3.2.2-复数代数形式的乘除运算

Click to edit Master title style,Click to edit Master text styles,Second level,Third level,Fourth level,Fifth level,*,Click to edit Master title style,Click to edit Master text styles,Second level,Third level,Fourth level,Fifth level,*,Click to edit Master title style,Click to edit Master text styles,Second level,Third level,Fourth level,Fifth level,*,Click to edit Master title style,Click to edit Master text styles,Second level,Third level,Fourth level,Fifth level,*,Click to edit Master title style,Click to edit Master text styles,Second level,Third level,Fourth level,Fifth level,*,Click to edit Master title style,Click to edit Master text styles,Second level,Third level,Fourth level,Fifth level,*,Click to edit Master title style,Click to edit Master text styles,Second level,Third level,Fourth level,Fifth level,*,Click to edit Master title style,Click to edit Master text styles,Second level,Third level,Fourth level,Fifth level,*,Click to edit Master title style,Click to edit Master text styles,Second level,Third level,Fourth level,Fifth level,*,Click to edit Master title style,Click to edit Master text styles,Second level,Third level,Fourth level,Fifth level,*,Click to edit Master title style,Click to edit Master text styles,Second level,Third level,Fourth level,Fifth level,*,Click to edit Master title style,Click to edit Master text styles,Second level,Third level,Fourth level,Fifth level,*,3.2.2,复数代数形式的乘除运算,已知两复数,z,1,=a+bi,z,2,=c+di,（,a,b,c,d,是实数）,即,:,两个复数相加,(,减,),就是,实部与实部,虚部与虚部分别相加,(,减,).,(1),加法法则：,z,1,+z,2,=(a+c)+(b+d)i;,(2),减法法则：,z,1,-z,2,=(a-c)+(b-d)i.,(a,+bi),(c,+,d,i,),=,(ac),+,(bd),i,x,o,y,Z,1,(a,b),Z,2,(c,d),Z(a+c,b+d),z,1,+z,2,=OZ,1,+OZ,2,=OZ,符合向量加法的平行四边形法则,.,1.,复数,加法,运算的几何意义,?,x,o,y,Z,1,(a,b),Z,2,(c,d),复数,z,2,z,1,向量,Z,1,Z,2,符合向量减法的三角形法则,.,2.,复数,减法,运算的几何意义,?,复平面中点,Z,1,与点,Z,2,间的距离,|z,1,-z,2,|,表示：,_,_.,已知两复数,z,1,=a+bi,，,z,2,=c+di(a,，,b,，,c,，,dR),3.,复数,模,的几何意义：,Z,1,(a,，,b),o,x,y,Z,2,(c,，,d),特别地，,|z|,表示：,_.,复平面中点,Z,与原点间的距离,如：,|z+(1+2i)|,表示：,_,_.,点,(-1,，,-2),的距离,点,Z(,对应复数,z),到,探究点,1,复数乘法运算,我们规定，复数乘法法则如下：,设,z,1,=a+bi,z,2,=c+di,是任意两个复数，那么它们的乘积为：,(,a+bi,)(,c+di,)=ac+adi+bci+bdi,2,=ac+adi+bci-bd,=(ac-bd)+(ad+bc)i.,即,(,a+bi,)(,c+di,)=(ac-bd)+(ad+bc)i,注意：,两个复数的积是一个确定的复数,.,探究点,2,复数乘法的运算律,复数的乘法是否满足交换律，结合律以及乘法对加法的分配律？,请验证乘法是否满足交换律,?,对任意复数,z,1,=a+bi,z,2,=c+di,则,z,1,z,2,=(,a+bi,)(,c+di,)=ac+adi+bci+bdi,2,=ac+adi+bci-bd=(ac-bd)+(ad+bc)i,而,z,2,z,1,=(,c+di,)(,a+bi,)=ac+bci+adi+bdi,2,=(ac-bd)+(ad+bc)i,所以,z,1,z,2,=z,2,z,1,(,交换律,),乘法运算律,对任意,z,1,z,2,z,3,C,有,z,1,z,2,=z,2,z,1,(,交换律,),(z,1,z,2,)z,3,=z,1,(z,2,z,3,),(,结合律,),z,1,(z,2,+z,3,)=z,1,z,2,+z,1,z,3,(,分配律,),例,1,计算,(1-2i)(3+4i)(-2+i).,解,:,(1-2i)(3+4i)(-2+i),=(11-2i)(-2+i),=-20+15i.,分析：,类似两个多项式相乘，把,i,2,换成,-1,例,2,计算,:(1)(3+4i)(3-4i);,(2)(1+i),2,.,解,:,(1)(3+4i)(3-4i),=3,2,-(4i),2,=9-(-16),=25.,(2)(1+i),2,=1+2i+i,2,=1+2i-1,=2i.,1.,计算,2.,已知,则,=,变式训练：,【,总结提升,】,（,1,）实数集中的乘法公式在复数集中仍然成立；,（,2,）复数的混合运算也是先乘方，再乘除，最后加减，有括号应先处理括号里面的,探究点,3,共轭复数的定义,一般地，当两个复数的实部相等，虚部互为相反数时，这两个复数叫做互为,共轭复数,.,虚部不等于的两个共轭复数也叫做,共轭虚数,.,实数,的共轭复数是它本身,.,思考,：若,z,1,，,z,2,是共轭复数，那么,(1),在复平面内，它们所对应的点有怎样的位置关系？,(2)z,1,z,2,是一个怎样的数？,记法：,复数,z=,a+b,i,的共轭复数记作,=,a-b,i,解：,作图,y,x,(a,b),(a,-b),z,1,=a+b,i,o,y,x,(a,0),z,1,=a,o,x,y,z,1,=b,i,(0,b),(0,-b),o,得出结论：,在复平面内，共轭复数,z,1,z,2,所对应的点关于,实轴,对称,.,令,z,1,=a+b,i,则,z,2,=a-b,i,则,z,1,z,2,=(a+b,i,)(a-b,i,),=a,2,-ab,i,+ab,i,-b,2,i,2,=a,2,+b,2,结论：,任意两个互为共轭复数的乘积是一个实数,.,探究点,4,复数除法的法则,类比实数的除法是乘法的逆运算，我们规定复数的除法是乘法的逆运算,.,试探求复数除法的法则,.,复数除法的法则是,:,方法,:,在进行复数除法运算时,通常先把,在作根式除法时,分子分母都乘以分母的“有理化因式”,从而使分母“有理化”,.,这里分子分母都乘以分母的“实数化因式”,(,共轭复数,),从而使分母“实数化”,.,先写成分式形式,然后分母实数化,分子分母同时乘以分母的共轭复数,结果化简成代数形式,B,2.,若复数,z=1+i(i,为虚数单位,),是,z,的共轭复数，,则,+,的虚部为（,）,A.0 B.-1 C.1 D.-2,A,A,B,5.,已知方程,x,2,-,2,x,+2=0,有两虚根为,x,1,x,2,求,x,1,4,+,x,2,4,的值,.,注,:,在复数范围内方程的根与系数的关系仍适用,.,1.,复数相乘类似于多项式相乘，只要在所得的结果中把,i,2,换成,1,，并且把实部和虚部分别合并,.,2.,实数系中的乘法公式在复数系中仍然成立,.,3.,当两个复数的实部相等，虚部互为相反数时，这两个复数叫做互为,共轭复数,.,虚部不等于的两个共轭复数也叫做,共轭虚数,.,实数,的共轭复数是它本身,.,4.,复数代数形式的除法实质：,分母实数化,.,