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The basic arithmetic operations
The sum and the product of two complex numbers is a complex number, because in the definition above \(a + c\), \(b + d\), \(ac - bd\), and \(ad + bc\) are real numbers. Other notations often used:
\[a + b\iu = a + \iu b = a + b\ju = a + \ju b.\]
Complex numbers can be visualized by expressing them as points or vectors on the complex plane:
A complex number of the form \(a + 0\iu\) is identified with the real number \(a\) and we denote \(a + 0\iu = a\). In this sense, each real number is a complex number, that is, \(\R\subset\C\). Similarly we denote \(0 + b\iu = b\iu\). Especially,
\[1 + 0\iu = 1 \qquad\text{and}\qquad 0 + 1\iu = 1\iu = \iu.\]
Hence, \(\iu\) itself is also a complex number. Some terminology for a complex number \(z=a+b\iu\):
- \(a = \re z\) is the real part of \(z\) and \(b = \im z\) is the imaginary part of \(z\).
- If \(b = 0\), then \(z\) is said to be real.
- If \(b \ne 0\), then \(z\) is imaginary.
- If \(a = 0\) and \(b \ne 0\), then \(z\) is purely imaginary.
Two complex numbers \(z\) and \(w\) are equal if their real parts are equal and their imaginary parts are equal, i.e., \(z = w\) if and only if \(\re z = \re w\) and \(\im z = \im w\).
Let us compute the product of a real number \(t=t+0\iu\) and a complex number \(a+b\iu\) using the definition:
\[t(a+b\iu)=(t+0\iu)(a+b\iu)=(ta-0\cdot b)+(tb+0\cdot a)\iu=ta+tb\iu.\]
We notice that it is equivalent to just multiply both the real and imaginary parts by \(t\). In general, the sum, the negative, the difference, and multiplication with a real number work in exactly the same way than the corresponding operations of planar vectors, see the picture below.
Example 2.2.3
- \(\re(-2-3\iu)=-2\) and \(\im(-2-3\iu)=-3\)
- \((3-2\iu)-(-5+3\iu)=8-5\iu\)
The definition of the product of complex numbers implies the following important property of \(\iu\).
Theorem 2.2.4
\(\iu^2=\iu\cdot \iu=-1\).
Hide/show proof
Write \(\iu=0+1\iu\) and compute the product using the definition. We have \(a=c=0\) and \(b=d=1\), and therefore
\[\iu^2 = \iu\cdot \iu = (0+1\iu(0+1\iu)=(0\cdot0-1\cdot1)+(0\cdot1+1\cdot0)\iu=-1+0\iu=-1.\qedhere\]
Let’s try what happens if we multiply \(a+b\iu\) and \(c+d\iu\) just like we would multiply real numbers, using the result \(\iu^2 = -1\):
\[\begin{split}\begin{aligned}
(a+b\iu)(c+d\iu)&=ac+ad\iu+bc\iu+bd\iu^2\\ &=ac+ad\iu+bc\iu-bd\\ &=(ac-bd)+(ad+bc)\iu.
\end{aligned}\end{split}\]
The last expression equals to the definition of the product of complex numbers. We conclude that we can always multiply two complex numbers as we would multiply two real binomials, taking into account that \(\iu^2=-1\).
We will get back to the geometric interpretation of the product later.
Example 2.2.5
\((-3-2\iu)(5+\iu)=-15-3\iu-10\iu-2\iu^2=-13-13i\).
For real numbers \(a=a+0\iu\) and \(b=b+0\iu\), the real and complex sums are equal. The same applies to the difference and the product. The next result implies that we can define the quotient of complex numbers such that for real numbers this quotient equals to the quotient of real numbers.
Theorem 2.2.6
For each complex number \(z\ne0\) there exists a unique complex number, denoted by \(z^{-1}\), such that \(zz^{-1}=1\). The number \(z^{-1}\) is called the reciprocal of \(z\).
Hide/show proof
We will prove the existence first. Let \(z=a+b\iu\ne0\). Define
\[w=\frac{1}{a^2+b^2}(a-b\iu).\]
A direct computation
\[zw = \frac{1}{a^2 + b^2}(a + b\iu)(a - b\iu) = \frac{1}{a^2 + b^2}(a^2 - ab\iu + ab\iu - b^2\iu^2) = \frac{a^2 + b^2}{a^2 + b^2} = 1\]
shows that \(w\) is a reciprocal of \(z\). Hence, \(z^{-1} = w\) exists.
