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Introduction to Complex Numbers and Complex Algebra

Complex numbers were introduced so that every polynomial equation has a complete set of solutions. The real numbers are not enough for this. The equation \(x^2+1=0\) has no real solution at all, and the cubic below has only one real root, \(x=2\), because its quadratic factor \(x^2+x+1\) has no real roots: \[ x^2+1=0\quad\text{or}\quad x^3-x^2-x-2=(x-2)(x^2+x+1)=0 \] So the real numbers do not form a large enough set to contain the solutions of every polynomial equation. The complex numbers fix this.

Notation:  A complex number is written \(\z\in\C\), in one of two equivalent ways: \[ \underbrace{\z=a+ib}_{\text{mathematicians prefer this notation}} \qquad\text{or}\qquad \underbrace{\z=a+jb}_{\text{electrical engineers prefer this notation}} \] In both forms \(a\) and \(b\) are real numbers. We call \(a\) the real part of \(\z\) and \(b\) its imaginary part, and we write \[ a\equiv\operatorname{Re}(a+jb),\qquad b\equiv\operatorname{Im}(a+jb). \] The symbol \(j\) stands for a root of \(z^2+1=0\), so that \(j^2=-1\). These notes use \(j\), as electrical engineers do.

Note that the imaginary part of \(\z\) is itself a real number, namely \(b\) and not \(jb\). Also, every real number is a complex number with imaginary part zero, because \(a\in\R\) gives \(a+j0\in\C\). In symbols, \(\R\subset\C\).

Definitions

A complex number \(\z=a+jb\) can be drawn as the point \((a,b)\) in the plane. The horizontal axis is the real axis and the vertical axis is the imaginary axis, so \(\z\) is a point in the complex plane.

Algebraic Definitions

1. Summation  Let \(\z_1=a_1+jb_1\) and \(\z_2=a_2+jb_2\). Their sum is the complex number \(\z=\z_1+\z_2=a+jb\), where the real parts add and the imaginary parts add separately:

For example, with \(\z_1=2+j3\) and \(\z_2=5-j4\) we get \(\z_1+\z_2=(2+5)+j(3-4)=7-j\).

In the complex plane, adding two numbers is the same as adding the two arrows from the origin: the sum is the far corner of the parallelogram built on \(\z_1\) and \(\z_2\).

Properties of Complex Summation

For any \(\z_1,\z_2,\z_3\in\C\).

Addition of complex numbers obeys the same rules as addition of real numbers.

2. Multiplication  Let \(\z_1=a_1+jb_1\) and \(\z_2=a_2+jb_2\). Their product \(\z=\z_1\cdot\z_2\) is defined by

This is not an arbitrary rule. It is what you get by multiplying out the brackets as usual, \[ (a_1+jb_1)\cdot(a_2+jb_2)=a_1a_2+ja_1b_2+jb_1a_2+j^2b_1b_2, \] and then using \(j^2=-1\) to collect the real part \(a_1a_2-b_1b_2\) and the imaginary part \(a_1b_2+a_2b_1\).

Example  For \(\z_1=2+j3\) and \(\z_2=5-j4\): \[ \z=\z_1\cdot\z_2=(2\cdot5-3\cdot(-4))+j(2\cdot(-4)+3\cdot5)=22+j7. \]

Properties of Complex Multiplication

Let \(\z_1=a_1+jb_1\), \(\z_2=a_2+jb_2\) and \(\z_3=a_3+jb_3\). Multiplication is commutative, associative and distributes over addition:

Properties 1–3. Each follows by writing both sides out in terms of real and imaginary parts and comparing them; the calculation is routine.

More very important Definitions

Two quantities are attached to every complex number. The complex conjugate of \(\z=a+jb\) flips the sign of the imaginary part, and the modulus (or magnitude) is the distance from the origin:

Geometrically, \(\z^*\) is the mirror image of \(\z\) in the real axis, and \(|\z|\) is the length of the arrow from the origin to \(\z\).

For example, with \(\z=2+j3\) we have \(\z^*=2-j3\) and \(|\z|=\sqrt{2^2+3^2}=\sqrt{13}\).

Properties of Complex Conjugation

Conjugation passes through sums and products:

Verification. All three are checked by writing \(\z_1\) and \(\z_2\) in terms of their real and imaginary parts and comparing both sides.

Triangle Inequality

The next result says that going straight to \(\z_1+\z_2\) is never longer than going to \(\z_1\) first and then moving by \(\z_2\).

Proof  The picture explains why the inequality is true. The three numbers \(|\z_1+\z_2|\), \(|\z_1|\) and \(|\z_2|\) are the lengths of the three sides of a triangle, and no side of a triangle is longer than the other two together.

