3.2.2 复数代数形式的乘除运算（学案）-【成才之路】2020-2021学年高中新课程数学同步学习指导（人教A版选修1-2）

现行旧教材·高中新课程学习指导 所以 cosx > 0 sinx > 0 sinx > cosx { ,解得 π4 < x < π2 , 所以 π 2 < x + π 4 < 3π 4 , 故 cos(x + π 4 )∈( - 2 2 ,0), 所以 2 2cos(x + π 4 ) + 3∈(1, 3), 故 | z1 + z2 | ∈(1, 3). 课堂达标·固基础 1. D　 ∵ z1 = 2 + ai,z2 = b + i,∴ z1 + z2 = 2 + ai + (b + i) = 0, ∴ (2 + b) + (a + 1)i = 0, ∴ a = - 1,b = - 2,∴ a + bi = - 1 - 2i. 故选 D. 2. A 3. A　 原式 = [(a - b) - (a + b)] + [ - (a + b) + (a - b)]i = - 2b - 2bi. 4. 3　 (1 - i) - (2 + i) + (4 - i) + 3i = 1 - i - 2 - i + 4 - i + 3i = (1 - 2 + 4) + ( - i - i - i + 3i) = 3. 5. (1)解法一:设 z = x + yi(x,y∈R), ∵ z + 1 - 3i = 5 - 2i,∴ x + yi + (1 - 3i) = 5 - 2i, 即 x + 1 = 5 且 y - 3 = - 2,解得 x = 4,y = 1, ∴ z = 4 + i. 解法二:∵ z + 1 - 3i = 5 - 2i,∴ z = (5 - 2i) - (1 - 3i) = 4 + i. (2)设 z = x + yi(x,y∈R),则 | z | = x2 + y2 , 又 | z | + z = 1 + 3i,∴ x2 + y2 + x + yi = 1 + 3i, 由复数相等的定义得 x 2 + y2 + x = 1 y = 3{ ,解得 x = - 4 y = 3{ . 所以 z = - 4 + 3i. 3. 2. 2　 复数代数形式的乘除运算 新知导学 　 　 1. (ac - bd) + (ad + bc)i 2. z2 ·z1 　 z1 z2 + z1 z3 3. (1)a = c 且 b = - d　 (2)a = c 且 b = - d≠0 4. ac + bd c2 + d2 + bc - ad c2 + d2 i 预习自测 1. C　 z = - 3 - 2i,故z 对应的点( - 3, - 2) 位于第三象限. 故选 C. 2. D　 由 z(1 + i) = 2i,得 z = 2i 1 + i = 2i(1 - i) (1 + i)(1 - i) = 2i(1 - i) 2 = i(1 - i) = 1 + i. 故选 D. 3. D　 解法一:∵ z = 2 + i,∴ z = 2 - i, ∴ z·z = (2 + i)(2 - i) = 5. 故选 D. 解法二:∵ z = 2 + i,∴ z·z = | z | 2 = 5. 故选 D. 4. 3 - 2i　 8 - i 2 + i = (8 - i)(2 - i) (2 + i)(2 - i) = 15 - 10i 5 = 3 - 2i. 5. (1)( - 1 + i)(2 + i) i3 = - 2 - i + 2i - 1 - i = - 3 + i - i = ( - 3 + i)i - i·i = - 1 - 3i. (2)(1 + 2i) 2 + 3(1 - i) 2 + i = 1 + 4i - 4 + 3 - 3i 2 + i = i 2 + i = i(2 - i) (2 + i)(2 - i) = 1 + 2i 5 . (3) 1 - i (1 + i)2 + 1 + i (1 - i)2 = 1 - i 2i - 1 + i 2i = 1 - i - 1 - i 2i = - 2i 2i = - 1. (4) 1 - 3i ( 3 + i)2 = 1 - 3i 3 + 2 3i - 1 = 1 - 3i 2 + 2 3i = (1 - 3i) 2 2(1 + 3i)(1 - 3i) = 1 - 2 3i - 3 8 = - 1 + 3i 4 . 互动探究·攻重难 　 　 典例试做 1:(1)(2 + i)(1 + 2i)(2 - i) - 5i = (2 + i)(2 - i)(1 + 2i) - 5i = (4 - i2 )(1 + 2i) - 5i = 5(1 + 2i) - 5i = 5 + 10i - 5i = 5 + 5i. (2)(1 - i)2 (1 + i)2 + 4 = [(1 - i)(1 +