A complex number is an ordered pair \((a, b)\in\mathbb R\times \mathbb R\).
For two complex numbers \(x=(a, b), y=(c, d)\in\mathbb R\times \mathbb R\) we define
addition \({\color{blue}+}\) by setting \[x+y=(a+c, b+d),\]
multiplication \({\color{blue}\cdot}\) by setting \[x\cdot y=(ac-bd, ad+bc).\]
These operations addition \({\color{blue}+}\) and multilpication \({\color{blue}\cdot}\) turn the set of all complex numbers into a field with \((0,0)\) and \((1, 0)\) playing, respectively, the role of \(0\) and \(1\). This field will be denoted by \(\mathbb C\).
Proof. We have to verify the field axioms.
Addition axioms (A)
(A1) if \(x,y \in \mathbb{C}\), then \(x+y \in \mathbb{C}\),
(A2) \(x+y=y+x\) for all \(x,y \in \mathbb{C}\),
(A3) \((x+y)+z=x+(y+z)\) for all \(x,y,z \in \mathbb{C}\),
(A4) \(\mathbb{C}\) contains the element \(0\) such that \(x+0=x\) for all \(x \in \mathbb{C}\),
(A5) to every \(x \in \mathbb{C}\) corresponds an element \((-x) \in \mathbb{C}\) such that \[x+(-x)=0.\]
Multiplication axioms (M)
(M1) if \(x,y \in \mathbb{C}\), then their product \(xy \in \mathbb{C}\),
(M2) \(xy=yx\) for all \(x,y \in \mathbb{C}\),
(M3) \((xy)z=x(yz)\) for all \(x,y,z \in \mathbb{C}\),
(M4) \(\mathbb{C}\) contains the element \(1\neq0\) such that \(1\cdot x=x\) for all \(x \in \mathbb{C}\),
(M5) if \(0\neq x \in \mathbb{C}\) then there is an element \({x}^{-1} =\frac{1}{x}\in \mathbb{C}\) such that \[x \cdot {x}^{-1}=1.\]
Distributive law (D)
(D1) \(x(y+z)=xy+xz\) holds for all \(x,y,z \in \mathbb{C}\).
Let \(x=(a, b), y=(c,d), z=(e,f)\). We will use the field structure of \(\mathbb R\).
Proof of (A1). By the definition of addition \[x+y=(a, b)+(c, d)=(a+c, b+d)\in\mathbb C.\]
Proof of (A2). \[x+y=(a+c, b+d)=(c+a)+(d+b)=y+x.\]
Proof of (A3). \[\begin{aligned} (x+y)+z&=(a+c, b+d)+(e, f)\\ &=(a+c+e, b+d+f)\\ &=(a, b)+(c+e, d+f)=x+(y+z). \end{aligned}\]
Proof of (A4). \[x+0=(a, b)+(0,0)=(a, b)=x.\]
Proof of (A5). Set \(-x=(-a, -b)\) and note that \[x+(-x)=(a-a, b-b)=(0,0)=0.\]
Proof of (M1). By the definition of multiplication \[x\cdot y=(a, b)\cdot(c, d)=(ac-bd, ad+bc)\in\mathbb C.\]
Proof of (M2). \[x\cdot y=(ac-bd, ad+bc)=(ca-db, da+cb)=y\cdot x.\]
Proof of (M3). \[\begin{aligned} (x\cdot y)\cdot z&=(ac-bd, ad+bc)\cdot(e, f)\\ &=(ace-bde-adf-bcf, acf-bdf+ade+bce)\\ &=(a, b)\cdot(ce-df, cf+de)=x\cdot (y\cdot z). \end{aligned}\]
Proof of (M4). \[1\cdot x=(1, 0)\cdot (a, b)=(a, b)=x.\]
Proof of (M5). If \(x\neq0\) then \((a, b)\neq(0, 0)\), which means that at least one of the real numbers \(a, b\) is different from \(0\). Hence \(a^2+b^2>0\) and we define \[\frac{1}{x}=\bigg(\frac{a}{a^2+b^2}, \frac{-b}{a^2+b^2}\bigg).\] Then \[x\cdot \frac{1}{x}=(a, b)\cdot\bigg(\frac{a}{a^2+b^2}, \frac{-b}{a^2+b^2}\bigg)=(1, 0).\]
