Algebra 2

Complex numbers and the imaginary unit i

A complex number has a real part and an imaginary part: a+bia + bi, where i=−1i = \sqrt{-1}, so i2=−1i^2 = -1. Add and subtract by combining real parts with real parts and imaginary parts with imaginary parts. Multiply like binomials, then replace i2i^2 with −1-1. To divide, multiply the top and bottom by the conjugate of the bottom, which turns the denominator into a real number.

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The key idea

No real number squared gives a negative, so mathematicians named a new number ii with i2=−1i^2 = -1. Every complex number is a real part plus a real multiple of ii. Real numbers are complex numbers too: they just have b=0b = 0.

i=−1,i2=−1,−k=ik  (k>0)i = \sqrt{-1}, \qquad i^2 = -1, \qquad \sqrt{-k} = i\sqrt{k} \ \ (k > 0)

Powers of ii repeat every 4 steps, because i4=(i2)2=1i^4 = (i^2)^2 = 1:

The cycle of powers of i
Poweri1i^1i2i^2i3i^3i4i^4i5i^5
Valueii−1-1−i-i11ii (the cycle restarts)

The conjugate of a+bia + bi is a−bia - bi. Their product is always a real number, which is why it is the tool for division:

(a+bi)(a−bi)=a2−b2i2=a2+b2(a + bi)(a - bi) = a^2 - b^2i^2 = a^2 + b^2

Worked examples

Example 1: subtracting complex numbers

Problem Simplify (5−2i)−(3−7i)(5 - 2i) - (3 - 7i).

  1. The minus sign applies to both parts of the second number.
    (5−2i)−(3−7i)=5−2i−3+7i(5 - 2i) - (3 - 7i) = 5 - 2i - 3 + 7i
  2. Group real parts and imaginary parts.
    5−2i−3+7i=(5−3)+(−2+7)i5 - 2i - 3 + 7i = (5 - 3) + (-2 + 7)i
  3. Combine.
    (5−3)+(−2+7)i=2+5i(5 - 3) + (-2 + 7)i = 2 + 5i

Answer 2+5i2 + 5i

Example 2: multiplying

Problem Simplify (3+4i)(2−i)(3 + 4i)(2 - i).

  1. Multiply every term by every term, like FOIL.
    (3+4i)(2−i)=6−3i+8i−4i2(3 + 4i)(2 - i) = 6 - 3i + 8i - 4i^2
  2. Replace i2i^2 with −1-1. Then −4i2-4i^2 becomes +4+4.
    6−3i+8i−4i2=6+5i+46 - 3i + 8i - 4i^2 = 6 + 5i + 4
  3. Combine the real parts.
    6+5i+4=10+5i6 + 5i + 4 = 10 + 5i

Answer 10+5i10 + 5i

Example 3: dividing with the conjugate

Problem Write 7+i1−2i\frac{7 + i}{1 - 2i} in the form a+bia + bi.

  1. The conjugate of the bottom, 1−2i1 - 2i, is 1+2i1 + 2i. Multiply top and bottom by it. This does not change the value, because you are multiplying by 1.
    7+i1−2i=(7+i)(1+2i)(1−2i)(1+2i)\frac{7 + i}{1 - 2i} = \frac{(7 + i)(1 + 2i)}{(1 - 2i)(1 + 2i)}
  2. Top: 7+14i+i+2i2=7+15i−27 + 14i + i + 2i^2 = 7 + 15i - 2.
    (7+i)(1+2i)=5+15i(7 + i)(1 + 2i) = 5 + 15i
  3. Bottom: a2+b2a^2 + b^2 with a=1a = 1, b=2b = 2.
    (1−2i)(1+2i)=5(1 - 2i)(1 + 2i) = 5
  4. Divide each part by 5.
    5+15i5=1+3i\frac{5 + 15i}{5} = 1 + 3i

Answer 1+3i1 + 3i

Example 4 (test-hard): a quadratic with complex roots

Problem Solve x2−6x+13=0x^2 - 6x + 13 = 0.

  1. Use the quadratic formula with a=1a = 1, b=−6b = -6, c=13c = 13. Start with the discriminant.
    b2−4ac=36−52=−16b^2 - 4ac = 36 - 52 = -16
  2. The discriminant is negative, so the roots are not real. Its square root is −16=4i\sqrt{-16} = 4i.
    x=6±4i2x = \frac{6 \pm 4i}{2}
  3. Divide both parts by 2.
    x=3±2ix = 3 \pm 2i
  4. Check x=3+2ix = 3 + 2i: (3+2i)2=9+12i−4=5+12i(3 + 2i)^2 = 9 + 12i - 4 = 5 + 12i, and 5+12i−6(3+2i)+13=5+12i−18−12i+13=05 + 12i - 6(3 + 2i) + 13 = 5 + 12i - 18 - 12i + 13 = 0.

