Exercise 4.1 answers: Complex Numbers and Quadratic Equations

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1. Multiply the numbers in front of i: 5 × (−3/5) = −3.
2. Multiply the two i's: i × i = i² = −1.
3. So the product is −3 × (−1) = 3.

Answer3

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1. 9 = 4+4+1, so i⁹ = (i⁴)² × i = 1 × i = i.
2. 19 = 4+4+4+4+3, so i¹⁹ = (i⁴)⁴ × i³ = 1 × i³.
3. i³ = i² × i = −i.
4. Add: i + (−i) = 0.

Answer0

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1. 39 = 4×9 + 3, so i³⁹ = (i⁴)⁹ × i³ = 1 × (−i) = −i.
2. i⁻³⁹ = 1 ÷ i³⁹ = 1 ÷ (−i).
3. Multiply top and bottom by i: 1/(−i) = i / (−i²) = i / 1 = i.

Answeri

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1. Multiply out the first bracket: 3×7 + 3×7i = 21 + 21i.
2. Multiply out the second bracket: i×7 + i×7i = 7i + 7i².
3. i² = −1, so 7i² = −7.
4. Add everything: 21 + 21i + 7i − 7 = 14 + 28i.

Answer14 + 28i

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1. Remove the bracket, flipping the sign of each term inside it: 1 − i + 1 − 6i.
2. Add the plain numbers: 1 + 1 = 2.
3. Add the i-numbers: −1i − 6i = −7i.

Answer2 − 7i

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1. Subtract the plain parts: 1/5 − 4 = −19/5.
2. Subtract the i-parts: 2/5 − 5/2 = 4/10 − 25/10 = −21/10.

Answer−19/5 − 21/10 i

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1. Add the first two brackets. Plain parts: 1/3 + 4 = 13/3. i-parts: 7/3 + 1/3 = 8/3.
2. So the sum so far is 13/3 + 8/3 i.
3. Now subtract (−4/3 + i). Plain parts: 13/3 − (−4/3) = 17/3.
4. i-parts: 8/3 − 1 = 5/3.

Answer17/3 + 5/3 i

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1. Square (1 − i) first: 1 − 2i + i² = 1 − 2i − 1 = −2i.
2. Now square that result: (−2i)² = 4i² = −4.

Answer−4

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1. Use (a+b)³ = a³ + 3a²b + 3ab² + b³ with a = 1/3, b = 3i.
2. a³ = 1/27. 3a²b = 3×(1/9)×3i = i. 3ab² = 3×(1/3)×9i² = 9×(−1) = −9. b³ = 27i³ = −27i.
3. Add the plain parts: 1/27 − 9 = −242/27.
4. Add the i-parts: i − 27i = −26i.

Answer−242/27 − 26i

Watch this explained “Cubes, and a fourth power”, 10:53 into Which school algebra identities survive the enlargement, and why

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1. Use (a+b)³ = a³ + 3a²b + 3ab² + b³ with a = −2, b = −1/3 i.
2. a³ = −8. b² = (−1/3 i)² = −1/9, so 3ab² = 3×(−2)×(−1/9) = 2/3.
3. 3a²b = 3×4×(−1/3 i) = −4i. b³ = b²×b = (−1/9)×(−1/3 i) = 1/27 i.
4. Plain parts: −8 + 2/3 = −22/3.
5. i-parts: −4i + 1/27 i = −107/27 i.

Answer−22/3 − 107/27 i

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1. For z = 4 − 3i, the inverse is z⁻¹ = 1/z = (conjugate of z)/|z|²
2. Conjugate: z̄ = 4 + 3i
3. |z|² = 4² + (−3)² = 16 + 9 = 25
4. z⁻¹ = (4 + 3i)/25 = 4/25 + 3i/25

Answer4/25 + 3i/25

Watch this explained “A few more, worked through”, 12:55 into Multiplying, then inverting: how division becomes possible

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1. The inverse of a + ib is (a − ib)/(a² + b²).
2. Here a = √5 and b = 3, so a − ib = √5 − 3i.
3. a² + b² = (√5)² + 3² = 5 + 9 = 14.
4. So the inverse is (√5 − 3i)/14 = √5/14 − 3/14 i.
5. Check: (√5 + 3i)(√5 − 3i) = 5 + 9 = 14, and 14/14 = 1.

Answer√5/14 − 3/14 i

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1. For z = −i, the inverse is z⁻¹ = 1/(−i)
2. Multiply by i/i: [1/(−i)] × [i/i] = i/(−i²)
3. = i/(−(−1))
4. = i/1 = i

Answeri

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1. The top is a number times its own conjugate: (3)²+(√5)² = 9+5 = 14.
2. The bottom: (√3+i√2) − (√3−i√2) = 2i√2.
3. Divide: 14/(2i√2) = 7/(i√2).
4. Multiply top and bottom by −i (since 1/i = −i): 7×(−i)/√2 = −7i/√2 = −7√2/2 i.

Answer0 − 7√2/2 i

Watch this explained “Clearing a denominator”, 12:14 into Which school algebra identities survive the enlargement, and why