Definition 1.3.1. A complex number is an ordered pair \((x,y)\) of real numbers. The set \(\mathbb C\) of all complex numbers is, therefore, \(\mathbb R^2.\)
We define the distinguished elements of \(\mathbb C\) by
For all \((x,y),(u,v)\in\mathbb C\), define
\[(x,y)+(u,v)=(x+u,y+v),\] \[-(x,y)=(-x,-y),\]and
\[(x,y)\cdot(u,v)=(xu-yv,xv+yu).\]We write
\[\mathcal C = \left( \mathbb C, 0_{\mathbb C}, 1_{\mathbb C}, +, -, \cdot \right)\]for the resulting \(\mathcal L_{\mathrm{field}}\)-structure.
Theorem 1.3.2. The structure \(\mathcal C\) is a field. Equivalently,
\[\mathcal C\vDash T_{\mathrm{field}}.\]Proof. The associativity and commutativity of addition follow coordinatewise from the corresponding properties of addition in \(\mathbb R\). Moreover,
\[(x,y)+(0,0)=(x,y)\]and
\[(x,y)+(-x,-y)=(0,0).\]Therefore, \((\mathbb C,+)\) is an abelian group.
For multiplication, let \((x,y),(u,v),(r,s)\in\mathbb C\). Direct calculation gives
whereas
\[\begin{aligned} (x,y)\left((u,v)(r,s)\right) &= (x,y)(ur-vs,us+vr)\\ &= (xur-xvs-yus-yvr,\;xus+xvr+yur-yvs). \end{aligned}\]Hence multiplication is associative. It is also commutative, since
\[(x,y)(u,v)=(xu-yv,xv+yu)=(ux-vy,uy+vx)=(u,v)(x,y).\]Furthermore,
\[(x,y)(1,0)=(x,y).\]To verify distributivity, observe that
\[\begin{aligned} (x,y)\left((u,v)+(r,s)\right) &= (x,y)(u+r,v+s)\\ &= (xu+xr-yv-ys,\;xv+xs+yu+yr)\\ &= (xu-yv,xv+yu)+(xr-ys,xs+yr)\\ &= (x,y)(u,v)+(x,y)(r,s). \end{aligned}\]Finally, suppose that \((x,y)\neq(0,0)\). Then at least one of \(x\) and \(y\) is nonzero, and therefore \(x^2+y^2>0.\) Define
\[(x,y)^{-1}= \left( \frac{x}{x^2+y^2}, \frac{-y}{x^2+y^2} \right).\]Then
\[\begin{aligned} (x,y)(x,y)^{-1} &= \left( \frac{x^2+y^2}{x^2+y^2}, \frac{-xy+xy}{x^2+y^2} \right)\\ &= (1,0). \end{aligned}\]Also, \(0_{\mathbb C}\neq1_{\mathbb C}\), since \((0,0)\neq(1,0)\). Consequently, \(\mathcal C\vDash T_{\mathrm{field}}.\) \(\square\)
Theorem 1.3.3. We define the map \(\iota:\mathbb R\to\mathbb C\) by
\[\iota(x)=(x,0).\]Then \(\iota\) is an \(\mathcal L_{\mathrm{field}}\)-embedding of the real field into the complex field.
Proof. The map \(\iota\) is injective, since \(\iota(x)=\iota(y)\) implies \((x,0)=(y,0),\) and hence \(x=y\). Moreover,
\[\iota(0)=(0,0)=0_{\mathbb C}, \qquad \iota(1)=(1,0)=1_{\mathbb C},\]and, for all \(x,y\in\mathbb R\),
\[\iota(x+y)=(x+y,0)=(x,0)+(y,0)=\iota(x)+\iota(y),\] \[\iota(-x)=(-x,0)=-\iota(x),\]and
\[\iota(xy)=(xy,0)=(x,0)(y,0)=\iota(x)\iota(y).\]Therefore, \(\iota\) preserves the interpretations of all symbols of \(\mathcal L_{\mathrm{field}}\). \(\square\)
Remark 1.3.4. The image \(\iota\left[\mathbb R\right]\) therefore forms a subfield of \(\mathcal C\) isomorphic to \(\mathcal R\). We henceforth identify each \(x\in\mathbb R\) with its image \((x,0)\in\mathbb C\). Under this identification, we simply write \(\mathbb R\subseteq\mathbb C.\)
Definition 1.3.5. The imaginary unit is the complex number
\[i=(0,1).\]It satisfies
\[i^2=(0,1)(0,1)=(-1,0)=-1.\]Theorem 1.3.6. For every \(z\in\mathbb C\), there exist unique \(x,y\in\mathbb R\) such that
\[z=x+yi.\]Proof. Let \(z=(x,y)\in\mathbb C\). Then,
