Treat i like a letter — until you meet i², which becomes −1.
Evaluate
(2 + 3i)(4 − i)
FOIL gives 8 − 2i + 12i − 3i². Because i² = −1, that −3i² becomes +3, so the real part is 8 + 3 = 11 and the answer is 11 + 10i. A leaves the i² term alone — converting it is the whole job.
i = √−1, so i² = −1
the plain number
the coefficient of i
Every complex number is a + bi. Keep the two parts apart and the arithmetic is ordinary algebra.
Complex numbers appear whenever a quadratic has a negative discriminant — the roots exist, just not on the real number line.
(3 + 2i) + (1 − 5i)
the question
reals: 3 + 1 = 4
plain numbers
imaginaries: 2i − 5i = −3i
the i terms
= 4 − 3i
done
Exactly like collecting 3 + 2x and 1 − 5x. Nothing about i matters yet.
8 + 10i − 3i²
collect the middle
−3i² = −3(−1) = +3
i² = −1
= 11 + 10i
8 + 3 = 11
It leaves the imaginary group entirely and joins the plain numbers. Forgetting to move it is the single most common error here.
Simplify (1 + 2i) ÷ (3 − i)
The conjugate of 3 − i is 3 + i — only the middle sign flips.
multiply top and bottom by 3 + i
that is multiplying by 1
bottom: (3 − i)(3 + i) = 9 − i²
difference of squares
= 9 + 1 = 10
real, which was the point
top: (1 + 2i)(3 + i) = 3 + 7i + 2i²
FOIL
= 1 + 7i
3 − 2 = 1
= (1 + 7i) ÷ 10
or 0.1 + 0.7i
The conjugate always kills the i on the bottom, because (a − bi)(a + bi) = a² + b² — a real number every time.
a + bi = c + di (all real) means a = c and b = d. One complex equation gives two real ones.
If x and y are real and (x + 3i) − (2 − yi) = 4 − i, find x + y.
(x − 2) + (3 + y)i = 4 − i
group the left side as a + bi
real: x − 2 = 4 → x = 6
match the plain numbers
imaginary: 3 + y = −1 → y = −4
match the coefficients of i
x + y = 2
what was asked
If x and y are real and (1 + 2i)(x + yi) = 5i, find x and y.
x + yi = 5i ÷ (1 + 2i)
isolate the unknown number
= 5i(1 − 2i) ÷ 5 = i − 2i² = 2 + i
conjugate: (1 + 2i)(1 − 2i) = 5
x = 2, y = 1
check: (1 + 2i)(2 + i) = 2 + 5i + 2i² = 5i ✓
Subtracting 2 − yi adds yi, so the imaginary part is 3 + y, not 3 − y. The same care applies to squares: (3 − 2i)² = 9 − 12i + 4i² = 5 − 12i — the 4i² is −4, not +4.
Two roots of negatives multiplied: √−4 × √−9.
√−4 = 2i, √−9 = 3i
take out the i first
2i × 3i = 6i² = −6
then multiply
What is different: the rule √a × √b = √ab only holds when a and b are not both negative.
Writing √−4 × √−9 = √36 = 6 joins the roots before the i is taken out. Convert each root to a multiple of i, then multiply.
A pure-imaginary divisor: (6 + 4i) ÷ 2i.
multiply top and bottom by i
no real part to keep, so i is enough
(6i + 4i²) ÷ 2i² = (−4 + 6i) ÷ (−2)
i² = −1 top and bottom
= 2 − 3i
check: 2i(2 − 3i) = 4i + 6 ✓
What is different: the bottom has no real part. Splitting first also works: 62i + 4i2i = −3i + 2.
A quadratic with no real roots: solve x² + 2x + 10 = 0.
D = 2² − 4(1)(10) = −36
negative discriminant
x = (−2 ± √−36) ÷ 2 = (−2 ± 6i) ÷ 2
√−36 = 6i
x = −1 ± 3i
a conjugate pair
What is different: the roots come in a conjugate pair, so their sum (−2) and product (1 + 9 = 10) are real, as the coefficients say.
i, i², i³, i⁴ — then it repeats forever
| Power | Value | Why |
|---|---|---|
| i¹ | i | — |
| i² | −1 | the definition |
| i³ | −i | i² × i |
| i⁴ | 1 | (−1)(−1) |
Divide any exponent by 4 and the remainder tells you which of the four you land on.
Complex-number items ask you to multiply or square (replace i² with −1), divide by the conjugate, or find real x and y from an equation — match real parts and imaginary parts. A last check: the final answer should be in the form a + bi with no i² and no i underneath.
Multiply, square, subtract
(1 + 2i)(3 − i) − (2 − i)²
(1 + 2i)(3 − i) = 3 − i + 6i − 2i² = 5 + 5i. (2 − i)² = 4 − 4i + i² = 3 − 4i. Subtract: 5 + 5i − 3 + 4i = 2 + 9i. D forgets to distribute the minus over −4i and gets 2 + i. C treats both i² terms as +1; A squares term by term, (2 − i)² = 4 + i² = 3, and loses the middle −4i.
