资源描述
5.,2,复数四则运算,1/27,2/27,1,.,复数加法、减法,设,z,1,=a+b,i,z,2,=c+d,i,a,b,c,d,R,(1),运算,:,z,1,+z,2,=,(,a+c,),+,(,b+d,)i,z,1,-z,2,=,(,a-c,),+,(,b-d,)i,.,(2),法则,:,两个复数,和或差,依然是一个复数,它实部是原来两个复数,实部和,(,或差,),它虚部是原来两个复数,虚部和,(,或差,),.,名师点拨,1,.,一个要求,:,复数加减法法则是一个要求,减法是加法逆运算,;,特殊情形,:,当复数虚部为零时,与实数加法、减法法则一致,.,2,.,运算律,:,实数加法交换律、结合律在复数集中仍成立,.,实数移项法则在复数中依然成立,.,3,.,运算结果,:,两个复数和,(,差,),是唯一确定复数,.,4,.,适当推广,:,能够推广到多个复数进行加、减运算,.,5,.,虚数单位,i:,在进行复数加减运算时,可将虚数单位,i,看成一个字母,然后去括号,合并同类项即可,.,3/27,4/27,5/27,名师点拨,虚数单位,i,常见结论,.,(1),虚数,i,乘方及其规律,:i,4,n,=,1,i,4,n+,1,=,i,i,4,n+,2,=-,1,i,4,n+,3,=-,i(,n,N,+,),即,i,n,含有周期性,最小正周期为,4,.,(2)i,n,+,i,n+,1,+,i,n+,2,+,i,n+,3,=,0,.,(3)(1,i),2,=,2i,.,6/27,【做一做,2,】,已知,i,为虚数单位,复数,z=,2i(2,-,i),实部为,a,虚部为,b,则,log,a,b,等于,(,),A,.,0B,.,1,C,.,2D,.,3,解析,:,z=,2i(2,-,i),=,4i,-,2i,2,=,2,+,4i,则,a=,2,b=,4,所以,log,a,b=,log,2,4,=,2,.,故选,C,.,答案,:,C,7/27,3,.,共轭复数与复数除法,(1),共轭复数,:,当两个复数实部相等,虚部互为相反数时,这么两个数叫作,互为共轭复数,(2),复数除法,:,名师点拨,共轭复数运算性质,(2),复数模运算性质,8/27,【做一做,3,】,已知复数,z,对应点在第二象限,它模是,3,实部是,答案,:,B,答案,:,B,9/27,思索辨析,判断以下说法是否正确,正确在后面括号内画,“,”,错误画,“,”,.,(1),若复数,z,1,z,2,满足,z,1,-z,2,0,则,z,1,z,2,.,(,),(2),两个互为共轭复数复数和与积都是实数,.,(,),(3),若两个复数,z,1,z,2,满足,|z,1,+z,2,|=|z,1,-z,2,|,则,z,1,=z,2,=,0,.,(,),10/27,探究一,探究二,探究三,思维辨析,复数加减运算,若,z,1,+z,2,为虚数,求,m,取值范围,.,分析,:,先求,z,1,+z,2,再依据复数为虚数判断求出,.,所以,m,取值范围是,m|m,R,且,m,4,且,m,-,1,且,m,-,2,.,11/27,探究一,探究二,探究三,思维辨析,反思感悟,复数加减运算方法,1,.,复数实部与实部相加减,虚部与虚部相加减,.,2,.,把,i,看作一个字母,类比多项式加减中合并同类项,.,12/27,探究一,探究二,探究三,思维辨析,变式训练,1,已知复数,z,满足,z+,i,-,3,=,3,-,i,则,z,等于,(,),A.0B.2iC.6D.6,-,2i,解析,:,z+,i,-,3,=,3,-,i,z=,3,-,i,+,3,-,i,=,6,-,2i,.,答案,:,D,13/27,探究一,探究二,探究三,思维辨析,复数乘法、除法运算,【例,2,】,计算,:(1)(1,-,i)(1,+,i),+,(,-,1,+,i);,(2)(2,-,i)(,-,1,+,5i)(3,-,4i),+,2i;,(3)(,-,2,+,3i),(1,+,2i),+,i,5,;,分析,:,按照复数乘法与除法运算法则进行计算,.,解,:,(1)(1,+,i)(1,-,i),+,(,-,1,+,i),=,1,-,i,2,+,(,-,1,+,i),=,2,-,1,+,i,=,1,+,i,.,(2)(2,-,i)(,-,1,+,5i)(3,-,4i),+,2i,=,(,-,2,+,10i,+,i,-,5i,2,)(3,-,4i),+,2i,=,(,-,2,+,11i,+,5)(3,-,4i),+,2i,=,(3,+,11i)(3,