To prove the uniqueness, assume that \(u\) and \(v\) are reciprocals of \(z \not= 0\). Then
\[u = u \cdot 1 = u(zv) = (uz)v = 1 \cdot v = v,\]
i.e., \(u\) and \(v\) are equal. This proves uniqueness. Here we used associativity of the product of complex numbers, see Theorem 1.8.
Since for each non-zero complex number has the reciprocal, we can define the quotient in the following way:
For real numbers \(a=a+0\iu\) and \(b=b+0\iu\), the real and complex operations give the same result for each of the basic arithmetic operations (sum, difference, product, quotient). In addition, according to the next theorem, complex operations satisfy the same field axioms as the real operations. In this sense, complex operations extend the real operations onto the complex plane (or onto \(\R^2\)).
Theorem 2.2.8
For any complex numbers \(x\), \(y\), and \(z\),
- \(x + y = y + x\) and \(xy = yx\), (commutativity)
- \(x + (y + z) = (x + y) + z\) and \(x(yz) = (xy)z\), (associativity)
- \(x(y + z) = xy + xz\). (distributivity)
Hide/show proof
We can prove these properties by direct computations. We take commutativity of the product as an example. Denote \(x=x_1+x_2\iu\) and \(y=y_1+y_2\iu\). Then
\[\begin{split}\begin{aligned} xy &= (x_1+x_2\iu)(y_1+y_2\iu)=x_1y_1+x_1y_2\iu+x_2y_1\iu+x_2y_2\iu^2\\ &= y_1x_1+y_1x_2\iu+y_2x_1\iu+y_2x_2\iu^2 = (y_1+y_2\iu)(x_1+x_2\iu)=yx.\qedhere \end{aligned}\end{split}\]
Lemma 2.2.9
If \(z \not= 0\) and \(w \not= 0\) are complex numbers, then \((zw)^{-1} = z^{-1}w^{-1}\), that is,
\[\frac1z\cdot\frac1w=\frac{1}{zw}.\]
Hide/show proof
By definition, \(zz^{-1}=1\) and \(ww^{-1}=1\). Using associativity and commutativity of the product, we get
\[\left(zw\right)\left(z^{-1}w^{-1}\right)=z\left(wz^{-1}\right)w^{-1}=\left(zz^{-1}\right)\left(ww^{-1}\right)=1\cdot1=1.\]
Hence, \(z^{-1}w^{-1}\) is the reciprocal of \(zw\).
Let \(v\), \(w\ne0\), and \(z\ne0\) be complex numbers. Using the definition of the quotient, associativity, commutativity, and the lemma, we can compute
\[\frac{v}{w} = \left(v\frac1w\right)\left(z\frac1z\right) = (vz)\left(\frac1w\frac1z\right) = (vz)\left(\frac{1}{wz}\right) = \frac{vz}{wz}.\]
In practice, we write directly
\[\frac{v}{w} = \frac{vz}{wz}.\]
This gives a way to compute the quotient of complex numbers \(a+b\iu\) and \(c+d\iu\ne0\). We notice that \((c + d\iu)(c - d\iu) = c^2 + d^2\) (check!) is a non-zero real number, and
(1)\[\frac{a + b\iu}{c+d\iu} = \frac{(a + b\iu)(c-d\iu)}{(c + d\iu)(c-d\iu)} = \frac{1}{c^2+d^2}(a + b\iu)(c-d\iu)=\cdots.\]
The point is that \(1/(c^2+d^2)\) is a real number, and by continuing this computation we will get the quotient in the basic form \(x+y\iu\), \(x,y\in\R\).
Example 2.2.10
- Find the reciprocal of \(2 + 3\iu\) in the form \(a + b\iu\).
- Express \(\dfrac{3 - 4\iu}{-2 + \iu}\) in the form \(a + b\iu\).
- Solve \((2 - \iu)z = 1 + \iu\) for \(z\). Give \(z\) in the form \(z=a + bi\).
Hide/show solution
- \(\displaystyle(2+3\iu)^{-1}=\frac{1}{2+3\iu}=\frac{2-3\iu}{(2+3\iu)(2-3\iu)}=\frac{2-3\iu}{4-9\iu^2}=\frac{2-3\iu}{13}=\frac{2}{13}-\frac{3}{13}\iu\)
- \(\displaystyle\frac{3-4\iu}{-2+\iu}=\frac{(3-4\iu)(-2-\iu)}{(-2+\iu)(-2-\iu)}=\frac{-10+5\iu}{5}=-2+\iu\)
- \(\displaystyle z=\frac{1+\iu}{2-\iu}=\frac{(1+\iu)(2+\iu)}{(2-\iu)(2+\iu)}=\frac{1+3\iu}{5}=\frac15+\frac35\iu\)
