=|z_1+z_2-z_1|\le |z_1+z_2|+|z_1| \implies |z_2|-|z_1|\le |z_1+z_2|.$ We can then combine the two inequalities to get $||z_1|-|z_2||\le |z_1+z_2|$ . ◻

Geometrically, the Triangle Inequality says that (i) the summation of the lengths of any two sides of a triangle is always greater or equal to the length of the third side, and (ii) the difference between the lengths of any two sides is always smaller than or equal to the length of the third side.

The selection of $\theta_0\in(-\pi, \pi]$ is quite arbitrary. In some literature, $\theta_0\in(0, 2\pi]$ is selected. There is no essential difference between the two choices since both give the same $\theta$ at the end.↩︎

Guided review and additional examples

Learning goals

Move confidently between ordered-pair, Cartesian, and polar forms of a complex number.
Use conjugates and moduli to simplify quotients and estimate expressions.
Interpret multiplication geometrically as scaling followed by rotation.
Find all roots of a complex number without losing solutions.

Concept connection: algebra and geometry say the same thing

Writing $z=x+iy$ is best for addition and conjugation. Writing $z=re^{i\theta}$ is best for multiplication, division, and powers: moduli multiply and arguments add. The argument is not unique; $\theta$ and $\theta+2\pi k$ describe the same nonzero complex number. That periodicity is exactly why an equation such as $z^n=w$ has $n$ distinct roots when $w\ne0$ .

Worked example 1: simplify a quotient

Compute $\dfrac{3-4i}{1+2i}$ . Multiply numerator and denominator by the conjugate $1-2i$ :

$\frac{3-4i}{1+2i} =\frac{(3-4i)(1-2i)}{(1+2i)(1-2i)} =\frac{-5-10i}{5}=-1-2i.$

The denominator becomes the real number $|1+2i|^2=5$ . As a quick check, $(-1-2i)(1+2i)=3-4i$ .

Worked example 2: find every cube root of $8i$

Since $8i=8e^{i(\pi/2+2\pi k)}$ , its cube roots have modulus $2$ and arguments

$\frac{\pi/2+2\pi k}{3}=\frac{\pi}{6}+\frac{2\pi k}{3}, \qquad k=0,1,2.$

Thus the roots are $2e^{i\pi/6}$ , $2e^{i5\pi/6}$ , and $2e^{i3\pi/2}$ . Values with other integers $k$ repeat these three roots.

Check your understanding

Why is $\sqrt{z_1z_2}=\sqrt{z_1}\sqrt{z_2}$ not a safe identity when each square root means the principal square root?

Show the answer

Principal arguments are forced into one interval, so adding two principal arguments can cross the branch cut. For example, the principal square roots give $\sqrt{-1}\sqrt{-1}=i^2=-1$ , while $\sqrt{(-1)(-1)}=\sqrt{1}=1$ .

Definition of ℂ\mathbb C