Notes PDF
More Challenging Problems
1. Evaluate (a)(2 + 3i) + (4 – 7i) (b) (2 + 3i) + (4 – 7i)
Answer
1. 1. (a) 6 – 4i (b) -2 + 10i
2. Evaluate (2 + 3i)⋅(4 – 7i) (2 + 3i)2
Answer
1. (a) 29 – 2i (b) -5 + 12i
3. Evaluate (2 + 3i)/(4 – 7i)
Answer
1. -1/5 + (2/5)i
Solution
1. (2 + 3i)/(4 – 7i) = (2 + 3i)/(4 – 7i) ⋅ (4 + 7i)/(4 + 7i) = (2 + 3i)⋅(4 + 7i) / (42 + 72) = (-13 + 26i)/65 = -1/5 + (2/5)i
4. Find the square roots of 9 – 9i.
Answer
1. 3⋅21/4 (cos(3π/8) + i sin(3π/8) and -3 ⋅21/4 (cos(3π/8) + i sin(3π/8)
Solution
1. 9 – 9i has modulus (92 + 92)1/2 = 9√2, and argument 3π/4. Then one square root has modulus (9√2)1/2 = 3⋅21/4 and argument 3π/8, so is 3⋅21/4(cos(3π/8) + i sin(3π/8)) ≈ 1.365 + 3.296i. The other square root is -3⋅21/4(cos(3π/8) + i sin(3π/8)) ≈ -1.365 – 3.296i.
5. Solve z2 = 9 – 9i.
Answer
1. 3⋅21/4 (cos(3π/8) + i sin (3π/8)) and -3⋅21/4 (cos(3π/8) + i sin (3π/8)
Solution
1. The solutions are z = square roots of 9 + 9i, solved in problem 4