复数的三角形式及乘除运算

复数的三角形式及乘除运算 一、主要内容: 复数的三角形式，模与辐角的概念及几何意义，用三角形式进行复数乘除运算及几何意义. 二、学习要求: 1(熟练进行复数的代数形式与三角形式的互化，会求复数的模、辐角及辐角主值. 2(深刻理解复数三角形式的结构特征，熟练运用有关三角公式化复数为三角形式. 3(能够利用复数模及辐角主值的几何意义求它们的范围(最值). 4(利用复数三角形式熟练进行复数乘除运算，并能根据乘除运算的几何意义解决相关问题. 5(注意多种解题方法的灵活运用，体会数形结合、分类讨论等数学思想方法. 三、重点: 复数的代数形式向三角形式的转换，复数模及复数乘除运算几何意义的综合运用. 四、学习建议: 1(复数的三角形式是彻底解决复数乘、除、乘方和开方问题的桥梁，相比之下，代数形式在这些方面显得有点力不从心，因此，做好代数形式向三角形式的转化是非常有必要的. 前面已经学习过了复数的另两种表示.一是代数表示，即Z=a+bi(a,b?R).二是几何表示，复数Z既可以用复平面上的点Z(a,b)表示，也可以用复平面上的向量来表示.现在需要学习复数的三角表示.既用复数Z的模和辐角来表示，设其模为r，辐角为θ，则Z=r(cosθ+isinθ)(r?0). 既然这三种方式都可以表示同一个复数，它们之间一定有内在的联系并能够进行互化. 代数形式r=三角形式 R) Z=r(cosθ+isinθ)(r?0) Z=a+bi(a,b? 复数三角形式的结构特征是:模非负，角相同，余弦前，加号连.否则不是三角形式.三角形式中θ应是复数Z的一个辐角，不一定是辐角主值. 例1(下列各式是否是三角形式，若不是，化为三角形式: (1) Z=-2(cosθ+isinθ) (2) Z=cosθ-isinθ (3) Z=-sinθ+icosθ 123 (4) Z=-sinθ-icosθ (5) Z=cos60?+isin30? 45 分析:由三角形式的结构特征，确定判断的依据和变形的方向.变形时，可按照如下进行:首先确定复数Z对应点所在象限(此处可假定θ为锐角)，其次判断是否要变换三角函数名称，最后确定辐角.此步骤可简称为“定点?定名?定角”.这样，使变形的方向更具操作性，能有效提高解决此类问题的正确率. 解:(1)由“模非负”知，不是三角形式，需做变换:Z=Z(-cosθ-isinθ) 1 复平面上Z(-2cosθ,-2sinθ)在第三象限(假定θ为锐角)，余弦“-cosθ”已在前，不需再变换三角函数名称，1 因此可用诱导公式“π+θ”将θ变换到第三象限.?Z=Z(-cosθ-isinθ)=2[cos(π+θ)+isin(π+θ)] 1 (2)由“加号连”知，不是三角形式 复平面上点Z(cosθ,-sinθ)在第四象限(假定θ为锐角)，不需改变三角函数名称，可用诱导公式 2 “2π-θ”或“-θ”将θ变换到第四象限. ? Z=cosθ-isinθ=cos(-θ)+isin(-θ)或Z=cosθ-isinθ=cos(2π-θ)+isin(2π-θ) 22 考虑到复数辐角的不唯一性，复数的三角形式也不唯一. (3)由“余弦前”知，不是三角形式 复平面上点Z(-sinθ,cosθ)在第二象限(假定θ为锐角)，需改变三角函数名称，可用诱导公式 3 “+θ”将θ变换到第二象限. height and strength are in line with the actual needs of the installation; prior to appliance installation, demand for flats will not be flat wall surface, ground level according to the request object level adjustable leveling feet and lines when installing fixed. Coordinates and elevations are not allowed to carefully check level elevations, wall lines and installation of centerline and high accuracy of the horizontal line ... Always in a State of controlled, manufacturing process quality. Third section user service and the engineering warranty user operation maintenance manual of prepared due to electromechanical installation engineering has strong of professional sex, and systemic, for easy owners of using and management, engineering completion Shi, we will for electromechanical installation engineering prepared operation and the maintenance manual, by approved of official manual submitted owners, manual contains system description, and technology description and maintenance maintenance content, each a system independent book, specific following: (1) system description: detailed introduced all system main equipment of specifications model and function; Provide equipment, product brochures and so on. (2) technical specifications: technical descriptions provided by the contract are included in the installation of equipment and technical data and descriptions of the features of the part: (4) security: the correct procedures for various types of equipment, operation ?Z(-sinθ,cosθ)=cos(+θ)+isin(+θ) 3 同理(4)Z=-sinθ-icosθ=cos(π-θ)+isin(π-θ) 4 (5)Z=cos60?+isin30?=+i=(1+i)=?(cos+isin)=(cos+isin) 5 小结:对这类与三角形式很相似的式子，如何将之变换为三角形式，对于初学者来讲是个难点.有了“定点?定名?定角”这样一个可操作的步骤，应能够很好地解决此类问题. 例2(求复数Z=1+cosθ+isinθ(π<θ<2π)的模与辐角主值. 分析:式子中多3个“1”，只有将“1”消去，才能更接近三角形式，因此可利用三角公式消“1”. 2 解:Z=1+cosθ+isinθ=1+(2cos-1)+2i?sincos=2cos(cos+isin)........(1) ? π<θ<2π ? <<π, ?cos<0 ?(1)式右端=-2cos(-cos-isin)=-2cos[cos(π+)]+isin(π+)] ? r=-2cos, ArgZ=π++2kπ(k?Z) ? <<π ? π<π+<2π， ?argZ=π+. 小结:(1)式右端从形式上看似乎就是三角形式.不少同学认为r=2cos, argZ=或ArgZ= 错误之处在于他们没有去考虑θ角范围，因此一定要用“模非负，角相同，余弦前，加号连”来判断是否为三角形式.看了这道例题，你一定能解决如Z=1-cosθ+isinθ(π<θ<2π) ，Z=1+cosθ-isinθ(π<θ<2π)等类似问题. 12 例3(将Z=(π<θ<3π)化为三角形式，并求其辐角主值. 分析:三角形中只有正余弦，因此首先想到“化切为弦”.下一步当然是要分母实数化，再向三角形式转化. 