Guides / Complex numbers

Complex numbers: i² = −1 and the mistakes it causes

Arithmetic with complex numbers is ordinary algebra with one extra rule. Almost every wrong answer comes from applying that rule too early, too late, or with the wrong sign.

Why it works

No real number squares to a negative, so √−1 has no real value. Complex numbers add one new number, i, whose whole definition is that i² = −1. Everything else follows from treating i like a variable and using that one fact.

That is why adding and subtracting are easy: you combine like terms. Real parts with real parts, i terms with i terms. (3 + 2i) − (5 − 4i) = 3 − 5 + 2i + 4i = −2 + 6i. The minus sign distributes to both parts, just as it would with x.

Multiplying is where i² shows up, and replacing it with −1 turns an i term back into a real number. Division uses that on purpose: (a + bi)(a − bi) = a² − b²i² = a² + b², a real number. Multiplying the top and bottom of a fraction by the conjugate clears the i from the denominator without changing the fraction's value, because you are multiplying by 1.

A complex answer is written in the form a + bi: real part first, then the i part, with nothing left in a denominator.

Example 1: square roots of negatives

Simplify √−12 and √−4 · √−9.

1. √−12 = i√12Take out the −1 as i first: √(−12) = √(−1) · √12, and both of those are allowed because 12 is positive.
2. = 2i√3√12 = √4 · √3 = 2√3.

1. √−4 · √−9 = 2i · 3iRewrite each root with i before doing anything else.
2. = 6i²Multiply: 2 · 3 = 6 and i · i = i².
3. = −6Replace i² with −1.

Example 2: multiply with FOIL, then replace i²

Multiply (2 + 3i)(4 − i).

1. 8 − 2i + 12i − 3i²FOIL: 2·4, 2·(−i), 3i·4, 3i·(−i). Four products, exactly as with (2 + 3x)(4 − x).
2. 8 + 10i − 3(−1)Combine the i terms. Replace i² with −1, in parentheses so the sign is kept.
3. 8 + 10i + 3−3 times −1 is +3.
4. 11 + 10iCombine the real parts.

Example 3: divide by multiplying by the conjugate

Simplify 3 + 4i1 − 2i.

1. 3 + 4i1 − 2i · 1 + 2i1 + 2iThe conjugate of 1 − 2i is 1 + 2i: same real part, opposite sign on the i term. Multiplying by (1 + 2i)/(1 + 2i) is multiplying by 1.
2. 3 + 6i + 4i + 8i²1 − 4i²FOIL the top. On the bottom, the middle terms +2i and −2i cancel.
3. 3 + 10i − 81 + 4Replace i² with −1 on top and bottom.
4. −5 + 10i5The denominator is real now. That was the point of the conjugate.
5. −1 + 2iDivide both parts by 5.

Check by multiplying back: (−1 + 2i)(1 − 2i) = −1 + 2i + 2i − 4i² = 3 + 4i. That is the numerator, so the division is right.

Example 4: complex roots from the quadratic formula

Solve x² − 4x + 13 = 0.

1. b² − 4ac = 16 − 52 = −36a = 1, b = −4, c = 13. The discriminant is negative, so the roots are complex.
2. x = 4 ± √−362−b = 4, and 2a = 2.
3. x = 4 ± 6i2√(−36) = i√36 = 6i.
4. x = 2 ± 3iDivide both terms of the numerator by 2.

Complex roots of a quadratic with real coefficients always come as a conjugate pair, 2 + 3i and 2 − 3i. If you get only one, or two that aren't conjugates, recheck the arithmetic.

Powers of i repeat every four

i¹ = i, i² = −1, i³ = i² · i = −i, i⁴ = i² · i² = 1. Then i⁵ = i⁴ · i = i and the cycle starts again. Since i⁴ = 1, every group of four in the exponent multiplies to 1 and disappears. Only the remainder matters.

1. 23 = 4 · 5 + 3Divide the exponent by 4. The remainder is 3.
2. i²³ = (i⁴)⁵ · i³ = 1 · i³Five groups of i⁴, each equal to 1.
3. i²³ = −ii³ = −i.

A remainder of 0 means the power equals i⁴ = 1, so i²⁴ = 1.

Common mistakes, and the exact line they happen on

1. Multiplying square roots of negatives under one root

1. √−4 · √−9 = √36 = 6This is the mistake. The rule √a · √b = √(ab) only holds when a and b are not negative. Writing each root with i first gives 2i · 3i = 6i² = −6. The sign is wrong, and nothing on this line looks suspicious.

The fix is the same every time: the moment you see a negative under a square root, rewrite it with i. Only then multiply.

2. Leaving i² in, or replacing it with +1

From (2 + 3i)(4 − i) = 8 + 10i − 3i²:

1. 8 + 10i − 3This is the mistake. i² was replaced with +1, as if it were a squared real number. The answer comes out as 5 + 10i instead of 11 + 10i.

Leaving the answer as 8 + 10i − 3i² is also wrong. It isn't in a + bi form, and it hides that the real part is 11, not 8. Write −3(−1) with the parentheses so the two minus signs are visible.

3. Putting the conjugate's sign change on the wrong part

Dividing by 1 − 2i:

1. 3 + 4i1 − 2i · −1 + 2i−1 + 2iThis is the mistake. Both signs were flipped, so −1 + 2i is just −(1 − 2i), the negative of the original. The new denominator is (1 − 2i)(−1 + 2i) = 3 + 4i. It still has an i, and the division is no closer to done.

The conjugate of a + bi is a − bi. The real part stays the same and only the sign in front of the i term changes. A quick test: the denominator after multiplying should be a positive real number, a² + b². Here that is 1 + 4 = 5.

4. Squaring a + bi without the middle term

1. (3 + 2i)² = 9 + 4i² = 5This is the mistake. Squaring each part separately drops 2 · 3 · 2i = 12i. The real answer is 9 + 12i + 4i² = 5 + 12i.

(a + bi)² is (a + bi)(a + bi), four products, just like (a + b)². The middle term is where the whole imaginary part of the answer lives, so dropping it turns a complex answer into a real one.

5. Being one step off in the cycle of powers

For i²³, a student lists the cycle starting from i⁰: 1, i, −1, −i.

1. 23 = 4 · 5 + 3The remainder is 3. Correct so far.
2. i²³ = −1This is the mistake. −1 is the third item in the list, but that list started at i⁰, so its third item is i². The remainder is the exponent, not a position in a list: i²³ = i³ = −i.

Avoid counting positions. Write i²³ = i³, then work out i³ = i² · i = −i from the definition.

Practice

1. Simplify √−2 · √−8.

   Show answer

   i√2 · i√8 = i²√16 = −4. Combining first under one root would give +4, which is wrong.

2. Multiply (3 − i)(2 + 5i).

   Show answer

   6 + 15i − 2i − 5i² = 6 + 13i + 5 = 11 + 13i.

3. Simplify 1 + 7i1 + 2i.

   Show answer

   Multiply top and bottom by 1 − 2i. Top: 1 − 2i + 7i − 14i² = 15 + 5i. Bottom: 1 + 4 = 5. Answer: 3 + i.

4. Simplify i⁵⁰.

   Show answer

   50 = 4 · 12 + 2, so i⁵⁰ = i² = −1.

5. Solve x² + 6x + 25 = 0.

   Show answer

   Discriminant 36 − 100 = −64, and √−64 = 8i. x = (−6 ± 8i)/2 = −3 ± 4i.