Answers: Arithmetic on complex numbers
These are the answers to Questions: Arithmetic on complex numbers.
Please attempt the questions before reading these answers!
Q1
1.1. \(\quad (5+7i)-(2+3i) = 3 + 4i\)
1.2. \(\quad (8+6i)+(2-4i) = 10 + 2i\)
1.3. \(\quad (4-i\sqrt{2})-(3+i\sqrt{7}) = 1 -(\sqrt{2}+\sqrt{7})i\)
1.4. \(\quad (\sqrt{8}+4i)-(\sqrt{5}+2i) = (2\sqrt{2} - \sqrt{5}) + 2i\)
1.5. \(\quad (\sqrt{7}+3i)+(2-i) = (\sqrt{7} + 2) + 2i\)
1.6. \(\quad (5 + i\sqrt{2}) - (7 - i) + (\sqrt{3} + 4i) = (\sqrt{3} - 2) + (\sqrt{2} + 5)i\)
Q2
2.1. \(\quad (2+3i)(4+5i) = -7 + 22i\)
2.2. \(\quad (3+i)(2-i) = 7 - i\)
2.3. \(\quad 4(6+3i) = 24 + 12i\)
2.4. \(\quad (1+i)^2 = 0 + 2i = 2i\)
2.5. \(\quad (3+2i)^3 = -9 + 46i\)
2.6. \(\quad (7-4i)^2(i-2) = -10 + 145i\)
2.7. \(\quad (1 - i\sqrt3)^3 = -8 + 0i = -8\)
2.8. \(\quad (5-2i)(5+2i) = 29 + 0i = 29\)
2.9. \(\quad (\sqrt{2} + i\sqrt{3})(\sqrt{8} - i\sqrt{3)} = 7 + i\sqrt{6}\)
Q3
3.1. \(\quad \dfrac{7-6i}{1+2i} = -1 - 4i\)
3.2. \(\quad \dfrac{4-i}{1+4i} = 0 - i = -i\)
3.3. \(\quad \dfrac{3}{5i} = 0 -\dfrac{3}{5}i = -\dfrac{3}{5}i\)
3.4. \(\quad \dfrac{4+2i}{3-i} = 1 + i\)
3.5. \(\quad \dfrac{9+i}{i} = 1 - 9i\)
3.6. \(\quad \dfrac{-2-2i}{-2+2i} = 0 + i = i\)
3.7. \(\quad \dfrac{1+5i}{-3i} = -\dfrac{5}{3} + \dfrac{1}{3}i\)
3.8. \(\quad \dfrac{-4}{1-i} = -2 - 2i\)
3.9. \(\quad \dfrac{1-3i}{1+2i} = -1 - i\)
Q4
4.1. \(\quad \dfrac{(6+4i)(3-i)}{2i} = 3 - 11i\)
4.2. \(\quad 3i(5-4i)+(6+2i) = 18 + 17i\)
4.3. \(\quad (2+3i)(1-i)-(5-4i) = 0 + 5i = 5i\)
4.4. \(\quad \dfrac{(5+2i)+(4-i)}{1+i} = 5 - 4i\)
4.5. \(\quad \dfrac{(2+i)^3}{(3+i)-(1+i)} = 1 + \dfrac{11}{2}i\)
4.6. \(\quad (\dfrac{6-3i}{2(1-i)})^2 = \dfrac{9}{2} + \dfrac{27}{8}i\)
Version history and licensing
v1.0: initial version created 11/24 by Charlotte McCarthy as part of a University of St Andrews VIP project.