解:====cos2θ+isin2θ features of the part: (4) security: the correct procedures for various types of equipment, operation of theical descriptions provided by the contract are included in the installation of equipment and technical data and descriptions ent of specifications model and function; Provide equipment, product brochures and so on. (2) technical specifications: technquipmce content, each a system independent book, specific following: (1) system description: detailed introduced all system main eof official manual submitted owners, manual contains system description, and technology description and maintenance maintenan provedg completion Shi, we will for electromechanical installation engineering prepared operation and the maintenance manual, by apal installation engineering has strong of professional sex, and systemic, for easy owners of using and management, engineerinchanicality. Third section user service and the engineering warranty user operation maintenance manual of prepared due to electromeallation of centerline and high accuracy of the horizontal line ... Always in a State of controlled, manufacturing process qud instnd lines when installing fixed. Coordinates and elevations are not allowed to carefully check level elevations, wall lines andemand for flats will not be flat wall surface, ground level according to the request object level adjustable leveling feet aheight and strength are in line with the actual needs of the installation; prior to appliance installation, 2 ?π<θ<3π, ?<2θ<6π, ?π<2θ-4π<2π，? argZ=2θ-4π 小结:掌握三角变形是解决这类问题的根本.但在此之前的解题方向一定要明确，即要分析式子结构.比较其与三角形式的异同，从而决定变形的方向，采用正确的方法.要求学生做好每道例题后的反思，并能由此及彼，举一反三，达到熟练解决一类问题的目的，如1-itgθ, tgθ+i, i-ctgθ等. 2(复数Z的模|Z|的几何意义是:复平面上点Z到原点距离，复数模|Z-Z|的几何意义是:复平面上两点Z，121Z之间距离.辐角几何意义是:以x轴正半轴为角始边，以向量所在射线为终边的角记为ArgZ.在[0，2π)范围2 内的辐角称辐角主值，记为argZ. 要求学生不仅要理解以上所说各几何意义，还要运用几何意义去解决相关问题. 例4(若Z?c，|Z-2|?1，求,,,的最大，最小值和argZ范围. 解:法一，数形结合 由|Z-2|?1，知,的轨迹为复平面上以(2，0)为圆心，1为半径的圆面(包括圆周)，|Z|表示圆面上任一点到原点的距离. 显然1?|Z|?3, ?|Z|=3, |Z|=1, maxmin 另设圆的两条切线为OA，OB，A，B为切点，由|CA|=1，|OC|=2知 ?AOC=?BOC=，?argZ?[0,]?[π,2π) 法二:用代数形式求解|Z|的最大，最小值，设Z=x+yi(x,y?R) 22 则由|Z-2|?1得(x-2)+y?1, ? |Z|=?=, 222 ? (x-2)+y?1, ?(x-2)?1, ?-1?x-2?1, ?1?x?3, ? 1?4x-3?9, ?1?|Z|?3. 小结:在一题多解的基础上，分析比较各种方法的异同，如何做好方法的选择.各种方法的本质和优势，通过分析与比较都一目了然. 例5(复数Z满足arg(Z+3)=π,求|z+6|+|z-3i|最小值. 分析:由两个复数模的和取最小值，联想到一个点到两个定点距离和的最小 值，将之转化为几何问题来解决应比较简便. 解法一:由arg(Z+3)=π,知Z+3的轨迹是一条射线OA，?xOA=π,而 |Z+6|+|Z-3i|=|(z+3)-(-3)|+|(Z+3)-(3+3i)| 将B(-3,0)与C(3,3)连结，BC连线与OA交点为D，取Z+3为D点，表示复数时， |Z+6|+|Z-3i|=|BD|+|DC|=|BC|=3, ? 所求最小值=3. curityontract are included in the installation of equipment and technical data and descriptions of the features of the part: (4) section; Provide equipment, product brochures and so on. (2) technical specifications: technical descriptions provided by the cnd funent book, specific following: (1) system description: detailed introduced all system main equipment of specifications model as, manual contains system description, and technology description and maintenance maintenance content, each a system independowner romechanical installation engineering prepared operation and the maintenance manual, by approved of official manual submittedong of professional sex, and systemic, for easy owners of using and management, engineering completion Shi, we will for electas strnd the engineering warranty user operation maintenance manual of prepared due to electromechanical installation engineering huracy of the horizontal line ... Always in a State of controlled, manufacturing process quality. Third section user service agh accrdinates and elevations are not allowed to carefully check level elevations, wall lines and installation of centerline and hiall surface, ground level according to the request object level adjustable leveling feet and lines when installing fixed. Coowill not be flat wheight and strength are in line with the actual needs of the installation; prior to appliance installation, demand for flats : the correct procedures for various types of equipment, operation3 法二:由arg(Z+3)=π, 知Z+3的轨迹是射线OA，则Z轨迹应是平行于OA，且过点(-3，0)的射线BM， ? |Z+6|+|Z-3i|就表示射线BM上点到点P(-6,0)和点Q(0,3)距离之和，连结PQ与射线BM交于点N，取E为N点表示复数时， |Z+6|+|Z-3i|=|PN|+|NQ|=|PQ|=3, ?所求最小值=3. 小结:两种方法的本质相同，都是将数学式子利用其几何意义转化成几何问题进行解决.