Complex Numbers and Their Algebra

A num­ber whose square is neg­a­tive#

The equa­tion $x^2 + 1 = 0$ has no real solu­tion, since no real num­ber has a neg­a­tive square. Math­e­mati­cians faced this prob­lem long ago, and their answer was bold: allow a new num­ber $i$ with $i^2 = -1$. That sin­gle step gives us the com­plex num­bers $a + ib$. Don't let the word "imag­i­nary" put you off; these num­bers fol­low very sen­si­ble rules. We will define them, learn their arith­metic (adding, mul­ti­ply­ing, divid­ing with the help of the con­ju­gate, and tak­ing pow­ers of $i$), and meet the mod­u­lus.

Com­plex num­bers#

A com­plex num­ber has the form $z = a + ib$, where $a, b$ are real and $i = \sqrt{-1}$. We call $a = \operatorname{Re} z$ the real part and $b = \operatorname{Im} z$ the imag­i­nary part. Two com­plex num­bers are equal only when both parts match: $a + ib = c + id$ exactly when $a = c$ and $b = d$. This one fact turns a sin­gle com­plex equa­tion into two real ones.

Arith­metic#

• $(a + ib) + (c + id) = (a + c) + i(b + d)$; sub­trac­tion like­wise.
• $(a + ib)(c + id) = (ac - bd) + i(ad + bc)$, using $i^2 = -1$.
• All the old rules still hold. Addi­tion and mul­ti­pli­ca­tion are com­mu­ta­tive and asso­cia­tive, mul­ti­pli­ca­tion dis­trib­utes over addi­tion, $0$ and $1$ are the iden­ti­ties, and $-z$ is the addi­tive inverse.

Pow­ers of $i$ go round in a cycle of four: $i, -1, -i, 1$, and then back again. So $i^{4k} = 1$, $i^{4k+1} = i$, $i^{4k+2} = -1$ and $i^{4k+3} = -i$. For any power, divide the index by four and look only at the remain­der. For exam­ple, $i^{39} = i^{36} i^3 = -i$, and $\displaystyle i^{-5} = \tfrac{1}{i^5} = \tfrac{1}{i} = -i$.

Square roots of neg­a­tives. We write $\sqrt{-9} = 3i$. Now a warn­ing that catches many stu­dents: the rule $\sqrt{a}\sqrt{b} = \sqrt{ab}$ fails when both num­bers are neg­a­tive. Indeed $\sqrt{-4}\sqrt{-9} = (2i)(3i) = -6$, not $\sqrt{36}$. Always write each root in terms of i first, then mul­ti­ply.

Con­ju­gate, mod­u­lus, divi­sion#

The con­ju­gate of $z = a + ib$ is $\bar z = a - ib$; you just flip the sign of the imag­i­nary part. The beauty of it is that $z\bar z = a^2 + b^2$, a non-neg­a­tive real num­ber. The mod­u­lus is $\lvert z \rvert = \sqrt{a^2 + b^2}$.

That is exactly what we need for divi­sion. Mul­ti­ply the top and bot­tom by the con­ju­gate of the denom­i­na­tor, and the denom­i­na­tor becomes real, much like ratio­nal­is­ing a surd:

$$\displaystyle \frac{1}{z} = \frac{\bar z}{\lvert z \rvert^2}, \qquad \frac{2 + 3i}{1 - 2i} = \frac{(2 + 3i)(1 + 2i)}{1 + 4} = \frac{-4 + 7i}{5}.$$

A few prop­er­ties save a lot of work: $\lvert z_1 z_2 \rvert = \lvert z_1 \rvert \lvert z_2 \rvert$, $\displaystyle \left\lvert \tfrac{z_1}{z_2} \right\rvert = \tfrac{\lvert z_1 \rvert}{\lvert z_2 \rvert}$, $\overline{z_1 z_2} = \bar z_1 \bar z_2$ and $\overline{z_1 + z_2} = \bar z_1 + \bar z_2$.

Prac­tice ques­tions#

1. Write in the form $a + ib$: $(5 - 3i) + (-2 + 7i)$; $(4 + i)(2 - 3i)$; $i^{27}$; $i^{-18}$; $(2 - i)^3$.
2. Find real $x, y$: $(x + y) + i(x - y) = 6 + 2i$; $(3 + 2i)x - (1 - i)y = 5 + 5i$.
3. Find the con­ju­gate, mod­u­lus and mul­ti­plica­tive inverse of $-5 + 12i$.
4. Sim­plify $\displaystyle \frac{3 - 4i}{2 + i}$ and $\displaystyle \frac{(1 + i)^2}{3 - i}$.
5. Eval­u­ate $\sqrt{-16} + 3\sqrt{-25} - \sqrt{-49}$.
6. Show that $i^{n} + i^{n+1} + i^{n+2} + i^{n+3} = 0$ for every inte­ger $n$.
7. Find $\displaystyle \left(\tfrac{1 + i}{1 - i}\right)^{50}$.
8. If $z_1 = 2 - i$ and $z_2 = 1 + 3i$, ver­ify $\lvert z_1 z_2 \rvert = \lvert z_1 \rvert \lvert z_2 \rvert$ and find $\displaystyle \operatorname{Re}\left(\tfrac{z_1}{z_2}\right)$.
9. Prove $z + \bar z$ is real and $z - \bar z$ is purely imag­i­nary.
10. Find the least pos­i­tive inte­ger $m$ with $\displaystyle \left(\tfrac{1 + i}{1 - i}\right)^m = 1$.

Answers#