复数代数形式的加减运算及其几何意义(教案).pdf

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1、学习必备欢迎下载新授课:3.2.1 复数代数形式的加减运算及其几何意义教学目标重点:复数代数形式的加法、减法的运算法则难点:复数加法、减法的几何意义.知识点:1.掌握复数代数形式的加、减运算法则;2.理解复数代数形式的加、减运算的几何意义.能力点:培养学生渗透转化、数形结合的数学思想方法,提高学生分析问题、解决问题以及运算的能力教育点:通过探究学习,培养学生互助合作的学习习惯,培养学生对数学探索和渴求的思想.在掌握知识的同时,形成良好的思维品质和锲而不舍的钻研精神.自主探究点：如何运用复数加法、减法的几何意义来解决问题.考试点：会计算复数的和与差;能用复数加、减法的几何意义解决简单问题.易错易

2、混点：复数的加法与减法的综合应用.拓展点：复数与其他知识的综合.一、引入新课复习引入1.虚数单位i：它的平方等于1,即2i1;2.对于复数i,zaba bR：当且仅当0b时,z是实数a;当0b时,z为虚数;当0a且0b时,z为纯虚数;当且仅当0ab时,z就是实数0.3.复数集与其它数集之间的关系：NZQRC.4.复数几何意义：我们把实数系扩充到了复数系,那么复数之间是否存在运算呢?答案是肯定的,这节课我们就来研究复数的加减运算.【设计意图】通过复习回顾复数概念、几何意义等相关知识,使学生对这一知识结构有个清醒的初步认知,逐渐过渡到对复数代数形式的加减运算及其几何意义的学习情境,为探究本节课的新

3、知识作铺垫.二、探究新知复数i,zaba bR复平面内的点,a bZ一一对应一一对应复数i,zaba bR复平面内的向量=,OZa b学习必备欢迎下载探究一:复数的加法1.复数的加法法则我们规定,复数的加法法则如下:设1izab,2i(,)zcd a b c dR是任意两个复数,那么：12(i)(i)()()izzabcdacbd提出问题：(1)两个复数的和是个什么数,它的值唯一确定吗?(2)当=0,0bd时,与实数加法法则一致吗?(3)它的实质是什么?类似于实数的哪种运算方法?学生明确:(1)仍然是个复数,且是一个确定的复数;(2)一致;(3)实质是实部与实部相加,虚部与虚部相加,类似于实数

4、运算中的合并同类项【设计意图】加深对复数加法法则的理解,且与实数类比,了解规定的合理性:将实数的运算通性、通法扩充到复数,有利于培养学生的学习兴趣和创新精神2.复数加法的运算律实数的加法有交换律、结合律,复数的加法满足这些运算律吗?对任意的123,z zzC,有1221zzzz（交换律）,123123()()zzzzzz（结合律）.【设计意图】引导学生根据实数加法满足的运算律,大胆尝试推导复数加法的运算律,学生先独立思考,然后小组交流.提高学生的建构能力及主动发现问题,探究问题的能力3.复数加法的几何意义复数与复平面内的向量有一一对应关系,那么请同学们猜想一下,复数的加法也有这种对应关系吗?设

5、12,OZ OZ分别与复数i,iab cd 对应,则有12(,),(,)OZa b OZc d,由平面向量的坐标运算有12(,)OZOZac bd.这说明两个向量12OZOZ与的和就是与复数()+()iacbd对应的向量.因此,复数的加法可以按照向量加法的平行四边形法则来进行.这就是复数加法的几何意义.如图所示：2(,)Zc d1(,)Za b由图可以看出,以1OZ、2OZ 为邻边画平行四边形12OZ ZZ,其对角线OZ 所表示的向量OZ就是复数()+()iacbd对应的向量.【设计意图】通过向量的知识,让学生体会从数形结合的角度来认识复数的加减法法则,训练学生的形象思ZOyx文档编码:CW3

6、T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4

7、 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3

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9、 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3

10、T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4

11、 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3

12、T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3学习必备欢迎下载维能力,也培养了学生的数形结合思想.另外,当两复数的对应向量共线时,可直接运算;当不共线时,可类比向量加法的平行四边形,也培养了学生的类比思想.探究二:复数的减法类比复数的加法法则,你能试着推导复数减法法则吗?1.复数的减法法则我们规定,复数的减法是加法的逆运算,即把满足(i)(i)icdxyab的复数ixy 叫做复数iab 减去icd 的差,记作(i)(i

