De Moivre’s theorem, n-th roots and square roots of complex numbers
Week 6 · De Moivre’s theorem, n-th roots of a complex number, square roots of a complex number and quadratic equations with complex coefficients · 3 lessons
3
Square roots of a complex number in the form x + yi, and quadratic equations with complex coefficients
Pure Mathematics Sec 3 Term 1, Unit 2, Lesson 2
GoalFind the two square roots of a + bi in algebraic form and solve a quadratic with complex coefficients.
New words
square root of a + bi · a number x + yi whose square is a + bi; there are always two, one the negative of the otherdiscriminant · b² − 4ac of the quadratic az² + bz + c = 0complex coefficients · the numbers a, b, c of the quadratic are themselves complex
Explanation
Sometimes you need square roots in the form x + yi, not in trigonometric form. Look for x + yi with (x + yi)² = a + bi. Expanding gives x² − y² + 2xy·i = a + bi, and comparing real and imaginary parts: x² − y² = a and 2xy = b. A third equation comes from moduli: the modulus of (x + yi)² is x² + y², and it must equal the modulus of a + bi, so x² + y² = √(a² + b²). Add and subtract it with the first equation to get x² and y², take square roots, and decide the signs from 2xy = b: if b is positive, x and y have the same sign; if b is negative, opposite signs. The answers always come as a pair ±(x + yi). The table shows three numbers with the same modulus 13. For a quadratic az² + bz + c = 0 with complex coefficients, the formula z = (−b ± √(b² − 4ac))/(2a) still works: find the discriminant (in Arabic المميز), take its complex square root with the method above, and finish. Unlike real quadratics, the two roots need not be conjugates of each other.
number w
x² − y²
x² + y²
x² and y²
square roots
5 + 12i
5
13
9 and 4
±(3 + 2i)
−5 + 12i
−5
13
4 and 9
±(2 + 3i)
−5 − 12i
−5
13
4 and 9
±(2 − 3i)
Worked examples
Find the square roots of 15 − 8i. Here x² − y² = 15 and 2xy = −8. The modulus is √(225 + 64) = 17, so x² + y² = 17. Adding: 2x² = 32, so x² = 16 and x = ±4; then y² = 1. Because 2xy is negative the signs are opposite: 4 − i or −4 + i. Check: (4 − i)² = 16 − 1 − 8i = 15 − 8i. The square roots are ±(4 − i).
Solve z² − (4 + i)z + 5 + 5i = 0. Discriminant: (4 + i)² − 4(5 + 5i) = 15 + 8i − 20 − 20i = −5 − 12i. Its square roots (third row of the table): ±(2 − 3i). Then z = (4 + i ± (2 − 3i))/2 gives z = (6 − 2i)/2 = 3 − i or z = (2 + 4i)/2 = 1 + 2i.
Solve (1 + i)z² − (2 + 4i)z + 4i = 0. Divide by the leading coefficient first: (2 + 4i)/(1 + i) = (2 + 4i)(1 − i)/2 = 3 + i and 4i/(1 + i) = 4i(1 − i)/2 = 2 + 2i. So z² − (3 + i)z + 2 + 2i = 0. The discriminant is (3 + i)² − 4(2 + 2i) = 8 + 6i − 8 − 8i = −2i, whose square roots are ±(1 − i) because (1 − i)² = −2i. Then z = (3 + i ± (1 − i))/2: z = 2 or z = 1 + i.
Class activity
Fast check. Square 3 + 2i, and also square −3 − 2i. Why do both give 5 + 12i? What does that tell you about the two square roots of a number?
Practice
1
Find the square roots of 21 + 20i.
x² − y² = 21 and x² + y² = 29, so x² = 25 and y² = 4, same signs: ±(5 + 2i). Check: 25 − 4 + 20i.
2
Find the square roots of −8 − 6i.
x² − y² = −8 and x² + y² = 10, so x² = 1 and y² = 9, opposite signs: ±(1 − 3i). Check: 1 − 9 − 6i.
3
Solve z² − (5 − i)z + 8 − i = 0.
The discriminant is (5 − i)² − 4(8 − i) = 24 − 10i − 32 + 4i = −8 − 6i, with square roots ±(1 − 3i) (previous item). So z = (5 − i ± (1 − 3i))/2: z = 3 − 2i or z = 2 + i.
4
Solve iz² + (1 − 2i)z − 1 + i = 0. Divide by i first (remember 1/i = −i).
(1 − 2i)/i = (1 − 2i)(−i) = −2 − i and (−1 + i)/i = (−1 + i)(−i) = 1 + i, so z² − (2 + i)z + 1 + i = 0. The discriminant is (2 + i)² − 4(1 + i) = 3 + 4i − 4 − 4i = −1, with square roots ±i. So z = (2 + i ± i)/2: z = 1 + i or z = 1.
5
A quadratic with real coefficients has the root 2 + 3i. What is its other root? Does the same hold for Example 2, whose roots are 3 − i and 1 + 2i?
A real quadratic has conjugate roots, so the other root is 2 − 3i. In Example 2 the roots 3 − i and 1 + 2i are not conjugates, because that quadratic does not have real coefficients.