如果纯粹用代数方法求解，难度会很大.对有关最值问题，尤其是模(距离)和辐角主值最值问题，用数形结合方法显然较为简便. 例6(已知|Z-2i|?1,求arg(Z-4i)最大值. 解:?|Z-2i|?1,?点Z轨迹是以(0，2)为圆心，1为半径的圆面，在其上任取一点Z，连Z与点(0，4)得一以(0，4)为起点，Z为终点的向量，将起点平移到原点，则θ为其对应的辐角主值，显然arg(Z-4i)最大值为π. 3(两个复数相乘，积的模等于模的积，辐角为两辐角之和，其几何意义是模的伸缩及对应向量的旋转. 两个复数相除，商的模等于模的商(除数不为零)，辐角为两辐角之差，其几何意义同乘法. 由复数三角形式乘除运算的几何意义，可解决向量或图形的旋转问题，如等腰、等边三角形、直角三角形，平行四边形顶点间的几何何关系利用复数的乘除运算来表示. 复数三角形式较之代数形式，在乘除运算中非常方便，可顺利解决多项相乘(乘方)，相除及乘除混合运算. 例7(若与分别表示复数Z=1+2i, Z=7+i, 求?ZOZ并判断ΔOZZ的形状. 122112 解:欲求?ZOZ，可计算 21 ==== ??ZOZ= 且=， 21 2222 由余弦定理，设|OZ|=k, |OZ|=2k(k>0)|ZZ|=k+(2k)-2k?2k?cos=3k 1212 ? |ZZ|=k, 12 222 而k+(k)=(2k)，?ΔOZZ为有一锐角为60?的直角三角形. 12 小结:此题中利用除法几何意义来解决三角形中角的大小问题，十分方便. features of the part: (4) security: the correct procedures for various types of equipment, operation of theical descriptions provided by the contract are included in the installation of equipment and technical data and descriptions ent of specifications model and function; Provide equipment, product brochures and so on. (2) technical specifications: technquipmce content, each a system independent book, specific following: (1) system description: detailed introduced all system main eof official manual submitted owners, manual contains system description, and technology description and maintenance maintenan provedg completion Shi, we will for electromechanical installation engineering prepared operation and the maintenance manual, by apal installation engineering has strong of professional sex, and systemic, for easy owners of using and management, engineerinchanicality. Third section user service and the engineering warranty user operation maintenance manual of prepared due to electromeallation of centerline and high accuracy of the horizontal line ... Always in a State of controlled, manufacturing process qud instnd lines when installing fixed. Coordinates and elevations are not allowed to carefully check level elevations, wall lines andemand for flats will not be flat wall surface, ground level according to the request object level adjustable leveling feet aheight and strength are in line with the actual needs of the installation; prior to appliance installation, 4 例8(已知直线l过坐标原点，抛物线C的顶点在原点，焦点在x轴正半轴上，若点A(-1,0)和B(0,8)关于l的对称点都在C上，求直线l与抛物线C的方程. 解:如图，建立复平面x0y，设向量、对应复数分别为 x+yi, x+yi. 1122 由对称性，|OA'|=|OA|=1, |OB'|=|OB|=8, ? x+yi=(x+yi)8i=-8y+8xi 221111 222 ? 设抛物线方程为y=2px(p>0)则有y=2px, y=2px, 1122 22 ? x=, y=p, 又|OA'|=1， 11 22 ?()+p=1, ?p=或-(舍) 2 ?抛物线方程为y=x，直线方程为:y=x. 小结:对于解析几何的许多问题，若能借助于复数的向量来表示，常常有意想不到的功效.尤其涉及到特殊位置，特殊关系的图形时，尤显其效. 五、易错点 1(并不是每一个复数都有唯一确定的辐角主值.如复数零的模为0，辐角主值不确定. 2(注意ArgZ与argZ的区别.ArgZ表示复数Z的辐角，而argZ表示复数Z的辐角主值. ArgZ=argZ+2kπ(k?Z),argZ?[0,2π), 辐角主值是[0,2π)内的辐角，但辐角不一定是辐角主值. 3(复数三角形式的四个要求:模非负，角相同，余弦前，加号连，缺一不可.任何一个不满足，就不是三角形式. 4(注意复数三角形式的乘除运算中，向量旋转的方向. 六、练习 1(写出下列复数的三角形式 (1) ai(a?R) (2) tgθ+i(<θ<π) (3) -(sinθ-icosθ) n 2(设Z=(-3+3i), n?N，当Z?R时，n为何值, 2 3(在复平面上A，B表示复数为α,β(α?0),且β=(1+i)α，判断ΔAOB形状，并证明S=|d|. ΔAOB 参考答案: 1((1)ai= ontract are included in the installation of equipment and technical data and descriptions of the features of the part: (4) section; Provide equipment, product brochures and so on. (2) technical specifications: technical descriptions provided by the cnd funent book, specific following: (1) system description: detailed introduced all system main equipment of specifications model as, manual contains system description, and technology description and maintenance maintenance content, each a system independowner romechanical installation engineering prepared operation and the maintenance manual, by approved of official manual submittedong of professional sex, and systemic, for easy owners of using and management, engineering completion Shi, we will for electas strnd the engineering warranty user operation maintenance manual of prepared due to electromechanical installation engineering huracy of the horizontal line ... Always in a State of controlled, manufacturing process quality. Third section user service agh accrdinates and elevations are not allowed to carefully check level elevations, wall lines and installation of centerline and hiall surface, ground level according to the request object level adjustable leveling feet and lines when installing fixed. Coowill not be flat wheight and strength are in line with the actual needs of the installation; prior to appliance installation, demand for flats : the correct procedures for various types of equipment, operationcurity5 (2)tgθ+i(<θ<π)=-[cos(π-θ)+isin(π-θ)] (3)-(sinθ-icosθ)=[cos(+θ)+isin(+θ)] 2(n为4的正整数倍 3(法一:?