\[\begin{aligned} x+yi &= (x,0)+(y,0)(0,1)\\ &= (x,0)+(0,y)\\ &= (x,y)\\ &= z. \end{aligned}\]For uniqueness, suppose that \(x+yi=u+vi\) for some \(x,y,u,v\in\mathbb R\). In terms of ordered pairs, this equality is \((x,y)=(u,v).\) Consequently, \(x=u\) and \(y=v\). \(\square\)
Definition 1.3.7. If \(z=x+yi,\) where \(x,y\in\mathbb R\), then \(x\) is called the real part of \(z\), and \(y\) is called the imaginary part of \(z\). We write
\[\operatorname{Re}(z)=x \qquad\text{and}\qquad \operatorname{Im}(z)=y.\]The complex number \(\overline z = x + (- y)i\) is called the conjugate of \(z.\)That is, \(\operatorname{Re} (\overline z) = \operatorname{Re} (z)\) and \(\operatorname{Im} (\overline z) = - \operatorname{Im} (z).\)
Proposition 1.3.8. Let \(z,w\in \mathbb C.\) Then
- \(\overline{z+w} = \overline z + \overline w,\)
- \(\overline {zw} = \overline z \cdot \overline w,\)
- \(z+\overline z = 2 \operatorname{Re}(z),\)
- \(z\overline z \in \mathbb R_{>0},\) if \(z\ne 0.\)
Proof. Let \(z=(x,y)\) and \(w=(u,v)\) where \(x,y,u,v\in \mathbb R.\)
(1) Since \(z+w = (x,y) + (u,v) = (x+u,y+v),\) \(\overline {z+w} = (x+u,-(y+v)).\) Also \(\overline z + \overline w = (x,-y) + (u,-v) = (x+u,(-y)+(-v)) = (x+u, -(y+v)).\) Hence \(\overline{z+w} = \overline z + \overline w.\)
(2) Since \(zw = (xu-yv,xv+yu),\) \(\overline {zw} = (xu-yv,-(xv+yu)).\) Also \(\overline z \cdot \overline w = (x,-y)(u,-v) = (xu-yv,-xv-yu) = (xu-yv,-(xv+yu)).\) Hence \(\overline {zw} = \overline z \cdot \overline w.\)
(3) \(z+\overline z = (x,y) + (x,-y) = (2x,0) = 2x = 2\operatorname{Re}(z).\)
(4) \(z\overline z = (x,y)(x,-y) = (x^2+y^2,0)=x^2+y^2.\) Since \(z\ne 0,\) at least one of \(x\) and \(y\) is nonzero so that \(0<x^2+y^2 = z\overline z.\) \(\square\)
Definition 1.3.9. If \(z\in \mathbb C,\) its modulus \(|z|\) is algebraically defined by
\[|z|^2 = z\overline z.\]The existence and uniqueness of \(|z|\) follows from Theorem 1.1.11 and (4) of Proposition 1.3.8.
Note that if \(z\) is real, then \(\overline z = z,\) hence \(|z| = \left(z^2\right)^{\! 1/2}.\) Thus \(|z| = z\) if \(0\le z,\) or \(|z| = -z\) if \(z<0.\)
Proposition 1.3.10. If \(z,w\in \mathbb C,\) then
- \(|zw| = |z| |w|,\)
- \(|\operatorname {Re} (z) | \le |z|,\)
Proof. Let \(z = (x,y)\) and \(w = (u,v)\) where \(x,y,u,v\in \mathbb R.\)
(1) Since \(zw = (xu-yv,xv+yu),\) \((zw)\overline {(zw)} = x^2u^2 + y^2v^2 + x^2v^2 + y^2u^2 = (x^2+y^2)(u^2+v^2).\) Hence \(|zw| = \left((x^2+y^2)(u^2+v^2)\right)^{\!1/2}=\left(x^2+y^2\right)^{\!1/2}\left(u^2+v^2\right)^{\!1/2} = |z||w|.\)
(2) We let \(a=\left( x^2\right)^{1/2}\) and \(b = \left( x^2+y^2\right)^{1/2}.\) Then we have \(a^2\le b^2.\) Since \(a\) and \(b\) are nonnegative, \(a\le b.\) Thus \(|\operatorname {Re} (z) | \le |z|.\)\(\square\)
Theorem 1.3.11. If \(z,w\in \mathbb C,\) then
\[|z+w| \le |z| + |w|.\]Proof. Since we are more familiar with the field operations than moduli, it is reasonable to translate the terms into the field language:
\[|z+w|^2 = (z+w)\overline {(z+w)},\quad \text{and} \quad (|z|+|w|)^2= z\overline z + 2|z||w| + w\overline w.\]Since \(\overline {(z+w)} = \overline z + \overline w,\) we have \((z+w)\overline {(z+w)} = z\overline z + z\overline w + \overline z w + w\overline w.\) Thus, it is enough to examine the relation between