13、)abcd.根据复数相等的定义,有,cxa dyb,因此,xac ybd,所以i()()ixyacbd,即(i)(i)()()iabcdacbd.这就是复数的减法法则,所以两个复数的差是一个确定的复数.【设计意图】复数的减法运算法则是通过转化为加法运算而得到的,渗透了转化的数学思想方法,是学生体会数学思想的素材.让学生自己动手推导减法法则,有利于培养学生的创新能力和互助合作的学习习惯.考查学生的类比思想,提高学生主动发现问题,探究问题的能力2.复数减法的几何意义设12,OZ OZ分别与复数i,iab cd对应,则这两个复数的差12zz与向量12OZOZ（即21Z Z）对应,这就是复数减法的几何

14、意义.如图所示.【设计意图】两个复数的差12zz（即12OZOZ）与连接两个终点1Z,2Z,且指向被减数的向量对应,这与平面向量的几何解释是一致的;它不仅又一次让我们看到了向量这一工具的功能,也使数和形得到了有机的结合 注意：只有将差向量平移至以原点为起点时,其终点才能对应该复数.三、理解新知1.复数的加减法法则:设1izab,2i(,)zcd a b c dR是任意两个复数,规定：12()()izzacbd;12()()izzacbd.yx2Z1ZO文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4

15、P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1

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17、P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1

18、S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4

19、P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1

20、S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4

21、P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3学习必备欢迎下载2.复数加、减法的几何意义：(1)复数的加法按照向量加法的平行四边形法则;(2)复数的减法按照向量减法的三角形法则.3.几点说明:(1)复数的加(减)法法则规定的合理性:它既与实数运算法则,运算律相同,又与向量完美地结合起来;(2)复数的加(减)法实质是:复数的实部与实部、虚部与虚部分别相加减;(3)多个复数相加减:可将各个复数的实部与实部、虚部与虚部分别相加减(4)复平面内的两点间距离公式:12dzz.其中12,z z是复平面内的两点1Z和2Z所对应的复数,d为点1Z和点2Z间的距

22、离.即两个复数差的模的几何意义是:两个复数所对应的两个点之间的距离【设计意图】加深对复数加(减)法法则的理解,从不同的角度总结,既学到知识,又学到了数学方法,使知识更加系统化,学生的思维将上升到一个更高的层面,为准确地运用新知,作必要的铺垫.培养学生的归纳概括能力,使学生对所学的知识有一个整体的认识,解决问题时可以信手拈来.四、运用新知例1.计算：(1)(23i)(5i);(2)(12i)(12i);(3)(23i)(52i);(4)(56i)(2i)(34i);解:(1)(23i)(5i)(25)(31)i32i;(2)(12i)(12i)(11)(22)i0;(3)(23i)(52i)(2

23、5)(32)i35i;(4)(56i)(2i)(34i)(523)(614)i11i.【设计意图】直接运用复数的加、减法运算法则进行,就是将它们的实部、虚部分别相加、减,实数范围的运算律在复数范围内仍然成立.变式训练:计算(12i)(23i)(34i)(45)i(19992000i)(20002001i).解：（解法一）原式(1 2345619992000)(2345620002001)i1000 1000i.（解法二）(12i)(23i)1i;(34i)(45i)1i;(19992000i)(20002001i)1i.将上列1000个式子累加,得1000(1i)10001000i.【设计意图

24、】复数的加减法,相当于多项式加减中的合并同类项的过程;如果根据给出复数求和的特征从局部入手,抓住了式子中相邻两项之差是一个常量这一特点,适当地进行组合,从而可简化运算.进一步巩固复数加减运算,并带有一定的规律性.例2.(1)设12,OZ OZ分别与复数1253i,14izz对应,计算12zz,并在复平面内作出12OZOZ,文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 H

25、R1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9

26、V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 H

27、R1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9

28、V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 H

29、R1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9

30、V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3学习必备欢迎下载Z2Z1OyxZZ2

31、Z1Oyx(2)设12,OZ OZ分别与复数1213i,2izz对应,计算12z z+,并在复平面内作出12OZOZ.解：图1图2(1)12=(5+3i)(14i)(51)(34)i4izz.（如图1所示）;(2)12(13i)(2i)(12)(31)i34iz z+.（如图2所示）.【设计意图】由复数的几何意义知,复数1z,2z 所对应的的点分别为12,Z Z.12OZOZ就是表示向量21Z Z,而12OZOZ可利用平行四边形法则作出.变式训练:已知复数213(5)izaa,221(21)i()zaaaaR分别对应向量12,OZ OZ（O为坐标原点）,若向量12Z Z对应的复数为纯虚数,求a