α?0，β=(1+i)α ? =1+i=(cos+isin), ??AOB=, ?分别表示复数α，β-α， 由β-α=αi，得=i=cos+isin， ??OAB=90?, ?ΔAOB为等腰直角三角形. 法二:?||=|α|, ||=|β-α|=|αi|=|α|, ?||=|| 222222 又||=|β|=|(1+i)α|=|α|,||+||=|α|+|α|=2|α|=|| 2 ?ΔAOB为等腰直角三角形，?S=||?||=|α|. ΔAOB 在线测试 选择题 2 1(若复数z=(a+i)的辐角是，则实数a的值是( ) A、1 B、-1 C、- D、- 2 2(已知关于x的实系数方程x+x+p=0的两虚根a， b满足|a-b|=3， 则p的值是( ) A、-2 B、- C、 D、1 θ<，则复数的辐角主值为( ) 3(设π< A、2π-3θ B、3θ-2π C、3θ D、3θ-π 4(复数cos+isin经过n次乘方后，所得的幂等于它的共轭复数，则n的值等于( ) A、3 B、12 C、6k-1(k?Z) D、6k+1(k?Z) of theical descriptions provided by the contract are included in the installation of equipment and technical data and descriptions ent of specifications model and function; Provide equipment, product brochures and so on. (2) technical specifications: technquipmce content, each a system independent book, specific following: (1) system description: detailed introduced all system main eof official manual submitted owners, manual contains system description, and technology description and maintenance maintenan provedg completion Shi, we will for electromechanical installation engineering prepared operation and the maintenance manual, by apal installation engineering has strong of professional sex, and systemic, for easy owners of using and management, engineerinchanicality. Third section user service and the engineering warranty user operation maintenance manual of prepared due to electromeallation of centerline and high accuracy of the horizontal line ... Always in a State of controlled, manufacturing process qud instnd lines when installing fixed. Coordinates and elevations are not allowed to carefully check level elevations, wall lines andemand for flats will not be flat wall surface, ground level according to the request object level adjustable leveling feet aheight and strength are in line with the actual needs of the installation; prior to appliance installation, features of the part: (4) security: the correct procedures for various types of equipment, operation6 |z|z+3|-1-3| 5(z为复数，()()的图形是( ) =() A、直线 B、半实轴长为1的双曲线 C、焦点在x轴，半实轴长为的双曲线右支 D、不能确定 答案与解析 答案:1、B 2、C 3、B 4、C 5、C 解析: 22 1(?z=(a+i)=(a-1)+2ai， argz=，? ，?a=-1，本题选B. 2(求根a，b=(Δ=1-4p<0) ?|a-b|=||=3， ? 4p-1=9， p=，故本题应选C. ==cos3θ+isin3θ. 3( ? π<θ<，?3π<3θ<，?π<3θ-2π<，故本题应选B. n 4(由题意，得(cos+isin)=cos+isin=cos-isin 由复数相等的定义 ，得 解得=2kπ-，(k?Z)，?n=6k-1.故本题应选C. 5(依题意，有 |z-3|=|z+3|-1，? |z+3|-|z-3|=1.由双曲线定义，此方程表示焦点(?3，0)，2a=1， a= 的双曲线右支，故本题应选C. 复数三角形式的运算?疑难问题解析 1(复数的模与辐角: (1)复数模的性质:,z?z,=,z,?,z, 1212 : the correct procedures for various types of equipment, operationcurityontract are included in the installation of equipment and technical data and descriptions of the features of the part: (4) section; Provide equipment, product brochures and so on. (2) technical specifications: technical descriptions provided by the cnd funent book, specific following: (1) system description: detailed introduced all system main equipment of specifications model as, manual contains system description, and technology description and maintenance maintenance content, each a system independowner romechanical installation engineering prepared operation and the maintenance manual, by approved of official manual submittedong of professional sex, and systemic, for easy owners of using and management, engineering completion Shi, we will for electas strnd the engineering warranty user operation maintenance manual of prepared due to