Adding and Subtracting Complex Numbers
Chapter Seven
Syllabus topic Module 1, "1.1 Complex Numbers"
Pages 15 to 16 of 303
In one line
Add the real parts and add the imaginary parts, separately, and never let the two mix.
The rule
For z1 = a + ib and z2 = c + id:
(a + i b) + (c + i d) = (a + c) + i(b + d)
(a + i b) - (c + i d) = (a - c) + i(b - d)
That is all there is to it. Addition and subtraction treat the two parts as two separate sums, and nothing passes between them.
Why it is that simple, and why multiplication will not be
The reason the parts do not mix is that i is only a factor here, never something being multiplied by itself. Collecting ib and id gives i(b + d), and i has not been squared, so the rule i squared equals minus one never gets a chance to fire.
The moment you multiply two complex numbers, an i does meet another i, i squared appears, and a term that started imaginary lands in the real part. That is the whole difference between this chapter and the next one, and it is worth noticing now.
Worked
(3 + 4i) + (5 - 7i) = 8 - 3i
(3 + 4i) - (5 - 7i) = -2 + 11i
(-2 + i) + (2 - i) = 0
(1/2 + (2/3)i) + (1/2 + (1/3)i) = 1 + i
The third line is worth a second look. Two complex numbers can add to zero, exactly as two real numbers can, and the one that cancels another is its negative: the negative of a + ib is minus a minus ib.
In the fourth line the two parts are added as ordinary fractions. Nothing special happens to a fraction because it is inside a complex number.
The three properties, which are asked as a theory question
Addition of complex numbers is commutative, associative, and has an identity and inverses. Each follows immediately from the same property of real numbers, because all that is happening is two real additions side by side.
| Property | Statement | Why it holds |
|---|---|---|
| Commutative | z1 + z2 = z2 + z1 | Real addition is commutative, in both parts |
| Associative | (z1 + z2) + z3 = z1 + (z2 + z3) | Real addition is associative, in both parts |
| Identity | z + 0 = z | 0 means 0 + 0i |
| Inverse | z + (-z) = 0 | -z means (-a) + i(-b) |
A question asking you to "verify that addition of complex numbers is commutative" wants the two lines below, not a paragraph.
(a + i b) + (c + i d) = (a + c) + i(b + d)
(c + i d) + (a + i b) = (c + a) + i(d + b)
Adding and Subtracting Complex Numbers
Since real addition is commutative, a + c equals c + a and b + d equals d + b, so the two results are the same complex number.
Subtraction is addition of the negative
There is no separate theory of subtraction. Writing z1 minus z2 means z1 plus the negative of z2, and every rule above applies.
(7 + 2i) - (3 + 5i) = (7 + 2i) + (-3 - 5i) = 4 - 3i
Two consequences a student should know. Subtraction is not commutative, and the two answers differ by a sign throughout.
(7 + 2i) - (3 + 5i) = 4 - 3i
(3 + 5i) - (7 + 2i) = -4 + 3i
Adding several at once
Collect all the real parts, then all the imaginary parts. Doing it term by term is where sign errors come from.
(2 + 3i) + (4 - i) - (1 + 5i) + (-3 + 2i) = 2 - i
Check it by parts: the real parts are 2, 4, minus 1 and minus 3, which sum to 2. The imaginary parts are 3, minus 1, minus 5 and 2, which sum to minus 1. So the answer is 2 minus i.
What this looks like on the diagram
Both operations have a picture, and the chapter on addition on the Argand diagram draws it. In one sentence now, so the algebra does not feel arbitrary: adding two complex numbers is adding two vectors nose to tail, and subtracting them gives the vector that joins one point to the other. That is why the modulus of z1 minus z2 turns out to be the distance between the two points, which is the fact every locus question in this topic rests on.
Check yourself
| Question | Answer |
|---|---|
| (1 + i) + (1 - i) | (1 + i) + (1 - i) = 2 |
| (1 + i) - (1 - i) | (1 + i) - (1 - i) = 2i |
| (5 - 3i) + (-5 + 3i) | (5 - 3i) + (-5 + 3i) = 0 |
| (4 + 0i) - (0 + 4i) | (4 + 0i) - (0 + 4i) = 4 - 4i |
| (2 + i) + (3 - 4i) - (1 - i) | (2 + i) + (3 - 4i) - (1 - i) = 4 - 2i |
The first two together make a point worth keeping: adding a number to its conjugate kills the imaginary part and leaves twice the real part, and subtracting kills the real part and leaves twice the imaginary part times i. Both are used constantly from here on, and the conjugate gets its own chapter shortly.
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.