Proof of (D1). \[\begin{aligned} x\cdot(y+z)&=(a, b)\cdot(c+e, d+f)\\ &=(ac+ae-bd-bf, ad+af+bc+be)\\ &=(ac-bd, ad+bc)+(ae-bf, af+be)\\ &=x\cdot y+x\cdot z. \end{aligned}\] This completes the proof that \(\mathbb C\) is a field.$$\tag*{$\blacksquare$}$$
Remark. For any \(a, b\in \mathbb R\) we have \[ (a, 0)+(b, 0)=(a+b, 0) \quad \text{ and } \quad (a, 0)\cdot (b, 0)=(ab, 0).\]
The complex numbers from the set \(\{(a, 0): a\in\mathbb R\}\) have the same arithmetic properties as the corresponding real numbers \(\mathbb R\).
We can therefore identify \((a, 0)\) with \(a\). This identification gives us the real field \(\mathbb R\) as a subfield of the complex field \(\mathbb C\).
We have defined the complex numbers \(\mathbb C\) without any reference to the mysterious square root of \(-1\). We now show that the notation \((a, b)\) is equivalent to the more customary \(a + bi\).
We define the imaginary number by setting \(i=(0, 1)\).
One has that \(i^2=-1\).
Proof. Note that \(i^2=(0, 1)\cdot(0, 1)=(-1, 0).\)$$\tag*{$\blacksquare$}$$
We also have \[\mathbb C=\{a+ib: a, b\in\mathbb R\}.\]
Proof. It suffices to note that \[\begin{aligned} a+ib&=(a, 0)+(0,1)\cdot(b, 0)\\ &=(a, 0)+(0, b)=(a, b). \end{aligned}\]$$\tag*{$\blacksquare$}$$
If \(z\in\mathbb C\) and \(z=a+ib\) for some \(a,b \in\mathbb R\) then the complex number \[\overline{z}=a-ib\] is called the conjugate of \(z\). The numbers \(a\) and \(b\) are the real part and imaginary part of \(z\) respectively. We shall write \[a=\Re(z)={\rm Re}(z) \quad\text{ and } \quad b=\Im(z)={\rm Im}(z).\]
If \(z, w\in\mathbb C\) then
\(\overline{z+w}=\overline{z}+\overline{w}.\)
\(\overline{zw}=\overline{z} \cdot \overline{w}.\)
\(z+\overline{z}=2{\rm Re}(z)\) and \(z-\overline{z}=2i{\rm Im}(z)\).
\(z\overline{z}\) is a positive real number except when \(z=0\).
Proof. Let \(z=a+ib\) and \(w=c+id\).
Proof of (i). Note that \[\overline{z+w}=\overline{(a+c)+i(b+d)}=(a+c)-i(b+d)=\overline{z}+\overline{w}.\]
Proof of (ii). Note that \[\begin{aligned} \overline{z\cdot w}&=(ac-bd)-i (ad+bc) \quad \text{ and }\\ \overline{z} \cdot \overline{w}&=(a-ib)(c-id)=(ac-bd)-i (ad+bc). \end{aligned}\]
Proof of (iii). We have \[\begin{aligned} z+\overline{z}&=(a+ib)+(a-ib)=2a=2{\rm Re}(z),\\ z-\overline{z}&=(a+ib)-(a-ib)=2ib=2i{\rm Im}(z). \end{aligned}\]
Proof of (iv). We have \(z\cdot\overline{z}= (a+ib)(a-ib)=a^2+b^2>0\) if and only if \(z\neq0\). $$\tag*{$\blacksquare$}$$
If \(z\in\mathbb C\) its absolute value \(|z|\) is defined by setting \[|z|=\sqrt{z\cdot\overline{z}}.\]
Remark. This absolute value exists and is unique. Moreover, it coincides with the absolute value from \(\mathbb R\). If \(x\in \mathbb R\) then \(\overline{x}=x\) hence \(|x|=\sqrt{x\cdot \overline{x}}=\sqrt {x^2}\). Thus \[|x|= \begin{cases} x & \text{ if } x\ge 0,\\ -x& \text{ if } x< 0. \end{cases}\]
If \(z, w \in\mathbb C\) then
\(|z|>0\) if and only if \(z\neq 0\), and \(|0|=0\).
