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1、6.5復(fù)數(shù),,第六章 平面向量、復(fù)數(shù),,NEIRONGSUOYIN,內(nèi)容索引,基礎(chǔ)知識 自主學(xué)習(xí),題型分類 深度剖析,課時作業(yè),1,基礎(chǔ)知識 自主學(xué)習(xí),PART ONE,,知識梳理,1.復(fù)數(shù)的有關(guān)概念 (1)定義:形如abi(a,bR)的數(shù)叫做復(fù)數(shù),其中a叫做復(fù)數(shù)z的 ,b叫做復(fù)數(shù)z的 (i為虛數(shù)單位). (2)分類:,ZHISHISHULI,,,,實部,虛部,b0,b0,a0且b0,(3)復(fù)數(shù)相等:abicdi (a,b,c,dR). (4)共軛復(fù)數(shù):abi與cdi共軛 (a,b,c,dR). (5)模:向量 的模叫做復(fù)數(shù)zabi的模,記作 或 ,即|z|
2、|abi| (a,bR).,ac且bd,ac,bd,|abi|,|z|,2.復(fù)數(shù)的幾何意義 復(fù)數(shù)zabi與復(fù)平面內(nèi)的點 及平面向量 (a,b)(a,bR)是一一對應(yīng)關(guān)系. 3.復(fù)數(shù)的運算 (1)運算法則:設(shè)z1abi,z2cdi,a,b,c,dR.,Z(a,b),,,(2)幾何意義:復(fù)數(shù)加減法可按向量的平行四邊形或三角形法則進行.,1.復(fù)數(shù)abi的實部為a,虛部為b嗎?,【概念方法微思考】,提示不一定.只有當(dāng)a,bR時,a才是實部,b才是虛部.,2.如何理解復(fù)數(shù)的加法、減法的幾何意義?,提示復(fù)數(shù)的加法、減法的幾何意義就是向量加法、減法的平行四邊形法則.,,題組一思考辨析 1.判斷
3、下列結(jié)論是否正確(請在括號中打“”或“”) (1)方程x2x10沒有解.() (2)復(fù)數(shù)zabi(a,bR)中,虛部為bi.() (3)復(fù)數(shù)中有相等復(fù)數(shù)的概念,因此復(fù)數(shù)可以比較大小.() (4)原點是實軸與虛軸的交點.() (5)復(fù)數(shù)的模實質(zhì)上就是復(fù)平面內(nèi)復(fù)數(shù)對應(yīng)的點到原點的距離,也就是復(fù)數(shù)對應(yīng)的向量的模.(),,,1,2,3,4,5,6,,,,7,,基礎(chǔ)自測,JICHUZICE,,,,題組二教材改編,,,1,2,3,4,5,6,7,,1,2,3,4,5,6,,7,,1,2,3,4,5,6,4.P116A組T2若復(fù)數(shù)z(x21)(x1)i為純虛數(shù),則實數(shù)x的值為 A.1 B.0 C.1 D.1
4、或1,,7,,1,2,3,4,5,6,題組三易錯自糾 5.設(shè)a,bR,i是虛數(shù)單位,則“ab0”是“復(fù)數(shù)a 為純虛數(shù)”的 A.充要條件 B.充分不必要條件 C.必要不充分條件 D.既不充分也不必要條件,,7,,1,2,3,4,5,6,6.若復(fù)數(shù)z滿足iz22i(i為虛數(shù)單位),則z的共軛復(fù)數(shù) 在復(fù)平面內(nèi)對應(yīng)的點所在的象限是 A.第一象限 B.第二象限 C.第三象限 D.第四象限,,7,7.i2 014i2 015i2 016i2 017i2 018i2 019i2 020____.,,1,2,3,4,5,6,i,7,解析原式i2i3i4i1i2i3i4i.,2,題型分類深度剖析,PART TW
5、O,,題型一復(fù)數(shù)的概念,,自主演練,1.(2018麗水、衢州、湖州三地市質(zhì)檢)若復(fù)數(shù)z滿足iz32i(i為虛數(shù)單位),則復(fù)數(shù)z的虛部是 A.3 B.3iC.3 D.3i,,所以復(fù)數(shù)z的虛部是3.故選C.,,3.(2018杭州質(zhì)檢)設(shè)aR,若(13i)(1ai)R(i是虛數(shù)單位),則a等于,解析由題意得,(13i)(1ai)13a(3a)i為實數(shù), 3a0, a3,故選B.,,復(fù)數(shù)的基本概念有實部、虛部、虛數(shù)、純虛數(shù)、共軛復(fù)數(shù)等,在解題中要注意辨析概念的不同,靈活使用條件得出符合要求的解.,命題點1復(fù)數(shù)的乘法運算 例1(1)(2018全國)(1i)(2i)等于 A.3i B.3i C.3i D.
