Unit 2: Impossible Numbers

A question with no answer

What times itself makes -1?

Not 1, because 1 × 1 = 1. Not -1, because -1 × -1 = 1 too. Every number you know squares to something positive or zero.

For hundreds of years mathematicians said "that has no answer" and moved on. Then somebody tried something braver.

Make one up

They invented a new number and called it i, with exactly one rule:

i × i = -1

That is the whole definition. i is not a fraction, not a decimal, not on the number line at all. It is a new kind of number, and it behaves itself perfectly — you can add, subtract, multiply and divide with it just like anything else.

It goes round in a circle

The powers of i repeat forever: i, -1, -i, 1, then straight back to i.

Four steps and you are home. So to work out something like i^27, you only need to know where 27 lands in that cycle of four.

A real part and an imaginary part

A complex number looks like 3 + 2i. It carries two pieces at once: the real part 3 and the imaginary part 2.

They never mix, in the same way you cannot add 3 apples to 2 oranges. So adding complex numbers is easy: reals with reals, imaginaries with imaginaries.

(3 + 2i) + (1 + 5i) = 4 + 7i

Multiplying: everything meets everything

To multiply, each piece meets each piece — the same way you would expand brackets.

(2 + i)(3 + 4i) gives 6 + 8i + 3i + 4i^2.

But i^2 is -1, so that last bit is really -4. Tidy up and you get 2 + 11i.

The conjugate: a useful twin

Flip the sign in the middle and you get the conjugate. The twin of 3 + 2i is 3 - 2i.

Multiply a number by its conjugate and something lovely happens — the imaginary parts cancel out completely and you are left with a plain real number.

(3 + 2i)(3 - 2i) = 9 + 4 = 13

Only four answers exist

However high the power, i to anything lands on one of just four values: i, -1, -i, 1.

That is because i^4 = 1, so every block of four multiplies by 1 and changes nothing.

So i^27 is the same as i^3, because 24 of those 27 cancel themselves out.

Four in a row always cancel

Add any four consecutive powers of i and you get zero, because you are adding i - 1 - i + 1.

So a sum like i + i^2 + i^3 + … + i^100 collapses to nothing at all: twenty-five complete cycles, each worth zero.

Spotting the cycle is the whole skill. The arithmetic afterwards is trivial.

Subtracting keeps the parts apart

Subtraction works exactly like addition: reals with reals, imaginaries with imaginaries.

(7 + 3i) - (2 + 8i) = 5 - 5i

The only trap is the sign. That minus applies to both parts of the second bracket, not just the first.

Squaring a complex number

(3 + 2i)^2 is (3 + 2i)(3 + 2i), which expands the usual way:

9 + 6i + 6i + 4i^2

That last term is 4 × -1 = -4. So the answer is 5 + 12i.

The pattern from unit 7 still holds — square the first, double the product, square the last — you just have to remember what the squared i turns into.

The complex plane

A complex number carries two pieces, so one line cannot hold it. It needs a plane.

Across for the real part, up for the imaginary part. So 3 + 2i sits at the point 3 along and 2 up.

Ordinary numbers all lie flat on the across-line. Complex numbers fill in the rest of the page.

How big is a complex number?

You cannot say 3 + 2i is bigger than 1 + 5i — there is no left and right on a plane.

What you *can* measure is how far it is from the middle. That is its modulus, written |z|, and Pythagoras gives it:

|3 + 4i| = sqrt(9 + 16) = 5

Add a number to its twin

Add 3 + 2i to its conjugate 3 - 2i and the imaginary parts wipe each other out:

(3 + 2i) + (3 - 2i) = 6

You get twice the real part, and always a real number. Subtract instead and the *real* parts vanish, leaving 4i.

Multiply and you get the modulus squared

Multiplying a number by its conjugate gave a real answer back in lesson 2. Now you can see what that answer *is*:

(a + bi)(a - bi) = a^2 + b^2

That is the modulus, squared. So the conjugate trick and the distance formula are the same fact.

Now every square root works

With i in hand, sqrt(-9) is no longer impossible.

-9 is 9 × -1, so its square root is 3 × i, which we write 3i. Check it: 3i × 3i = 9 × i^2 = -9.

This was the point. Invent one number, and suddenly *every* number has a square root.

Dividing, with the conjugate trick

How do you divide by a complex number? You use the twin.

Multiply the top and the bottom by the conjugate of the bottom. The bottom turns real, and a real bottom is easy.

2/(1 + i) becomes 2(1 - i)/((1 + i)(1 - i)), which is 2(1 - i)/2, which is 1 - i.