-,4i),+,2i,=,(9,-,12i,+,33i,-,44i,2,),+,2i,=,53,+,21i,+,2i,=,53,+,23i,.,14/27,探究一,探究二,探究三,思维辨析,反思感悟,复数乘法能够把,i,看作字母,按多项式乘法法则进行,注意把,i,2,化成,-,1,进行最终结果化简,;,复数除法先写成份式形式,再把分子与分母都乘以分母共轭复数,并进行化简,.,15/27,探究一,探究二,探究三,思维辨析,变式训练,3,若复数,z,满足,z,(2,-,i),=,11,+,7i(i,为虚数单位,),则,z,为,(,),A.3,+,5iB.3,-,5i,C.,-,3,+,5iD.,-,3,-,5i,答案,:,A,解,:,设纯虚数,z=b,i(,b,R,且,b,0),16/27,探究一,探究二,探究三,思维辨析,共轭复数,【例,3,】,(1),设,z=,则,z,共轭复数为,(,),A.,-,1,+,3iB.,-,1,-,3i,C.1,+,3iD.1,-,3i,答案,:,D,17/27,探究一,探究二,探究三,思维辨析,分析,:,将方程左边化成,a+b,i,形式,利用复数相等充要条件来求解,.,解,:,设,z=x+y,i(,x,R,y,R,),由题意得,x,2,+y,2,-,3,y-,3,x,i,=,1,+,3i,z=-,1,或,z=-,1,+,3i,.,反思感悟,当已知条件中出现,z,或,等式时,解题常规思绪是设,z=a+b,i(,a,b,R,),则,=a-b,i,代入所给等式,.,利用复数相等充要条件转化为实数问题求解,即转化为方程组求解,.,18/27,探究一,探究二,探究三,思维辨析,变式训练,5,已知,a,b,R,i,是虚数单位,若,a-,i,与,2,+b,i,互为共轭复数,则,(,a+b,i),2,=,(,),A.5,-,4iB.5,+,4i,C.3,-,4iD.3,+,4i,解析,:,a-,i,与,2,+b,i,互为共轭复数,a=,2,b=,1,.,(,a+b,i),2,=,(2,+,i),2,=,3,+,4i,.,答案,:,D,19/27,探究一,探究二,探究三,思维辨析,不清楚判别式使用条件而造成失误,【典例】,已知关于,t,一元二次方程,t,2,+,(2,+,i),t+,2,xy+,(,x-y,)i,=,0(,x,y,R,),有实数解,求点,(,x,y,),轨迹方程,.,易错分析,:,根判别式只有在实系数一元二次方程中才能用,本题正确处理方法是设出方程根,利用复数相等充要条件化为方程组,然后消参数求解,.,20/27,探究一,探究二,探究三,思维辨析,解,:,设实根为,t,则,t,2,+,(2,+,i),t+,2,xy+,(,x-y,)i,=,0,即,(,t,2,+,2,t+,2,xy,),+,(,t+x-y,)i,=,0,由,得,t=y-x,代入,得,(,y-x,),2,+,2(,y-x,),+,2,xy=,0,即,(,x-,1),2,+,(,y+,1),2,=,2,.,所求点轨迹方程为,(,x-,1),2,+,(,y+,1),2,=,2,轨迹是以,(1,-,1),为圆心,为半径圆,.,纠错心得,对于复系数一元二次方程,ax,2,+bx+c=,0(,a,b,c,为复数,),讨论解情况时,需先设,x=m+n,i(,m,n,R,),将上述方程利用复数相等转化为实系数方程再进行处理,.,21/27,探究一,探究二,探究三,思维辨析,变式训练,已知关于,x,方程,x,2,+,(,k+,2i),x+,2,+k,i,=,0,有实根,求实数,k,取值所组成集合,.,解,:,设,x,0,为方程,x,2,+,(,k+,2i),x+,2,+k,i,=,0,实根,22/27,1 2 3 4 5,1,.,复数,(1,-,i),-,(2,+,i),+,3i,等于,(,),A.,-,1,+,iB.1,-,iC.iD.,-,i,解析,:,原式,=,(1,-,2),+,(,-,1,-,1,+,3)i,=-,1,+,i,.,答案,:,A,23/27,1 2 3 4 5,答案,:,B,24/27,1 2 3 4 5,=,i,1,009,+,i,6,=,i,4,252,+,1,+,i,4,+,2,=,i,1,+,i,2,=-,1,+,i,.,答案,:,-,1,+,I,25/27,1 2 3 4 5,答案,:,3,+,4i,26/27,1 2 3 4 5,27/27,
展开阅读全文
浙ICP备2021020529号-1 | 浙B2-20240490 