\(|\overline{z}|=|z|\).
\(|zw|=|z||w|\).
\(|{\rm Re}(z)|\le |z|\) and \(|{\rm Im}(z)|\le|z|\)
\(|z+w|\le |z|+|w|\).
Proof. Let \(z=a+ib\) and \(w=c+id\).
Proof of (i). From the previous theorem we have \[|z|^2=z\cdot\overline{z}= (a+ib)(a-ib)=a^2+b^2>0,\] which gives the desired claim.
Proof of (ii). Note that \(|z|^2=a^2+b^2=|\overline{z}|^2\).
Proof of (iii). Note that \[\begin{aligned} |z\cdot w|=(ac-bd)^2+ (ad+bc)^2 =(a^2+b^2)(c^2+d^2)=|z|^2|w|^2. \end{aligned}\]
Proof of (iv). We have \[\begin{aligned} |{\rm Re}(z)|=|a|\le \sqrt{a^2+b^2}=|z|, \ \text{ and } \ |{\rm Im}(z)|=|b|\le \sqrt{a^2+b^2}=|z|. \end{aligned}\]
Proof of (v). Note that \(\overline{z}w\) is the conjugate of \(z\overline{w}\) so that \(z\overline{w}+\overline{z}w=2{\rm Re}(z\overline{w})\). Hence \[\begin{aligned} |z+w|^2&=(z+w)(\overline{z}+\overline{w})=z\overline{z}+z\overline{w}+\overline{z}w+ w\overline{w}\\ &=|z|^2+2{\rm Re}(z\overline{w})+|w|^2\\ &\le |z|^2+2|{\rm Re}(z\overline{w})|+|w|^2\\ &\le|z|^2+2|z||w|+|w|^2=(|z|+|w|)^2. \end{aligned}\]
The proof of the theorem is completed. $$\tag*{$\blacksquare$}$$
An argument \(\arg(z)\) of \(z=a+ib\in \mathbb C\) is defined as the angle which the line segment from \((0,0)\) to \((a, b)\) makes with the positive real axis. The argument is not unique, but is determined up to a multiple of \(2 \pi\).
If \(r=|z|\) and \(\theta=\arg(z)\) is an argument of \(z\in\mathbb C\), we may write \[z=r(\cos \theta+i \sin \theta).\] Then for \(z_1, z_2, z\in\mathbb C\) it follows from trigonometric identities that \[\begin{aligned} \arg \left(z_{1} z_{2}\right)&=\arg (z_{1})+\arg (z_{2}),\\ \arg \left(z_{1} / z_{2}\right)&=\arg \left(z_{1}\right)-\arg \left(z_{2}\right) \quad \text{ if } \quad z_2\neq 0,\\ \arg (\overline{z})&=-\arg (z). \end{aligned}\] The argument of \(z\) is called principal if \(\arg(z)\in(-\pi, \pi]\).
