MATH198 Solution Sheet 1 1. Given that z 1 =3 - 2i and z 2 =2+ i (a) determine (i) |z 1 | , (ii) Re z 1 , (iii) Im z 2 . (i) |z 1 | =  3 2 +(-2) 2 = √ 13 . Note: Not  3 2 +(-2i) 2 . (ii) Re z 1 =3 (iii) Im z 2 =1 . By definition, Im(a + ib) ≡ b. (b) Determine in the form a + ib (i)2z 1 +3z 2 , (ii) z 1 z 2 , (iii) 1 z 1 , (iv) z 1 z 2 . (i) 2z 1 +3z 2 = 2(3 - 2i) + 3(2 + i)=6 - 4i +6+3i = 12 - i (ii) z 1 z 2 = (3 - 2i)(2 + i)=6+3i - 4i - 2i 2 =8 - i (iii) 1 z 1 = 1 3 - 2i = 1 3 - 2i 3+2i 3+2i = 3+2i 9+4 = 3 13 + 2 13 i (iv) z 1 z 2 = 3 - 2i 2+ i = (3 - 2i)(2 - i) (2 + i)(2 - i) = 4 - 7i 5 = 4 5 - 7 5 i (c) Find the complex conjugate of z 1 - z 2 . Taking the complex conjugate means reversing the sign of the imaginary part: z 1 - z 2 = (3 - 2i) - (2 + i)=1 - 3i ⇒ (z 1 - z 2 ) ⋆ =1+3i 2. Simplify (i) i 6 , (ii) (1 + i)(1 - i), (iii) i -5 . (i) i 6 = i 4 i 2 =1 × (-1) = -1 (ii) (1 + i)(1 - i)=1+ i - i - i 2 =2 (iii) i -5 = i -4 i -1 = 1 i = -i -i is a “better” answer than 1 i , (because -i fits the standard form a + ib).