32、的值.答案：1a.例3.已知关于x的方程：2(6i)9i0()xxaaR有实数根b.(1)求实数,a b的值;(2)若复数z满足i20zabz,求z的最小值解：(1)由题意,得2(6i)9i0bba,即2(69)()i0bbab.由复数相等的定义得26900bbab,解得3ab.(2)设i(,)zxy x yR,由i20zabz,得(3)(3)i2xyz,即222(3)(3)4()xyxy,整理得22(1)(1)8xy,即复数z在复平面内所对应的点Z(,)x y的轨迹是以C(1,1)为圆心,半径长为2 2的圆.又z的几何意义是Z(,)x y与原点O(0,0)的距离,文档编码:CW3T6L1S8

33、I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5

34、A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8

35、I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5

36、A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8

37、I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5

38、A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8

39、I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3学习必备欢迎下载如图,由平面几何知识知,min2 222zCACO.【设计意图】在问题(1)中由复数相等的概念,列方程组求出两个参数值,把复数问题实数化,既复习了概念,又锻炼了学生的计算能力和解决问题的能力;在问题(2)中由22(0)(0)zxy,把z转化为复数z所对应的点与原点的距离,解决此类问题的关键是利用复数的几何意义画出图形,在图形中寻求答案,把数转化成形,利用数形结合思想解决

40、即可变式训练:复数z的模为1,求1iz的最大值和最小值.答案:2+1,21.【设计意图】通过变式训练,便于学生全面的认识利用复数差的模的几何意义解决问题,提高学生理解、运用知识的能力.五、课堂小结(一)知识:1.复数代数形式的加法、减法的运算法则;2.复数加法、减法的几何意义.3.几点说明:(1)复数的加(减)法法则规定的合理性:它既与实数运算法则,运算律相同,又与向量完美地结合起来;(2)复数的加(减)法实质是:复数的实部与实部、虚部与虚部分别相加减;(3)多个复数相加减:可将各个复数的实部与实部、虚部与虚部分别相加减(4)复平面内的两点间距离公式:12dzz.其中12,z z是复平面内的两

41、点1Z和2Z所对应的复数,d为点1Z和点2Z间的距离.即两个复数差的模的几何意义是:两个复数所对应的两个点之间的距离(二)思想方法:类比的思想、转化的思想、数形结合的思想【设计意图】通过课堂小结,增强学生对复数代数形式的加法、减法的运算法则及几何意义的理解,及时查缺补漏,从而更好地运用知识,解题要有目的性,加强对数学知识、思想方法的认识与自觉运用深化对知识的理解,完善认识结构,领悟思想方法,强化情感体验,提高认识能力.引导学生自我反馈、自我总结,并对所学知识进行提炼升华,使知识系统化.让学生学会学习,学会内化知识的方法与经验,促进学习目标的完成.六、布置作业文档编码:CW3T6L1S8I4 H

42、R1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9

43、V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 H

44、R1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9

45、V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 H

46、R1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9

47、V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 H

48、R1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3文档编码:CW3T6L1S8I4 HR1C9Z4Z4D4 ZE4P5A7Q9V3学习必备欢迎下载必做题：1.计算：(1)(24i)(34i);(2)(34 i)(2i)(1.2.复数6+5i与3+4i对应的向量分别是OA与OB,其中O是原点,求向量AB,BA对应的复数,并指出其对应的复数位于第几象限.3.复平面上三点,A B C 分别对应复数1,2i,52i,则由,A B C 所构成的三角形ABC是三角形.4.求复数2i,3i所对应的两点之间的距离5.已知复数z 满足+2

49、8izz,求复数 z.6.已知平行四边形OABC的三个顶点,O A C对应的复数分别为0,32i,24i,试求:(1)AO表示的复数;(2)CA表示的复数;(3)B点对应的复数.答案:1.(1)5;(2)22i.2.9i,位于第三象限;9i,位于第一象限.3.直角三角形.4.5.5.158iz.6.(1)32i;(2)52i;(3)16i选做题：1.在复平面内,求满足方程z+izi4的复数z所对应的点的轨迹2.复数12z,z满足12zz1,12z+z2,求12zz.答案：1.提示:方程可以变形为z(i)zi4|,表示到两个定点(0,1)和(0,1)距离之和等于4的点的轨迹,故满足方程的动点轨迹

50、是椭圆2.提示:法一:数形结合思想,构造边长为1的正方形,则其中一条对角线的长度为2,则所求的另一条对角线的长度也等于2.法二:(向量法)设12z,z所对应的向量分别是a,b,将12z+z2两边平方得0a b,则212(zz)2,所以12zz2.【设计意图】设计必做题是引导学生先复习,再作业,培养学生良好的学习习惯,是让学生会用复数代数形式的加法、减法的运算法则进行计算;设计选做题意在培养学生深刻理解复数差的模的几何意义,增加问题的多样性、趣味性,训练学生思维的发散性、深刻性.让学生理解知识之间的联系,培养学生用整体的观点看问题,起到巩固旧知的作用七、教后反思1.本教案的亮点是:文档编码:CW