We say that a sequence of complex numbers \((z_n)_{n\in\mathbb N}\subseteq \mathbb C\) converges to \(z\in \mathbb C\) and write \(\lim_{n\to \infty}z_n=z\) if and only if \[\lim_{n\to \infty}|z_n-z|=0.\]
This is also equivalent to say that for every \(\varepsilon>0\) there exists an integer \(N_{\varepsilon}\in\mathbb N\) such that if \(n\ge N_{\varepsilon}\) then \[|z_n-z|<\varepsilon.\]
Obviously \(\lim_{n\to \infty}z_n=z\) iff \(\lim_{n\to \infty}{\rm Re} (z_n)={\rm Re} (z)\) and \(\lim_{n\to \infty}{\rm Im} (z_n)={\rm Im} (z)\).
We say that \(z_n\ _{\overrightarrow{n\to \infty}}\ \infty\) diverges iff \(\lim_{n\to \infty}|z_n|=\infty\).
We say that a sequence of complex numbers \((z_n)_{n\in\mathbb N}\subseteq \mathbb C\) is said to be a Cauchy sequence in \(\mathbb C\) (or simply Cauchy) iff \[\lim_{m, n\to\infty}|z_n-z_m|=0.\]
This is also equivalent to say that for every \(\varepsilon>0\) there exists an integer \(N_{\varepsilon}\in\mathbb N\) such that if \(m, n\ge N_{\varepsilon}\) then \[|z_n-z_m|<\varepsilon.\]
The complex plane \(\mathbb C\) with a metric given by \[d\left(z_{1}, z_{2}\right)=\left|z_{1}-z_{2}\right|\quad \text { for } \quad z_{1}, z_{2} \in \mathbb{C}\]
is a complete metric space.
If \(a \in \mathbb{C}\) and \(r>0\), we define the open disc \(D\left(a, r\right)\) of radius \(r\) centered at \(a\) to be the set of the form \[D(a, r)=\{z\in \mathbb{C}: |z-a|<r\}.\]
We write \(D=D(0, 1)\) for the open unit disc centered at the origin.
The closed disc \(\overline{D}\left(a, r\right)\) of radius \(r\) centered at \(a\) is defined by \[\overline{D}(a, r)=\{z\in \mathbb{C}: |z-a|\le r\}.\]
The punctured disc \(D'\left(a, r\right)\) of radius \(r\) centered at \(a\) is defined by \[D^{\prime}(a, r)=\{z\in \mathbb{C}: 0<|z-a|<r\}.\] We observe that it is an open set.
The circle \(C\left(a, r\right)\) of radius \(r\) centered at \(a\) is defined by \[C(a, r)=\{z\in \mathbb{C}: |z-a|=r\}=\overline{D}(a, r)\setminus D(a, r).\]
A set \(\Omega\subseteq \mathbb C\) is open if for every \(a\in \Omega\) there exists \(r>0\) such that \[D(a, r)\subseteq \Omega.\]
A set \(\Omega\) is closed if its complement \(\Omega^{c}=\mathbb{C}\setminus\Omega\) is open.
This property can be reformulated in terms of limit points. A point \(z \in \mathbb{C}\) is said to be a limit point of the set \(\Omega\) if there exists a sequence of points \((z_{n})_{n\in\mathbb N} \subseteq \Omega\) such that \(z_{n} \neq z\) and \(\lim _{n \rightarrow \infty} z_{n}=z\).
One can check that a set is closed if and only if it contains all its limit points. The closure of any set \(\Omega\) is the union of \(\Omega\) and its limit points, and is often denoted by \(\overline{\Omega}\).
Finally, the boundary of a set \(\Omega\) is equal to its closure minus its interior, and is often denoted by \(\partial \Omega\).
For instance the circle \(C(a, r)\) is the boundary of the disc \(D(a, r)\).
If \(a\) and \(b\) are complex numbers, \([a, b]\) denotes the closed line segment with endpoints \(a\) and \(b\).
