Chapter 3 · Complex Numbers and Equations · Concept 1
3-2 Operations with Complex Numbers (1)
Here is the good news about arithmetic with complex numbers: you already know how to do it. Treat i as an ordinary letter and multiply out exactly as you would with x. There is only one new instruction in the whole lesson.
Guiding question
How much of this is actually new?
Learning outcome. Perform addition, subtraction and multiplication on complex numbers and put them in simplest form.
Ordinary algebra, with one substitution
Work exactly as you would with any algebraic expression. Then, wherever i2 appears:
write i2 = −1
And the answer is always written in the form a + bi — real part first, imaginary part second, nothing left over.
Powers of i
A high power looks frightening and is not. Break it into i2 pieces:
i8 = (i2)4 = (−1)4 = 1
i7 = i × (i2)3 = i × (−1)3 = −i
When the power is odd, peel off one i first and square what is left.
There are only four possible answers, ever. Multiply by i and you move one place round the wheel; after four moves you are back where you began. So the only thing that matters about the power is what it leaves when divided by 4.
27 ÷ 4 leaves 3
i27 = −i
Four steps and you are back where you started, so only the remainder matters. i27 is the same as i3 — and 40 is no harder than 4.
1
i⁰, i⁴, i⁸, …
i
i¹, i⁵, i⁹, …
−1
i², i⁶, i¹⁰, …
−i
i³, i⁷, i¹¹, …
(c) Calculate i6 + i27 + i17.
- 1
Take the even power first — it splits cleanly.
i6 = (i2)3 = (−1)3 = −1Six is three lots of two, so it is (i²) three times.
Work with a partner. For the multiplication, expand it fully before either of you replaces an i². Doing both at once is where the sign errors live.
Calculate the following.
Part (h) is the interesting one.
2/i has an i on the bottom, which is not allowed in a finished answer. Multiply top and bottom by i: 2i/i² = 2i/(−1) = −2i. Then the bracket becomes (−2i + i)(−2i − i) = (−i)(−3i) = 3i² = −3. Getting rid of an i on the bottom is the whole of the next lesson.
Given that x is a real number, solve
(7x − 7i)(x2 + xi) = 14x2 i40
i + i² + i³ + i⁴ came to 0. What would i + i² + … + i¹⁰⁰ come to, and how can you say so without adding a hundred terms?
How many complete turns of the wheel is that?
Adding a real part to an imaginary one.
3 + 2i is finished. It does not become 5i or 5 — the two parts keep separate accounts, exactly as in lesson 3-1.
Forgetting i × i = −1 when expanding.
In (2+3i)(1−2i) the last term is −6i², which is +6. The book lists this as its first misconception.
Only distributing the minus to the first term.
(3 − 2i) − (2 + 5i) is 3 − 2i − 2 − 5i. The bracket loses BOTH its signs.
Leaving an i on the bottom.
2/i is not a finished answer. Multiply top and bottom by i to get −2i.
Working out a big power the long way.
Divide the power by 4 and take the remainder. i⁴⁰ is 1, and you never needed to multiply anything forty times.
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