Subtract: 3) (6 + 5i) (8 + 11i) – 4) [(4 2i) (3 + 5i)] (2 + i) – – – Multiply: 4) 4i(2 + 3i) 5) (2 + i)(3 + 2i) 6) (4 + 6i)(3 2i) – 7) (3 + 2i)(3 2i) – Divide: For quotients use the to clear from the denominator i ‘complex conjugate’ and write your answer in the form a + bi 8) 9) 10) 11) Simplifying powers of i: Powers of are cyclic. Any power of can be simplified i i to 1. i,-1, -i, If you know these first 4 powers of you can figure out any power of i, i. i1 = i -1 i2 = = i3 i2i1=-i 1 i4 = i2i2= Watch what happens as we continue: 1i i5 = i4i = = i = -1 i6 =i4i2 =1(-1) = i7 i4i3=-i
Subtract: 3) (6 + 5i) – (8 + 11i) 4) [(4 – 2i) – (3 + 5i)] – (2 + i) Multiply: 4) 4i(2 + 3i) 5) (2 + i)(3 + 2i) 6) (4 + 6i)(3 – 2i) 7) (3 + 2i)(3 – 2i) Divide: For quotients use the ‘complex conjugate’ to clear i from the denominator and write your answer in the form a + bi 8) 9) 10) 11) Simplifying powers of i: Powers of i are cyclic. Any power of i can be simplified to i,-1, -i, 1. If you know these first 4 powers of i, you can figure out any power of i. i1 = i i2 = -1 i3 = i2i1=-i i4 = i2i2= 1 Watch what happens as we continue: i5 = i4i = 1i = i i6 =i4i2 =1(-1) = -1 i7 = i4i3=-i