1. Page | 1 Practice Test on Topic 18 Complex Numbers Test on Topic 18 Complex Numbers 1.Express the following as complex numbers a + bi (a) (b) 2  25 (c)  2 +  8 (d) 4 +  9 (e)  4 +  16 (f)  36 +  25 2.Write each of the following as a complex number in the form a + bi (a)  81  36 (b) (2i – 1)(3i – 2) 3.Write each of the following as a complex number in the form a + bi. (a) (5 + 4i) + (3 + 2i) (c) 2(3 + 2i) + 5(4 – 3i) (b) (3 + i) – (7 + 2i) (d) 4(3 + 4i) – 2(6 – 6i) 4. Write each of the following as a complex number in the form a + bi. (a) 2i(3 + 4i) (c) (2 + i)(4 – 3i) (e) (2 + 3i) 2 (b) 5i(3 + 4i) (d) (5 + 4i)(5 – 4i) (f) 5. 6. The quadratic equation x 2 – 2x + 6 = 0 has two solutions which are both complex numbers. Use the quadratic formula to find these solutions. The quadratic equation x 2 – x + 4 = 0 has two solutions which are both complex numbers. Use the quadratic formula to find these solutions.

2. Practice Test on Topic 18 Complex Numbers Complex Numbers Solutions 1.Express the following as complex numbers a + bi (a)=0 + 3i (b) 2  25 = 2 – 5i (c)  2 +  8 = 2i  2 2i = 0  3 2i (d) 4 +  9 =4+ 9  1 = 4+ 9  1 =4 + 3i (e)  4 +  16 = 4  1 + 16  1 ====== 4  1 + 16  1 2i + 4i 6i (f)  36 +  25 = 6i + 5i = 0 + 11i 2.Write each of the following as a complex number in the form a + bi (b)  81  36 =9i +6i = 15i (b)(2i – 1)(3i – 2)=6i 2 – 4i – 3i + 2=– 6 – 7i + 2=– 4 – 7i 3.Write each of the following as a complex number in the form a + bi. (a) (5 + 4i) + (3 + 2i)=5 + 3 + 4i + 2i=8 + 6i (c) (3 + i) – (7 + 2i) =3 + i – 7 – 2i = 3 – 7 + i– 2i = – 4 – i (c) 2(3 + 2i) + 5(4 – 3i) (d) 4(3 + 4i) – 2(6 – 6i) ==== 6 + 6i + 20 – 15i 12 + 16i – 12 + 12i ==== 9 + 20 + 6i – 15i 12 – 12 + 16i + 12i ==== 29 – 9i 28i 4. Write each of the following as a complex number in the form a + bi. Page | 2

3. Practice Test on Topic 18 Complex Numbers Page | 3 (a) 2i(3 + 4i) =6i + 8i 2 Using the distributive law ====== 6i + 8(–1) 6i – 8 – 8 + 6i Using the property that i 2 = – 1 Rearranging terms (b) 5i(3 + 4i) =15i + 20i 2 ====== Using the distributive law 15i + 20(–1) Using the property that i 2 = – 1 15i – 20 – 20 + 15i Rearranging terms Using FOIL Using the property that i 2 = – 1 Using FOIL Using the property that i 2 = – 1 (c) (2 + i)(4 – 3i) = = (d) (5 + 4i)(5 – 4i) = = 8 – 6i + 4i – 3i 2 8 – 2i – 3(–1) 8 – 2i + 3 11 – 2i 25 – 20i + 20i – 16i 2 8 – 2(–1) 8+2 10 (e)(2 + 3i) 2 =(2 + 3i)(2 + 3i) ======== 4 + 6i + 6i + 9i 2 4 + 12i + 9(–1) 4 + 12i – 9 – 5 + 12i Using the distributive law Using the property that i 2 = – 1 Adding and subtracting like terms (f) First we convert all the numbers to be in terms of i = (-) i Using the distributive law Using the property that i 2 = – 1 = = – 4 – 12i

4. Practice Test on Topic 18 Complex Numbers Page | 4 5.The quadratic equation x 2 – 2x + 6 = 0 has two solutions which are both complex numbers. Use the quadratic formula to find these solutions. Solution: x 2 – 2x + 6 = 0 a=1b= – 2c=6 b 2 – 4ac =(–2) 2 – 4(1)(6) =4 – 24=– 20 So b 2  4ac =  20 = 20 i = 2 5 i x=  b  b 2  4ac 2a =  (  4 )  2 5 i 2(1) = 4  2 5 i 2 The two solutions arex= 4  2 5 i 2 =2 – 5 i andx= 4  2 5 i 2 = 2+ 5 i There are 2 complex number solutions they are x = 2 – 5 i and x = 2 + 5 i 6.The quadratic equation x 2 – x + 4 = 0 has two solutions which are both complex numbers. Use the quadratic formula to find these solutions. Solution: x 2 – 2x + 2 = 0 a=1 b= – 1 c=4 b 2 – 4ac =(– 1) 2 – 4 x 1 x (4) =1 – 16 = – 15 So b 2  4ac =  15 = 15  1 = 15  1 = 15 i = 2  15i 2 x So x or x ======  b  b 2  4ac 2a 2  15i 2 2  15i 2 There are two complex numbers as solutions they are x = 2  15i 2 and x = 2  15i 2