electromechanical installation engineering huracy of the horizontal line ... Always in a State of controlled, manufacturing process quality. Third section user service agh accrdinates and elevations are not allowed to carefully check level elevations, wall lines and installation of centerline and hiall surface, ground level according to the request object level adjustable leveling feet and lines when installing fixed. Coowill not be flat wheight and strength are in line with the actual needs of the installation; prior to appliance installation, demand for flats 7 (2)辐角的性质:积的辐角等于各因数辐角的和( 商的辐角等于被除数的辐角减去除数的辐角所得的差( 一个复数n次幂(n?N)的辐角等于这个复数辐角的n倍( 注意:(1)辐角与辐角主值的区别，特别是解题过程中的不同点(如下面两个问题: 若arg(2-i)=α，arg(3-i)=β，求α+β的值((α+β?(3π，4π)) 若arg(2-i)=α，arg(3-i)=β，求arg[(2-i)(3-i)]的值( (2)两个复数乘积的辐角主值不一定等于两辐角主值的和，商的辐角主值不一定等于辐角主值的差. 2(关于数的开方 (1)复数的开方法则:r(cosθ+isinθ)的n次方根是 几何意义: 设对应于复平面上的点，则有: 所以，复数z的n次方根，在复平面内表示以原点为中心的正n边形的n个顶点( (2)复数平方根的求法( 求-3-4i的平方根( 解法一 利用复数代数形式(设-3-4i的平方根为x+yi(x，y?R)，则有 222 (x+yi)=-3-4i, 即(x-y)+2xyi=-3-4i， 由复数相等条件，得 ?-3-4i的平方根是?(1-2i)( 法二 利用复数的三角形式( features of the part: (4) security: the correct procedures for various types of equipment, operation of theical descriptions provided by the contract are included in the installation of equipment and technical data and descriptions ent of specifications model and function; Provide equipment, product brochures and so on. (2) technical specifications: technquipmce content, each a system independent book, specific following: (1) system description: detailed introduced all system main eof official manual submitted owners, manual contains system description, and technology description and maintenance maintenan provedg completion Shi, we will for electromechanical installation engineering prepared operation and the maintenance manual, by apal installation engineering has strong of professional sex, and systemic, for easy owners of using and management, engineerinchanicality. Third section user service and the engineering warranty user operation maintenance manual of prepared due to electromeallation of centerline and high accuracy of the horizontal line ... Always in a State of controlled, manufacturing process qud instnd lines when installing fixed. Coordinates and elevations are not allowed to carefully check level elevations, wall lines andemand for flats will not be flat wall surface, ground level according to the request object level adjustable leveling feet aheight and strength are in line with the actual needs of the installation; prior to appliance installation, 8 3(复数集中的方程( 2 关于实系数的一元二次方程的解法:设ax+bx+c=0(a?0，a，b，c?R，x，x为它的两个根) 1222 (1)当?=b-4ac?0时，方程有两个实数根 当?=b-4ac,0时，方程有一对共轭虚根 2 (4)二次三项式的因式分解:ax+bx+c=a(x-x)(x-x) 122 关于复系数的一元二次方程的解法:设ax+bx+c=0(a?0，a、b、c?C，且至少有一个虚数，xx为它的两12个根) 2 (4)二次三项式的因式分解ax+bx+c=a(x-x)(x-x)仍然适用( 12 关于二项方程的解法 nn 形如ax+a=0(a，a?C且a?0)的方程叫做二项方程，任何一个二项方程都可以化成x=b(b?C)的形式，n00nn 因此都可以通过复数开方来求根( 可以充分利用复数z的整体性质，复数z的三种表示方法及其转换来解方程( 2 已知方程x-4x+p=0两虚数根为α、β，且|α-β|=2求实数p的值( 解法1 ?实系数一元二次方程虚根共轭设α=a+bi， β=a-bi，(a，b?R) ?α+β=2a=4，?a=2 又?|α-β|=2, ?|2bi|=2得b=?1 即两根为2+i，2-i由韦达定理得:p=(2+i)(2-i)=5 由韦达定理可得:α+β=4，αβ=p 法22222 于是|α-β|=|(α-β)|=|(α+β)-4αβ|=|4-4p|=4， 即|4-p|=1 2 又??=4-4p,0 p,4， ?p-4=1， 得p=5 说明 注意实系数一元二次方程有两个实根与有两个虚根的区别( 一等式成立(若有两个虚根则 22上述等式不成立(因为,α-β,?(α-β)(因此在解题时要重视复数与实数之间的区别与联系，要避免出现混淆与干扰( 22 已知方程2x+3ax+a-a=0有模为1的根，求实数a的值( 分析 已知方程有模为1的根，此根可能是实数，也可能是虚数，故求实数a要注意分域讨论( 222 解 (1)若所给方程有实根则?=(3a)-4×2(a-a)=a+8a,0， 即a,-8或a,0 由条件得根必为1或-1， 2 ?将x=1代入原方程可得a+2a+2=0a无实数解( 2 (2)若所给方程有虚根则?