Dividing Complex Numbers
Chapter Ten
Syllabus topic Module 1, "1.1 Complex Numbers"
Pages 22 to 23 of 303
In one line
Multiply above and below by the conjugate of the denominator, which makes the denominator real, and then divide the two parts separately.
The method
There is one idea and it is the one the previous chapter set up: a complex number times its conjugate is real. So to divide, make the denominator real first.
(a + i b)/(c + i d) = ((a + i b)(c - i d))/((c + i d)(c - i d))
The bottom is now real, because it is c squared plus d squared. Expanding the top gives the general result.
(a + i b)/(c + i d) = (a c + b d)/(c^2 + d^2) + i(b c - a d)/(c^2 + d^2)
Nobody remembers that and nobody should. What you remember is the method: multiply top and bottom by the conjugate of the bottom. The formula then comes out every time.
Worked, in full
Divide 3 + 4i by 1 minus 2i.
Step one, multiply above and below by 1 plus 2i, the conjugate of the denominator.
(3 + 4i)/(1 - 2i) = ((3 + 4i)(1 + 2i))/((1 - 2i)(1 + 2i))
Step two, expand the bottom. It is a number times its conjugate, so it is real.
(1 - 2i)(1 + 2i) = 1 - 4i^2 = 5
Step three, expand the top as an ordinary product.
(3 + 4i)(1 + 2i) = 3 + 6i + 4i + 8i^2 = -5 + 10i
Step four, divide each part by 5.
(3 + 4i)/(1 - 2i) = (-5 + 10i)/5 = -1 + 2i
Step five, and this is the step worth making a habit: check it by multiplying back.
(-1 + 2i)(1 - 2i) = -1 + 2i + 2i - 4i^2 = 3 + 4i
That is the original numerator, so the division is right. Checking a division by multiplying takes ten seconds and catches every sign error.
The two special cases you should do by eye
Dividing by i. There is no need for a conjugate.
1/i = -i
(3 + 4i)/i = -i(3 + 4i) = 4 - 3i
Multiplying by one over i, which is minus i, is quicker than anything else. Note what it did to the number: 3 + 4i became 4 minus 3i, which is a quarter turn clockwise.
Dividing by a real number. Divide both parts and stop.
(6 - 9i)/3 = 2 - 3i
The reciprocal of a complex number
Taking a = 1 and b = 0 in the general rule gives the reciprocal, which is worth knowing in its own form.
1/(c + i d) = (c - i d)/(c^2 + d^2)
In words: the reciprocal of z is the conjugate of z divided by the modulus of z squared. Two consequences:
Dividing Complex Numbers
z = x + i y
1/z = conjugate(z)/(z conjugate(z))
And if the modulus is one, the reciprocal is simply the conjugate, which is why numbers on the unit circle are so convenient. The roots of unity, later in this module, all have this property.
Worked: a compound expression
Simplify the expression below into standard form.
(2 + i)/(1 - i) + (1 - i)/(2 + i)
Do the two divisions separately and add.
(2 + i)/(1 - i) = ((2 + i)(1 + i))/2 = (1 + 3i)/2
(1 - i)/(2 + i) = ((1 - i)(2 - i))/5 = (1 - 3i)/5
(1 + 3i)/2 + (1 - 3i)/5 = 7/10 + (9/10)i
The common denominator at the end is ten, and the parts are handled independently once both fractions are in standard form. Trying to combine the two fractions before clearing their denominators is possible but is where the marks go.
Division and the modulus
There is a fact here that the chapter on the modulus will use: the modulus of a quotient is the quotient of the moduli.
abs((3 + 4i)/(1 - 2i)) = abs(3 + 4i)/abs(1 - 2i)
That is 5 divided by the square root of 5, which is the square root of 5, and the answer minus one plus 2i does indeed have modulus the square root of 5. It is a useful check on a division: work out the two moduli, divide them, and see whether your answer has that size.
Check yourself
| Question | Answer |
|---|---|
| (1 + i)/(1 - i) | (1 + i)/(1 - i) = i |
| (2 - 3i)/i | (2 - 3i)/i = -3 - 2i |
| 1/(3 + 4i) | 1/(3 + 4i) = 3/25 - (4/25)i |
| (5 + 5i)/(1 + 2i) | (5 + 5i)/(1 + 2i) = 3 - i |
| Check the last one by multiplying back | (3 - i)(1 + 2i) = 5 + 5i |
The third one is the reciprocal rule with c = 3 and d = 4, so the denominator is 25, and the conjugate on top gives 3 minus 4i.
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.