Answer x=3+2ix = 3 + 2i or x=3−2ix = 3 - 2i

Common mistakes

  • Leaving i2i^2 in the answer. i2i^2 is just −1-1. Fix: after multiplying, replace every i2i^2 and combine the real parts.
  • Dropping the sign on the second number when subtracting. −(3−7i)-(3 - 7i) is −3+7i-3 + 7i. Fix: distribute the minus to both parts.
  • Using a2−b2a^2 - b^2 for the conjugate product. (1−2i)(1+2i)(1 - 2i)(1 + 2i) is 1+4=51 + 4 = 5, not 1−41 - 4. Fix: the i2i^2 turns the minus into a plus, so it is a2+b2a^2 + b^2.
  • Multiplying square roots of negatives directly. −4⋅−9\sqrt{-4} \cdot \sqrt{-9} is 2i⋅3i=−62i \cdot 3i = -6, not 36=6\sqrt{36} = 6. Fix: rewrite each −k\sqrt{-k} as iki\sqrt{k} first.
  • Dividing only the real part by the denominator. 5+15i5\frac{5 + 15i}{5} is 1+3i1 + 3i, not 1+15i1 + 15i. Fix: divide both parts.

Quick methods

Practice

5 practice questions

  1. Simplify i27i^{27}.

    1. ii
    2. −i-i
    3. 11
    4. −1-1
    Show answer

    Answer: −i-i

    27=4⋅6+327 = 4 \cdot 6 + 3, so i27=i3=−ii^{27} = i^3 = -i. Picking ii or −1-1 usually means the remainder was counted wrong.

  2. Simplify (4−3i)(4+3i)(4 - 3i)(4 + 3i).

    1. 77
    2. 2525
    3. 16+9i16 + 9i
    4. 7+24i7 + 24i
    Show answer

    Answer: 2525

    Conjugates multiply to a2+b2=16+9=25a^2 + b^2 = 16 + 9 = 25. The answer 7 comes from treating −9i2-9i^2 as −9-9. 7+24i7 + 24i is (4+3i)2(4 + 3i)^2, a different product.

  3. Simplify (2+5i)+(−6+i)−(1−3i)(2 + 5i) + (-6 + i) - (1 - 3i).

    1. −5+9i-5 + 9i
    2. −5+3i-5 + 3i
    3. −3+9i-3 + 9i
    4. −5−9i-5 - 9i
    Show answer

    Answer: −5+9i-5 + 9i

    Real parts: 2−6−1=−52 - 6 - 1 = -5. Imaginary parts: 5+1+3=95 + 1 + 3 = 9. −5+3i-5 + 3i forgets that subtracting −3i-3i adds 3i3i.

  4. Write 103+i\frac{10}{3 + i} in the form a+bia + bi.

    1. 3−i3 - i
    2. 3+i3 + i
    3. 103+10i\frac{10}{3} + 10i
    4. 30−10i8\frac{30 - 10i}{8}
    Show answer

    Answer: 3−i3 - i

    Multiply top and bottom by 3−i3 - i: 10(3−i)9+1=30−10i10=3−i\frac{10(3 - i)}{9 + 1} = \frac{30 - 10i}{10} = 3 - i. The choice with 8 on the bottom used 9−19 - 1 instead of 9+19 + 1.

  5. What are the solutions of x2+4x+20=0x^2 + 4x + 20 = 0?

    1. −2±4i-2 \pm 4i
    2. 2±4i2 \pm 4i
    3. −2±8i-2 \pm 8i
    4. −4±8i-4 \pm 8i
    Show answer

    Answer: −2±4i-2 \pm 4i

    The discriminant is 16−80=−6416 - 80 = -64, and −64=8i\sqrt{-64} = 8i. So x=−4±8i2=−2±4ix = \frac{-4 \pm 8i}{2} = -2 \pm 4i. −4±8i-4 \pm 8i forgets to divide by 2, and 2±4i2 \pm 4i drops the minus on b.

Frequently asked questions

What is i in math?

ii is the imaginary unit, the number whose square is −1-1. It lets you take square roots of negative numbers: −25=5i\sqrt{-25} = 5i. It follows all the usual rules of algebra, with one extra fact: whenever you see i2i^2, replace it with −1-1.

Why do complex roots come in pairs?

When a quadratic has real coefficients, the quadratic formula gives −b±D2a\frac{-b \pm \sqrt{D}}{2a}. If D is negative, the plus and minus give p+qip + qi and p−qip - qi: a conjugate pair. So if 3+2i3 + 2i is a root, 3−2i3 - 2i is a root too.

Is a real number a complex number?

Yes. A real number like 7 is 7+0i7 + 0i, a complex number whose imaginary part is zero. Complex numbers include all real numbers, all imaginary numbers like 4i4i, and mixes like 2−3i2 - 3i.

Why do we multiply by the conjugate when dividing?

A complex number in the denominator is not in standard a+bia + bi form. Multiplying top and bottom by the conjugate makes the bottom a2+b2a^2 + b^2, a plain real number, and does not change the value. Then you can split the fraction into a real part and an imaginary part.

Try asking Ducky

  • “Why do I multiply by the conjugate and not just the bottom?”
  • “I got 10 - 5i for the multiplication one. Can you check my signs?”
  • “Quiz me on powers of i until I get five right in a row.”

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