If \(t_{1}\) and \(t_{2}\) are arbitrary real numbers with \(t_{1}<t_{2}\), then we may write \[[a, b]=\left\{a+\frac{t-t_{1}}{t_{2}-t_{1}}(b-a): t_{1} \leq t \leq t_{2}\right\}\]
The notation is extended as follows. If \(a_{1}, a_{2}, \ldots, a_{n+1}\) are points in \(\mathbb{C}\), a polygon from \(a_{1}\) to \(a_{n+1}\) (or a polygon joining \(a_{1}\) to \(a_{n+1}\) ) is defined as \[\bigcup_{j=1}^{n}\left[a_{j}, a_{j+1}\right],\] often abbreviated as \(\left[a_{1}, \ldots, a_{n+1}\right]\).
Let \((X, d)\) be a metric space and \(E \subseteq X\).
A set \(E\) is connected if \(E\) cannot be written as a disjoint union of two non-empty relative open subsets of \(E\).
Thus \(E=A \cup B\) with \(A \cap B=\varnothing\) and \(A, B\) open in \(E\) implies that either \(A=\varnothing\) or \(B=\varnothing\). Otherwise \(E=A \cup B\) is called a separation \(E\) into open sets.
Union \(E\) of two disjoint open discs \(A\) and \(B\) is not connected since \[E=A \cup B=(A \cap E) \cup(B \cap E),\] and \(A \cap E\) and \(B \cap E\) are non-empty, disjoint and relatively open in \(E\).
An open connected set in a metric space is called a region.
A maximal connected subset of \(E\) is called a component of \(E\)
For \(a \in E\), let \(C(a)\) be the union of all connected subsets of \(E\) containing \(a\). We observe that \(a \in C(a)\) since \(\{a\}\) is connected and \[E=\bigcup_{a \in E} C(a) .\]
\(C(a)\) is connected.
The components of \(E\) are either disjoint or identical.
The components of an open set are open.
By combining (i), (ii) and (iii), we conclude:
An open set in a metric space is a disjoint union of regions.
Proof of (i)
We first prove that \(C(a)\) is connected. The proof is by contradiction.
Let \(C(a)=A \cup B\) be a separation of \(C(a)\) into open sets.
We may assume that \(a \in A\) and \(b \in B\). Then, since \(b \in C(a)\) and \(C(a)\) is the union of all connected subsets of \(E\) containing \(a\), there exists \(E_{0} \subseteq E\) such that \(E_{0} \subseteq C(a)\) is connected and \(a, b \in E_{0}\).
Thus \[E_{0}=E_{0} \cap C(a)=E_{0} \cap(A \cup B)=\left(E_{0} \cap A\right) \cup\left(E_{0} \cap B\right)\] implies that either \(E_{0} \cap A=\varnothing\) or \(E_{0} \cap B=\varnothing\).
This is a contradiction since \(a \in E_{0} \cap A\) and \(b \in E_{0} \cap B\). $$\tag*{$\blacksquare$}$$
Thus every component of \(E\) is of the form \(C(a)\) with \(a \in E\).
Proof of (ii)
The components of \(E\) are either disjoint or identical. Let \(a, b \in E\).
Assume that \(C(a) \cap C(b) \neq \varnothing\). We prove that \(C(a)=C(b)\).
Let \(x \in C(a) \cap C(b)\). Then \(x \in C(a)\). Since \(C(a)\) is connected, we derive that \(C(a) \subseteq C(x)\). Then \(a \in C(x)\) which implies \(C(x) \subseteq C(a)\) since \(C(x)\) is connected. Thus \(C(a)=C(x)\).
Similarly \(C(b)=C(x)\) and hence \(C(a)=C(b)\).$$\tag*{$\blacksquare$}$$
Proof of (iii)
The components of an open set are open. Let \(E\) be an open set.
It suffices to show that \(C(a)\) with \(a \in E\) is open. Let \(x \in C(a)\).
