高中數(shù)學 第五章 數(shù)系的擴充與復數(shù)的引入 5.2 復數(shù)的四則運算 5.2.1 復數(shù)的加法與減法課件 北師大版選修22

《高中數(shù)學 第五章 數(shù)系的擴充與復數(shù)的引入 5.2 復數(shù)的四則運算 5.2.1 復數(shù)的加法與減法課件 北師大版選修22》由會員分享，可在線閱讀，更多相關《高中數(shù)學 第五章 數(shù)系的擴充與復數(shù)的引入 5.2 復數(shù)的四則運算 5.2.1 復數(shù)的加法與減法課件 北師大版選修22（16頁珍藏版）》請在裝配圖網(wǎng)上搜索。

1、2 2復數(shù)的四則運算復數(shù)的四則運算2 2.1 1復數(shù)的加法與減法MUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISH

2、ULI知識梳理理解并掌握復數(shù)代數(shù)形式的加、減運算法則,能熟練地進行復數(shù)的加、減運算.MUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂

3、演練ZHISHISHULI知識梳理復數(shù)的加法與減法設a+bi(a,bR)和c+di(c,dR)是任意兩個復數(shù),我們定義復數(shù)加法、減法如下:(a+bi)(c+di)=(ac)+(bd)i.也就是說,兩個復數(shù)的和(或差)仍然是一個復數(shù).它的實部是原來兩個復數(shù)的實部的和(或差),它的虛部是原來兩個復數(shù)的虛部的和(或差).MUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUIT

4、ANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理MUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI

5、知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二分析根據(jù)復數(shù)的加法與減法法則進行運算. 方法總結(jié)復數(shù)的加減法運算,就是實部與實部相加減作實部,虛部與虛部相加減作虛部,同時也把i看作字母,類比多項式加減中的合并同類項.MUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂

6、演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二【變式訓練1】 計算下列各題:(1)(3-2i)-(10-5i)+(2+17i);(2)(1-2i)-(2-3i)+(3-4i)-(4-5i)+(2 015-2 016i).解:(1)原式=(3-10+2)+(-2+5+17)i=-5+20i

7、.(2)原式=(1-2+3-4+2 013-2 014+2 015)+(-2+3-4+5-2 014+2 015-2 016)i=1 008-1 009i.MUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳

8、理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二MUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透

9、析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理題型一題型二MUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練

10、ZHISHISHULI知識梳理題型一題型二【變式訓練2】 已知x,yR,z1=(3x+y)+(y-4x)i,z2=(4y-2x)-(5x+3y)i.設z=z1-z2,且z=13-2i,求z1,z2.解:z=z1-z2=(3x+y)+(y-4x)i-(4y-2x)-(5x+3y)i=(3x+y)-(4y-2x)+(y-4x)+(5x+3y)i=(5x-3y)+(x+4y)i.因為z=13-2i,所以z1=(32-1)+(-1-42)i=5-9i,z2=4(-1)-22-52+3(-1)i=-8-7i.MUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN隨

11、堂演練ZHISHI SHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理1 2 3 4 5 61(6-2i)-(3i+1)等于()A.3-3iB.5-5iC.7+iD.5+5i解析:(6-2i)-(3i+1)=(6-1

12、)+(-2-3)i=5-5i.故選B.答案:BMUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理1

13、2 3 4 5 62復數(shù)(1-i)-(2+i)+3i等于()A.-1+iB.1-iC.iD.-i答案:AMUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITAN

14、GYANLIAN隨堂演練ZHISHISHULI知識梳理1 2 3 4 5 63若z1=2+i,z2=3+ai(aR),z1+z2所對應的點在實軸上,則a等于()A.3B.2C.1D.-1解析:z1+z2=(2+i)+(3+ai)=5+(a+1)i,z1+z2對應的點在實軸上,即z1+z2為實數(shù),因此a+1=0,a=-1.答案:DMUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典

15、例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理1 2 3 4 5 64若實數(shù)x,y滿足(1+i)x+(1-i)y=2,則xy的值是. 解析:xR,yR,(1+i)x+(1-i)y=(x+y)+(x-y)i.又(1+i)x+(1-i)y=2,答案:1MUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI S

16、HULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理1 2 3 4 5 6MUBIAODAOHANG目標導航DIANLI TOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHI SHULI知識梳理目標導航DI

17、ANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGYANLIAN隨堂演練ZHISHISHULI知識梳理1 2 3 4 5 66計算:(1)(3+5i)+(3-4i);(2)(-3+2i)-(4-5i);(3)(5-6i)+(-2-2i)-(3+3i).解:(1)(3+5i)+(3-4i)=(3+3)+(5-4)i=6+i.(2)(-3+2i)-(4-5i)=(-3-4)+2-(-5)i=-7+7i.(3)(5-6i)+(-2-2i)-(3+3i)=(5-2-3)+-6+(-2)-3i=-11i.