=a+8,0， 即-8,a,0 : the correct procedures for various types of equipment, operationcurityontract are included in the installation of equipment and technical data and descriptions of the features of the part: (4) section; Provide equipment, product brochures and so on. (2) technical specifications: technical descriptions provided by the cnd funent book, specific following: (1) system description: detailed introduced all system main equipment of specifications model as, manual contains system description, and technology description and maintenance maintenance content, each a system independowner romechanical installation engineering prepared operation and the maintenance manual, by approved of official manual submittedong of professional sex, and systemic, for easy owners of using and management, engineering completion Shi, we will for electas strnd the engineering warranty user operation maintenance manual of prepared due to electromechanical installation engineering huracy of the horizontal line ... Always in a State of controlled, manufacturing process quality. Third section user service agh accrdinates and elevations are not allowed to carefully check level elevations, wall lines and installation of centerline and hiall surface, ground level according to the request object level adjustable leveling feet and lines when installing fixed. Coowill not be flat wheight and strength are in line with the actual needs of the installation; prior to appliance installation, demand for flats 9 2 即a-a-2=0， ?a=-1或a=2(舍) 2 已知方程x-(2i-1)x+3m-i=0有实数根，求实数m( 分析 求实数m的范围，若用判别式来判断是错误的，因为此方程的系数是复数( 利用求根公式或用韦达定理或选用复数相等，解方程组来求实数m均可以(现仅介绍一种方法( 2 解 ?x，m?R，方程变形可得，(x+x+3m)-(2x+1)i=0 复数例题讲解与分析 2 例1(已知x, y互为共轭复数，且(x+y)-3xyi=4-6i,求x, y. [思路1]:确定一个复数即分别确定它的实部、虚部或模与一个辐角，设z=a+bi或三角形式，化虚为实。 222 [解法1]:设x=a+bi(a,b?R), 则y=a-bi, 代入原等式得:(2a)-3(a+b)i=4-6i. 或 或 或， ? 或 或 或。 2 [思路2]:“x, y互为共轭”含义,?x+y?R, xy?R,则(x+y)-3xyi=4-6i . [解法2]:?x=，?x+y?R, xy?R, ?由两复数相等可得: ， 22 ?由韦达定理可知:x,y同是方程:z+2z+2=0或z-2z+2=0的两根， 分别解两个一元二次方程则得x,y……(略)。 2 例2(已知z?C,|z|=1且z?-1,则复数( ) A、必为纯虚数 B、是虚数但不一定是纯虚数 C、必为实数 D、可能是实数也可能是虚数 [思路分析]:选择题，从结论的一般性考虑，若z=?1，显然A、B选项不成立，分析C、D选项，显然穷举验证不能得出一般结论只能推演 22 解:[法1] 设z=a+bi, a,b?R, a+b=1,a?0. 则===?R，故，应选C。 ical descriptions provided by the contract are included in the installation of equipment and technical data and descriptions ent of specifications model and function; Provide equipment, product brochures and so on. (2) technical specifications: technquipmce content, each a system independent book, specific following: (1) system description: detailed introduced all system main eof official manual submitted owners, manual contains system description, and technology description and maintenance maintenan provedg completion Shi, we will for electromechanical installation engineering prepared operation and the maintenance manual, by apal installation engineering has strong of professional sex, and systemic, for easy owners of using and management, engineerinchanicality. Third section user service and the engineering warranty user operation maintenance manual of prepared due to electromeallation of centerline and high accuracy of the horizontal line ... Always in a State of controlled, manufacturing process qud instnd lines when installing fixed. Coordinates and elevations are not allowed to carefully check level elevations, wall lines andemand for flats will not be flat wall surface, ground level according to the request object level adjustable leveling feet aheight and strength are in line with the actual needs of the installation; prior to appliance installation, features of the part: (4) security: the correct procedures for various types of equipment, operation of the10 [法2] 设z=cosθ+isinθ (θ?R,且θ?kπ+)， 则===?R。 2 [法3] ?z?=|z|, ?当|z|=1时有z?=1, ?===?R. [法4] ?当|z|=1时有z?=1， ? ==?R. [法5] ?复数z为实数的充要条件是z=, 而()=, 又?|z|=1时，=， ? ==， ??R。 [评注]:复习中，概念一定要结合意义落实到位，一方面深化理解(比如复数定义:“形如a+bi (a,b?R)的数叫复数”深入理解就有凡是复数都能写成这样，求一个复数，使用一个复数都可通过这一形式将问题化虚为实;……。) 2 同时对一些概念的等价表达式要熟知。(比如:z=a+bi?R b=0(a,b?R) z= z?0; 2z=a+bi是纯虚数 a=0且b?0 (a,b?R) z+=0 (z?0) z<0;…….) 在面对具体问题时要有简捷意识(比如该例方法1，有同学可能会在算到时不注意及时化简分母又直接按两复数相除的运算法则进行)，多方理解挖掘题目立意。 2 例3(求使关于x的方程x+(m+2i)x+2+mi=0至少有一个实根的实数m. [思路分析]:根的判别式只适用实系数的一元二次方程，虚系数有实根用两复数相等，化虚为实。 解:设x为方程的一个实根，则有 0 2 x+mx+2+(2x+m)i=0 ，解得:m=?2。 000 例4(设 z?C， arg(z+2)=, arg(z-2)=, 求z。 : the correct procedures for various types of equipment, operationcurityontract are included in the installation of equipment and technical data and descriptions of the features of the part: (4) section; Provide equipment, product brochures and so on. (2) technical specifications: technical descriptions provided by the cnd funent book, specific following: (1) system description: detailed introduced all system main equipment of specifications model as, manual contains system description, and technology description and maintenance maintenance content, each a system independowner romechanical installation engineering prepared operation and the maintenance manual, by approved of official manual submittedong of professional sex, and systemic, for easy owners of using and management, engineering completion Shi, we will for electas strnd the engineering warranty user operation maintenance manual of prepared due to electromechanical installation engineering huracy of the horizontal line ... Always in a State of controlled, manufacturing process quality. Third section user service agh accrdinates and elevations are not allowed to carefully check level elevations, wall lines and installation of centerline and hiall surface, ground level according to the request object level adjustable leveling feet and lines when installing fixed. Coowill not be flat wheight and strength are in line with the actual needs of the installation; prior to appliance installation, demand for flats 11 [思路分析]:常规思路，设z=a+bi, 由已知列关于a,b的方程求解;数形结合思想，由题设可知z+2对应的点A在射线OA上，?AOX=，z-2对应的点B应在射线OB上， ?BOX=，z对应的点Z应在AB中点上，|AB|=4，AB//Ox轴，?AOB=， 故而易得:z=-1+i. 解:(略) 例5(设x,y?R, z=2-x+xi, z=y-1+(-y)i, 21 1002k-2k 已知|z|=|z|，arg=, (1)求()=? (2)设z=, 求集合A={x|x=z+z, 12 k?Z}中元素的个数。 [思路分析]:理解已知，|z|=|z|，arg=含义,?=i, 即z=zi?两复数相等?x, y. 1212 (1)解:?|z|=|z|, ?||=1, 12 又arg=, ? =||(cos+isin)=i, 即z=zi, 12 ? 2-x+xi=[y-1+(-y)i]i 即，解得 x=y=+, 10010050 ? ()=(+i)=(-+i)==--i. 0 [简评] 1 本题的解法体现了等价转化和整体的思想方法，如果把两个已知条件割裂开来去考虑，则需要解关于x, y的二元二次方程组，其运算肯定很麻烦; 03322 2 在计算题中对1的立方根之一:w=-+i的特性要熟知即 w==1, ==w,1+w+w=0, 1++=0, 关于此点设计问题是命题经常参考的着眼点。 features of the part: (4) security: the correct procedures for various types of equipment, operation of theical descriptions provided by the contract are included in the installation of equipment and technical data and descriptions ent of specifications model and function; Provide equipment, product brochures and so on. (2) technical specifications: technquipmce content, each a system independent book, specific following: (1) system description: detailed introduced all system main eof official manual submitted owners, manual contains system description, and technology description and maintenance maintenan provedg completion Shi, we will for electromechanical installation engineering prepared operation and the maintenance manual, by apal installation engineering has strong of professional sex, and systemic, for easy owners of using and management, engineerinchanicality. Third section user service and the engineering warranty user operation maintenance manual of prepared due to electromeallation of centerline and high accuracy of the horizontal line ... Always in a State of controlled, manufacturing process qud instnd lines when installing fixed. Coordinates and elevations are not allowed to carefully check level elevations, wall lines andemand for flats will not be flat wall surface, ground level according to the request object level adjustable leveling feet aheight and strength are in line with the actual needs of the installation; prior to appliance installation, 12 322k-2k3 (2) [思路分析]:由(1)知 z=+i，z的特性:z=-1==w, ……，z+z , |z|=1, =; z=cos+isin, z 2k2-k2k2k可怎么理解呢, (z)+(z), z+, …… 2k-2kk-k 解[法1]:令w=-+i，则z+z=w+w, 3 ?w=1,而k?z, ?k= 2k-2k3m3-m 当k=3m时，z+z=(w)+(w)=2, 2k-2k3m-3m-1-1 当k=3m+1时，z+z=w?w+w?w=w+w=w+=-1, 2k-2k3m2-3m-22-23-1-3-1 当k=3m+2时，z+z=w?w+w?w=w+w=w?w+w?w=w+w=-1, 综上可知，集合A中有2个元素。 [法2]:?|z|=1, ? =, 2k-2k2k2k ?z+z=z+=cos+isin+cos-isin=2cos = 由此可判定集合A中有2个元素。 