Then \(C(x)=C(a)\) by (ii). Since \(x \in E\) and \(E\) is open, then \(D(x, r) \subseteq E\) for some \(r>0\). In fact \(D(x, r) \subseteq C(x)\) since \(D(x, r)\) is connected containing \(x\). Thus \(x \in D(x, r) \subseteq C(a)\) and hence \(C(a)\) is open as desired. $$\tag*{$\blacksquare$}$$
Let \(E\) be a non-empty open subset of \(\mathbb{C}\). Then \(E\) is connected if and only if any two points in \(E\) can be joined by a polygonal path that lies in \(E\).
Proof \((\Longrightarrow)\).
Assume that \(E\) is connected. Since \(E \neq \varnothing\), let \(a \in E\). Let \(E_{1}\) be the subset of all elements of \(E\) that can be joined to \(a\) by a polygonal path. Let \(E_{2}\) be the complement of \(E_{1}\) in \(E\). Then \[E=E_{1} \cup E_{2} \quad \text { with } \quad E_{1} \cap E_{2}=\varnothing, \quad \text{ and } \quad a \in E_{1} .\]
It suffices to show that both \(E_{1}\) and \(E_{2}\) are open subsets of \(E\).
Then \(E_{2}=\varnothing\) since \(E\) is connected and \(a \in E_{1}\). Thus every point of \(E\) can be joined to \(a\) by a polygonal path that lies in \(E\). Hence any two points of \(E\) can be joined by a polygonal path that lies in \(E\) via \(a\).
First, we show that \(E_{1}\) is open. Let \(a_{1} \in E_{1}\), then \(a_{1} \in E\) and since \(E\) is open, we have \(D\left(a_{1}, r_{1}\right) \subseteq E\) for some \(r_{1}>0\).
Any point of \(D\left(a_{1}, r_{1}\right)\) can be joined to \(a_{1}\) and hence to \(a\) by a polygonal path that lies in \(E\) since \(a_{1} \in E_{1}\). Thus \[a_{1} \in D\left(a_{1}, r_{1}\right) \subseteq E_{1}.\]
Next, we show that \(E_{2}\) is open. Let \(a_{2} \in E_{2}\). Again we find \(r_{2}>0\) such that \(D\left(a_{2}, r_{2}\right) \subseteq E\) since \(E\) is open.
Now, as above, we see that no point of this disc can be joined to \(a\) as \(a_{2} \in E_{2}\) and hence \(a_{2} \in D\left(a_{2}, r_{2}\right) \subseteq E_{2}\). $$\tag*{$\blacksquare$}$$
Proof \((\Longleftarrow)\).
Now we assume that any two points of \(E\) can be joined by a polygonal path in \(E\) and we show that \(E\) is connected.
Let \(E=E_{1} \cup E_{2}\) be a separation of \(E\) into open sets.
Let \(a_{1} \in E_{1}\) and \(a_{2} \in E_{2}\) be such that \[\chi(t)=t a_{1}+(1-t) a_{2} \quad \text { with } \quad 0<t<1\] is an open segment from \(a_{2}\) to \(a_{1}\) lying in \(E\).
Let \[V=\left\{t \in(0,1) \mid \chi(t) \in E_{1}\right\} \quad \text { and } \quad W=\left\{t \in(0,1) \mid \chi(t) \in E_{2}\right\} .\]
We see that \(V\) and \(W\) are open in \((0,1)\). Further, we have separation of the open interval \(( 0,1 )\) into open sets \[(0,1)=V \cup W, \quad V \cap W=\varnothing .\]
Since \(a_{1} \in E_{1}\) and \(E_{1}\) is open, then \(D\left(a_{1}, r_{3}\right) \subseteq E_{1}\) for some \(r_{3}>0\). This implies \(V \neq \varnothing\). Similarly \(W \neq \varnothing\).
Hence the interval \((0,1)\) is not connected. This is a contradiction. $$\tag*{$\blacksquare$}$$