\[z\overline w + \overline z w, \quad \text{and} \quad 2|z||w|.\]Let \(u=z\overline w\) and \(v=\overline z w.\) Then \(\overline v= u,\) so that \(\text{LHS} = 2\operatorname{Re} (u).\) Also we observe that \(\text{RHS} = 2|z| |\overline w| = 2|z\overline w|= 2|u|.\) (2) of Proposition 1.3.10 implies \(\operatorname{Re} (u) \le |u|,\) and this proves \(\text{LHS} \le \text{RHS}.\) Therefore \(|z+w| \le |z| + |w|.\) \(\square\)
Theorem 1.3.12. If \(x_1,\dots,x_n\) and \(y_1,\dots,y_n\) are complex numbers, then
\[\left| \sum_{i=1}^n x_i\overline {y_i}\right|^2\le \sum_{i=1}^n |x_i|^2 \sum_{i=1}^n |y_i|^2.\]Proof. Let \(a_i=x_i\overline {x_i}, b_i=y_i\overline {y_i},\) and \(c_i=x_i\overline {y_i}.\) Then
\[\begin{align} \text{LHS} &= \left( \sum_{i=1}^n c_i\right) \left( \sum_{i=1}^n \overline {c_i}\right)\\ &= \sum_{\substack{1\le i,j\le n \\ i=j}}c_i\overline {c_j} + \sum_{1\le i<j\le n}(c_i\overline {c_j} + c_j\overline {c_i})\\ &=\sum_{i=1}^n c_i\overline {c_i} + \sum_{1\le i<j\le n}2\operatorname{Re} (c_i\overline {c_j}), \end{align}\]and
\[\begin{align} \text{RHS} &= \left( \sum_{i=1}^n a_i\right) \left( \sum_{i=1}^n {b_i}\right)\\ &= \sum_{i=1}^n a_i {b_i} + \sum_{1\le i<j\le n}(a_ib_j+a_jb_i). \end{align}\]Since
\[\begin{align} 2 \operatorname{Re}(c_i\overline{c_j}) &\le 2 |c_i\overline{c_j}|\\ &=2 \bigl(|x_i||y_j|\bigr)\bigl(|x_j||y_i|\bigr)\\ &\le |x_i|^2|y_j|^2+|x_j|^2|y_i|^2\\ &= a_i b_j+a_j b_i, \end{align}\]and
\[\sum_{i=1}^n c_i\overline {c_i} = \sum_{i=1}^n a_i {b_i},\]we finally have
\[\begin{align} \text{LHS} &=\sum_{i=1}^n c_i\overline {c_i} + \sum_{1\le i<j\le n}2\operatorname{Re} (c_i\overline {c_j})\\ &\le \sum_{i=1}^n a_i {b_i} + \sum_{1\le i<j\le n}(a_ib_j+a_jb_i)\\ &=\text{RHS}. \tag*{\(\square\)} \end{align}\]Exercises
Exercise 1. If \(z_1,\dots,z_n\) are complex, prove that
\[|z_1+z_2+\cdots +z_n| \le |z_1| + |z_2| + \cdots + |z_n|.\]Proof. The case \(n=1\) is immediate, since \(\vert{}z_1\vert{} \le \vert{}z_1\vert{}.\) The case \(n=2\) holds by Theorem 1.3.11.
Now we suppose that for some \(n\ge 3,\)
Then, applying Theorem 1.3.11, we obtain
\[\begin{aligned} \vert{}z_1+\cdots+z_n+z_{n+1}\vert{} &= \vert{}(z_1+\cdots+z_n)+z_{n+1}\vert{}\\ &\le \vert{}z_1+\cdots+z_n\vert{}+\vert{}z_{n+1}\vert{}\\ &\le \vert{}z_1\vert{}+\cdots+\vert{}z_n\vert{}+\vert{}z_{n+1}\vert{}. \end{aligned}\]Therefore, by mathematical induction,
\[\vert{}z_1+\cdots+z_n\vert{} \le \vert{}z_1\vert{}+\cdots+\vert{}z_n\vert{}\]for every positive integer \(n.\) \(\square\)
Exercise 2. If \(x\) and \(y\) are complex, prove that
\[\big\vert|x|-|y|\big\vert\le |x-y|.\]Proof. Let \(|x|=a,|y|=b,\) and \(|x-y|=c.\) Since \(|x\overline y| = \left( (x\overline y)(\overline x y\right)^{1/2} = ab,\) \(\operatorname{Re} (x\overline y)\le ab,\) by Proposition 1.3.10. Then we have
\[a^2+b^2-2ab\le a^2+b^2-2\operatorname{Re} (x\overline y).\]Since
\[\begin{align} a^2+b^2-2\operatorname{Re} (x\overline y) &= x\overline x + y\overline y -x\overline y -\overline x y\\ &= (x-y)(\overline x -\overline y)\\ &=(x-y)\overline {(x-y)} \\ &= c^2, \end{align}\]we have \((a-b)^2\le c^2.\) Taking square roots yields \(\big\vert|x|-|y|\big\vert\le |x-y|.\) \(\square\)
References
- Rudin, W. (1976). Principles of mathematical analysis (3rd ed.). McGraw-Hill Education.