6、3i,,題型二復(fù)數(shù)的運算,,,多維探究,解析(1i)(2i)22iii23i.,A.32i B.32i C.32i D.32i,,解析i(23i)2i3i232i,故選D.,命題點2復(fù)數(shù)的除法運算,,故選D.,A.i B.iC.1 D.i,由z28i,得b2,,,命題點3復(fù)數(shù)的綜合運算,,,解析對于兩個復(fù)數(shù)1i,1i, (1i)(1i)2,故不正確;,22(1i)2(1i)212i112i10,故正確.故選C.,(1)復(fù)數(shù)的乘法:復(fù)數(shù)乘法類似于多項式的四則運算. (2)復(fù)數(shù)的除法:除法的關(guān)鍵是分子分母同乘以分母的共軛復(fù)數(shù).,,,,題型三復(fù)數(shù)的幾何意義,,師生共研,,(2)(2018浙江重點中學(xué)
7、考試)已知復(fù)數(shù)z滿足(2i)z3ai(i是虛數(shù)單位).若復(fù)數(shù)z在復(fù)平面內(nèi)對應(yīng)的點在直線y2x4上,則實數(shù)a的值為______.,方法二因為復(fù)數(shù)z在復(fù)平面內(nèi)對應(yīng)的點在直線y2x4上, 不妨設(shè)zt(2t4)i(tR), 則(2i)t(2t4)i2t2t4(3t8)i3ai,,復(fù)平面內(nèi)的點、向量及向量對應(yīng)的復(fù)數(shù)是一一對應(yīng)的,要求某個向量對應(yīng)的復(fù)數(shù)時,只要找出所求向量的始點和終點,或者用向量相等直接給出結(jié)論即可.,,解析由已知得A(1,2),B(1,1),C(3,2),,5,(3,2)x(1,2)y(1,1)(xy,2xy),,3,課時作業(yè),PART THREE,基礎(chǔ)保分練,,1,2,3,4,5,6,
8、7,8,9,10,11,12,13,14,15,16,17,18,19,20,,,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,解析依題意得,z21i,,,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,
9、15,16,17,18,19,20,解析設(shè)zabi,a,bR, 則由z21216i,得a2b22abi1216i,,,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,,8.已知集合M1,m,3(m25m6)i,N1,3,若MN3,則實數(shù)m的值為______.,解析MN3,3M且1M, m1,3(m25m6)i3或m3, m25m60且m1或m3, 解得m6或m3,經(jīng)檢驗符合題意.,3或6,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,,9.(2019嘉興測試)若復(fù)數(shù)z43i,其中i是
10、虛數(shù)單位,則|z|__,z2______.,5,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,724i,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,10.若復(fù)數(shù)z滿足(3i)z2i(i為虛數(shù)單位),則z______;|z|___.,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,1,四,故復(fù)數(shù)z的共軛復(fù)數(shù)在復(fù)平面內(nèi)對應(yīng)的點位于第四象限.,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,1
11、8,19,20,11,解析由題意得17ai(45i)(3i)1711i,所以a11.,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,三,所以z13i, 所以z在復(fù)平面內(nèi)對應(yīng)的點為(1,3),位于第三象限,,,14.(2017浙江)已知a,bR,(abi)234i(i是虛數(shù)單位),則a2b2___,ab____.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,5,2,解析(abi)2a2b22abi. 解得a24,b21. 所以a2b25,ab2.,,解因為zbi(bR),,所以b2
12、,即z2i.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,,(2)若復(fù)數(shù)(mz)2所表示的點在第一象限,求實數(shù)m的取值范圍.,解因為z2i,mR, 所以(mz)2(m2i)2m24mi4i2 (m24)4mi, 又因為復(fù)數(shù)(mz)2所表示的點在第一象限,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,,16.若虛數(shù)z同時滿足下列兩個條件: z 是實數(shù); z3的實部與虛部互為相反數(shù). 這樣的虛數(shù)是否存在?若存在,求出z;若不存在,請說明理由.,1,2,3,4,5,6,7,8,9,10
13、,11,12,13,14,15,16,17,18,19,20,解存在.設(shè)zabi(a,bR,b0),,所以z12i或z2i.,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,17.若復(fù)數(shù) (i是虛數(shù)單位)在復(fù)平面內(nèi)對應(yīng)的點在第一象限,則實數(shù)a的取值范圍是 A.(,1) B.(1,) C.(1,1) D.(,1)(1,),,技能提升練,因為z在復(fù)平面內(nèi)對應(yīng)的點在第一象限,,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,,1,2,3,4,5,6,7,8,9,10,11,12,13,1
14、4,15,16,17,18,19,20,,拓展沖刺練,19.給出下列命題: 若zC,則z20; 若a,bR,且ab,則aibi; 若aR,則(a1)i是純虛數(shù); 若zi,則z31在復(fù)平面內(nèi)對應(yīng)的點位于第一象限. 其中正確的命題是____.(填上所有正確命題的序號),,解析由復(fù)數(shù)的概念及性質(zhì)知,錯誤;錯誤; 若a1,則a10,不滿足純虛數(shù)的條件,錯誤; z31(i)31i1,正確.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,,20.復(fù)數(shù)z1,z2滿足z1m(4m2)i,z22cos (4sin )i(m,,R),并且z1z2,求的取值范圍.,解由復(fù)數(shù)相等的充要條件可得,由此可得4cos24sin 44(1sin2)4sin 4,因為sin 1,1,所以4sin24sin 1,8. 所以的取值范圍是1,8.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,