Chapter 3 · Complex Numbers and Equations · Concept 1
3-3 Operations with Complex Numbers (2)
Adding, subtracting and multiplying were all ordinary algebra. Dividing is the one that resists: an i on the bottom of a fraction is not something you can leave. Clearing it takes one well-chosen multiplication — and it is the same trick you used on surds two chapters ago.
Guiding question
How do you get an i out of the bottom of a fraction?
Learning outcome. Perform division on complex numbers and put the answer in the form a + bi.
Complex conjugates
The numbers a + bi and a − bi are called complex conjugates. Only the sign of the imaginary part differs.
and their product is always REAL
(a + bi)(a − bi) = a2 + b2
(2 + 3i)(2 − 3i) = 4 − 9i2 = 4 + 9 = 13
The −9i² is what turns into +9. No i survives.
So that is how you divide
For a fraction with an i on the bottom, multiply top and bottom by the conjugate of the bottom. The bottom becomes real, and then the fraction can be split into a + bi.
1 − i2 + 3i = (1 − i)(2 − 3i)(2 + 3i)(2 − 3i) = −1 − 5i13
Multiplying top and bottom by the same thing does not change the value — it just moves the i off the bottom.
Multiplying a number by its conjugate gives four products. The two in the middle are equal and opposite, so they destroy each other — and what is left has no i in it at all.
(2 + 3i)(2 − 3i)
Four products. Two of them are about to disappear — press the button and watch which.
You have done this before.
In lesson 1-15 you cleared a root off the bottom by multiplying by (√a − √b) — the same expression with the sign flipped, for exactly the same reason. Rationalising a surd and clearing an i are one idea wearing two coats.
2(a) Calculate 2 + i2 − i and answer in the form a + bi.
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Look at the BOTTOM and write down its conjugate.
the bottom is 2 − i, so its conjugate is 2 + iThe conjugate of the bottom — not the top. Getting this the wrong way round is the commonest slip.
Work with a partner. One of you does the bottom and the other the top. The bottom should come out real every single time — if it does not, the wrong conjugate was used.
1. State the complex conjugate.
2. Calculate, answering in the form a + bi.
A free check on every one of these. Multiply your answer by the original bottom — you should get the original top back. It takes ten seconds and it catches a sign slip immediately.
( i(10 − 12i)(2 + i)(−1 + 4i) )2027
A real number is its own conjugate. Which complex numbers are the NEGATIVE of their own conjugate — and what does that tell you about the two kinds?
Write z = a + bi and set the conjugate equal to −z.
Multiplying by the conjugate of the TOP.
It is the bottom you are trying to clear, so it is the bottom whose conjugate you need.
Multiplying only the bottom.
Both top and bottom, or the value changes. That is the whole reason it is allowed.
Stopping at a stacked fraction.
(3 + 4i)/5 is correct but unfinished. The question says "in the form a + bi", so write 3/5 + (4/5)i.
Forgetting −i² = +1 in the bottom.
(2 − i)(2 + i) = 4 − i² = 5, not 3. If your bottom still has an i in it, something went wrong.
Thinking a real number has no conjugate.
It is its own. 6 = 6 + 0i, and flipping the sign of zero changes nothing.
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