To prove it algebraically, write \(\z_1=a_1+jb_1\) and \(\z_2=a_2+jb_2\), so that \(\z_1+\z_2=(a_1+a_2)+j(b_1+b_2)\). Squaring the modulus and expanding gives \begin{align*} |\z_1+\z_2|^2 &=(a_1+a_2)^2+(b_1+b_2)^2\\ &=(a_1^2+b_1^2)+(a_2^2+b_2^2)+2(a_1a_2+b_1b_2)\\ &\leq(a_1^2+b_1^2)+(a_2^2+b_2^2)+2\bigl(|a_1||a_2|+|b_1||b_2|\bigr), \end{align*} where the last step uses \(xy\leq|x||y|\). The first two brackets are \(|\z_1|^2\) and \(|\z_2|^2\). To handle the last term we claim that \[ |a_1||a_2|+|b_1||b_2|\leq\sqrt{a_1^2+b_1^2}\sqrt{a_2^2+b_2^2}=|\z_1||\z_2|. \] Both sides are non-negative, so it is enough to compare their squares: \[ \bigl(|a_1||a_2|+|b_1||b_2|\bigr)^2\leq(a_1^2+b_1^2)(a_2^2+b_2^2). \] Expanding both sides and cancelling the common terms \(a_1^2a_2^2+b_1^2b_2^2\), this is equivalent to \[ 2|a_1||a_2||b_1||b_2|\leq a_1^2b_2^2+b_1^2a_2^2 \iff0\leq\bigl(|a_1||b_2|-|b_1||a_2|\bigr)^2, \] which is true because a square is never negative. Putting the pieces together, \begin{align*} |\z_1+\z_2|^2 &\leq|\z_1|^2+|\z_2|^2+2|\z_1||\z_2|=(|\z_1|+|\z_2|)^2\\ \Rightarrow\quad|\z_1+\z_2|&\leq|\z_1|+|\z_2|.\quad\blacksquare \end{align*} Definition Division is defined through the multiplicative inverse. If \(\z_2\ne0\),

In practice we multiply the top and the bottom by the conjugate of the denominator, which makes the denominator real:

Example Compute \(\dfrac{2+j3}{3-j4}\). \begin{align*} \frac{2+j3}{3-j4} &=(2+j3)\cdot\frac{(3+j4)}{(3-j4)(3+j4)}=\frac{(2+j3)(3+j4)}{25}\\ &=\frac{(6-12)+j(8+9)}{25}=-\frac6{25}+j\frac{17}{25}. \end{align*}

Elementary Complex Functions

A complex function \(\w\) takes a complex number \(\z\) in its domain and returns a complex number \(\w(\z)\) in its range. Because both \(\z\) and \(\w(\z)\) live in planes, we picture a function as a map from one plane to another.

If \(\z=x+jy\), the output has a real part and an imaginary part, each depending on \(x\) and \(y\): \[ \w(\z)=\w(x+jy)=u(x,y)+j\,v(x,y). \]Example Squaring. Let \(\w(\z)=\z^2\) with the first quadrant of the \(\z\) plane as its domain, that is, \(0<x<\infty\) and \(0<y<\infty\). Then \[ \w(x+jy)=(x+jy)(x+jy)=x^2-y^2+j2xy, \] so \(u(x,y)=x^2-y^2\) and \(v(x,y)=2xy\). Since \(x\) and \(y\) are both positive, \(v>0\), while \(u\) can be any real number. The range is therefore \(-\infty<u<\infty\) and \(0<v<\infty\), the upper half plane.

Example Conjugation. Let \(\w(\z)=\z^*\) on the entire complex plane. Then \(\w(x+jy)=x-jy\), so \(u(x,y)=x\) and \(v(x,y)=-y\). This reflects the plane in the real axis, and the range is the whole complex plane.

Example The exponential function. The complex exponential is defined by the same power series as the real one: \[ \w(\z)=e^{\z}\equiv1+\z+\dfrac{\z^2}{2!}+\cdots+\dfrac{\z^n}{n!}+\cdots. \] For \(\z=x+jy\) it splits into a real exponential and a pure phase, \(e^{x+jy}=e^x\cdot e^{jy}\) (we use this fact without proof here). To understand \(e^{jy}\), put \(jy\) into the series. The powers of \(j\) repeat, \(j^2=-1\), \(j^3=-j\), \(j^4=1\), so the terms separate into a real part with the even powers and an imaginary part with the odd powers: \begin{align*} e^{jy}&=1+jy+\frac{(jy)^2}{2!}+\cdots+\frac{(jy)^n}{n!}+\cdots\\ &=\left(1-\frac{y^2}{2!}+\frac{y^4}{4!}-\cdots+\frac{(-1)^ky^{2k}}{(2k)!}+\cdots\right)\\ &\quad+j\left(y-\frac{y^3}{3!}+\cdots+\frac{(-1)^ky^{2k+1}}{(2k+1)!}+\cdots\right). \end{align*} The first bracket is the series of \(\cos y\) and the second is the series of \(\sin y\). This gives one of the most useful formulas in engineering mathematics:

Putting Euler’s formula back into \(e^{\z}=e^x\cdot e^{jy}\) gives the real and imaginary parts of the exponential: \begin{align*} \w(\z)=e^{\z}&=e^x\cdot e^{jy}=e^x(\cos y+j\sin y)\\ &=e^x\cos y+je^x\sin y\\ &=u(x,y)+jv(x,y), \end{align*} with \(u(x,y)=e^x\cos y\) and \(v(x,y)=e^x\sin y\). In particular \(|e^{\z}|=e^x>0\), so \(e^{\z}\) is never zero, for any complex \(\z\).

Polar representation  A point in the plane can be described by its distance \(r=|\z|\) from the origin and its angle \(\theta\) from the real axis, instead of by \(x\) and \(y\).

Factoring out the length \(r=\sqrt{x^2+y^2}\) leaves a bracket whose two entries are \(\cos\theta\) and \(\sin\theta\), and by Euler’s formula the bracket is \(e^{j\theta}\): \[ \z=x+jy=\sqrt{x^2+y^2}\underbrace{\left(\underbrace{\frac{x}{\sqrt{x^2+y^2}}}_{\cos\theta}+j\underbrace{\frac{y}{\sqrt{x^2+y^2}}}_{\sin\theta}\right)}_{e^{j\theta}} \]