例6(设复数z=cosθ+isinθ(0<θ<π), w=, 并且|w|=, argw<,求θ。(93年全国理) [思路分析]:欲用已知，需化简w, 解:w=== =tg2θ(sin4θ+icos4θ) ? |w|=|tg2θ| 由|w|=得 tg2θ=?. ? 0<θ<π, 故有(i)当tg2θ=时，得θ=或θ=. ction; Provide equipment, product brochures and so on. (2) technical specifications: technical descriptions provided by the cnd funent book, specific following: (1) system description: detailed introduced all system main equipment of specifications model as, manual contains system description, and technology description and maintenance maintenance content, each a system independowner romechanical installation engineering prepared operation and the maintenance manual, by approved of official manual submittedong of professional sex, and systemic, for easy owners of using and management, engineering completion Shi, we will for electas strnd the engineering warranty user operation maintenance manual of prepared due to electromechanical installation engineering huracy of the horizontal line ... Always in a State of controlled, manufacturing process quality. Third section user service agh accrdinates and elevations are not allowed to carefully check level elevations, wall lines and installation of centerline and hiall surface, ground level according to the request object level adjustable leveling feet and lines when installing fixed. Coowill not be flat wheight and strength are in line with the actual needs of the installation; prior to appliance installation, demand for flats : the correct procedures for various types of equipment, operationcurityontract are included in the installation of equipment and technical data and descriptions of the features of the part: (4) se13 此时 w=(cos+isin)，?argw=<,适合题意。 (ii)当 tg2θ=-时，得θ=π或θ=π，此时，w=(cosπ+isinπ). ?argw=π>, 不合题意，舍去， 故综合(i), (ii)知，θ=或θ=. 0 [简评] 1 复数与三角的综合题目是命题的一个方向，其中应用三角公式“1?cosa的升幂式”及“诱导公式”化复数代数形式为三角形式应用频率较高。 0 2 此题在w的化简中亦可利用 |z|=1, z?=|z|来化简: w====……以下略，这样可省去较为繁锁的三角变换。 5 例7(已知|z|=1,且z+z=1, 求z。 [思路分析1]:已知含未知数的等式求未知数，方程问题，设元化虚为实， 5 解:[法1]设z=cosθ+isinθ，则由z+z=1可得: 22 由(1)+(2)得:cos4θ=-……(以下略)。 55 [思路分析2]:复数的概念，运算都有几何意义，由z+z=1，若设z, z,1对应点为A，B，C则四边形OACB为平行四边形。 55 ?[法2]:设z,z,1在复平面上对应点分别为A，B，C，则由z+z=1，可知，四边形OACB为平行四边形， 55 又? |z|=|z|=1=|z| ? OACB为边长为1的菱形且?AOB=120?，? 易求得:z=+i或 z=-i。 5?i时，z 可以验证当z==i符合题意。 0 [简评]:1 数形结合思想方法应是处理复数有关问题的习惯思路，因复数中的概念，运算都有一定的几何含义，这源于z=a+bi本身就表示一个点，当a,b确定，z表示定点，当a,b不定则z就能表示一个动点轨迹，如 z=x+i就可表示双曲线。故在解题时变换角度从几何意义去分析，往往会事半功倍。 0 5555 2此题还可这样联系，由z+z=1得 z-1=-z，两边取模|z-1|=|-z|=|z|=1，从而知z应是圆|z|=1与 |z-1|=1的交点。 of theical descriptions provided by the contract are included in the installation of equipment and technical data and descriptions ent of specifications model and function; Provide equipment, product brochures and so on. (2) technical specifications: technquipmce content, each a system independent book, specific following: (1) system description: detailed introduced all system main eof official manual submitted owners, manual contains system description, and technology description and maintenance maintenan provedg completion Shi, we will for electromechanical installation engineering prepared operation and the maintenance manual, by apal installation engineering has strong of professional sex, and systemic, for easy owners of using and management, engineerinchanicality. Third section user service and the engineering warranty user operation maintenance manual of prepared due to electromeallation of centerline and high accuracy of the horizontal line ... Always in a State of controlled, manufacturing process qud instnd lines when installing fixed. Coordinates and elevations are not allowed to carefully check level elevations, wall lines andemand for flats will not be flat wall surface, ground level according to the request object level adjustable leveling feet aheight and strength are in line with the actual needs of the installation; prior to appliance installation, features of the part: (4) security: the correct procedures for various types of equipment, operation 14