2019-2020年高三數(shù)學(xué)一輪復(fù)習(xí)-第4章-第4課時-數(shù)系的擴(kuò)充與復(fù)數(shù)的引入課時訓(xùn)練-文-新人教版

1、2019-2020年高三數(shù)學(xué)一輪復(fù)習(xí) 第4章 第4課時 數(shù)系的擴(kuò)充與復(fù)數(shù)的引入課時訓(xùn)練 文 新人教版1(2014高考天津卷)i是虛數(shù)單位，復(fù)數(shù)eq f(7i,34i)()A1iB1iC.eq f(17,25)eq f(31,25)iDeq f(17,7)eq f(25,7)i解析：選A.根據(jù)復(fù)數(shù)的除法法則進(jìn)行運(yùn)算，eq f(7i,34i)eq f(7i34i,34i34i)eq f(2525i,25)1i.2在復(fù)平面內(nèi)，復(fù)數(shù)eq f(23i,34i)(i是虛數(shù)單位)所對應(yīng)的點(diǎn)位于()A第一象限B第二象限C第三象限D(zhuǎn)第四象限解析：選B.eq f(23i,34i)eq f(23i34i,34i34

2、i)eq f(18i,25)eq f(18,25)eq f(1,25)i，eq f(18,25)eq f(1,25)i對應(yīng)的點(diǎn)為eq blc(rc)(avs4alco1(f(18,25)，f(1,25)，在第二象限3若復(fù)數(shù)z滿足(2i)z|12i|，則z的虛部為()A.eq f(r(5),5)Beq f(r(5),5)iC1Di解析：選A.由zeq f(|12i|,2i)eq f(r(5)2i,2i2i)eq f(r(5)2i,5)eq f(2r(5),5)eq f(r(5),5)i，可知其虛部為eq f(r(5),5).4(2015洛陽統(tǒng)考)已知復(fù)數(shù)eq f(a3i,12i)是純虛數(shù)，則實(shí)數(shù)

3、a()A2B4 C6D6解析：選D.eq f(a3i,12i)eq f(a62a3i,5)，當(dāng)a6時，復(fù)數(shù)eq f(a3i,12i)為純虛數(shù)5(2014高考湖南卷)滿足eq f(zi,z)i(i為虛數(shù)單位)的復(fù)數(shù)z()A.eq f(1,2)eq f(1,2)iBeq f(1,2)eq f(1,2)iCeq f(1,2)eq f(1,2)iDeq f(1,2)eq f(1,2)i解析：選B.根據(jù)復(fù)數(shù)的乘、除法運(yùn)算法則求解eq f(zi,z)i，zizi，iz(i1)zeq f(i,i1)eq f(i1i,1i1i)eq f(1i,2)eq f(1,2)eq f(i,2).6(2014高考浙江卷)

4、已知i是虛數(shù)單位，計算eq f(1i,1i2)_.解析：直接利用復(fù)數(shù)的四則運(yùn)算法則進(jìn)行運(yùn)算eq f(1i,1i2)eq f(1i,12ii2)eq f(1i,2i)eq f(i1i,2i2)eq f(i1,2)eq f(1,2)eq f(1,2)i.答案：eq f(1,2)eq f(1,2)i7(2014高考江蘇卷)已知復(fù)數(shù)z(52i)2(i為虛數(shù)單位)，則z的實(shí)部為_解析：先進(jìn)行復(fù)數(shù)的運(yùn)算，再確定復(fù)數(shù)的實(shí)部因?yàn)閦(52i)22520i(2i)22520i42120i，所以z的實(shí)部為21.答案：218(2013高考重慶卷)已知復(fù)數(shù)zeq f(5i,12i)(i是虛數(shù)單位)，則|z|_.解析：e

5、q f(5i,12i)eq f(5i12i,12i12i)2i，所以|z|eq r(5).答案：eq r(5)9若eq f(3bi,1i)abi(a，b為實(shí)數(shù)，i為虛數(shù)單位)，求ab的值解析：由eq f(3bi,1i)eq f(3bi1i,1i1i)eq f(3b3bi,2)abi，得aeq f(3b,2)，beq f(3b,2)，解得b3，a0，所以ab3.1(2014高考新課標(biāo)全國卷)設(shè)復(fù)數(shù)z1，z2在復(fù)平面內(nèi)的對應(yīng)點(diǎn)關(guān)于虛軸對稱，z12i，則z1z2()A5B5 C4iD4i解析：選A.先求出z2，再代入計算z12i在復(fù)平面內(nèi)的對應(yīng)點(diǎn)的坐標(biāo)為(2,1)，又z1與z2在復(fù)平面內(nèi)的對應(yīng)點(diǎn)關(guān)于

6、虛軸對稱，則z2的對應(yīng)點(diǎn)的坐標(biāo)為(2,1)，即z22i，z1z2(2i)(2i)i245.2(2014高考江西卷)eq o(z,sup10() 是z的共軛復(fù)數(shù)，若zeq o(z,sup10() 2，(zeq o(z,sup10() )i2(i為虛數(shù)單位)，則z()A1iB1iC1iD1i解析：選D.將z，eq o(z,sup10() 看做兩個未知數(shù)，利用方程思想求解也可利用復(fù)數(shù)相等的條件求解法一：設(shè)zabi，a，b為實(shí)數(shù)，則eq o(z,sup10() abi.zeq o(z,sup10() 2a2，a1.又(zeq o(z,sup10() )i2bi22b2，b1.故z1i.法二：(zeq

7、o(z,sup10() )i2，zeq o(z,sup10() eq f(2,i)2i.又zeq o(z,sup10() 2，(zeq o(z,sup10() )(zeq o(z,sup10() )2i2，2z2i2，z1i.3(2013高考陜西卷)設(shè)z1，z2是復(fù)數(shù)，則下列命題中的假命題是()A若|z1z2|0，則eq o(z,sup10()1eq o(z,sup10()2B若z1eq o(z,sup10()2，則eq o(z,sup10()1z2C若|z1|z2|，則z1eq o(z,sup10()1z2eq o(z,sup10()2D若|z1|z2|，則zeq oal(2,1)zeq o

8、al(2,2)解析：選D.結(jié)合復(fù)數(shù)的模、共軛復(fù)數(shù)及復(fù)數(shù)的運(yùn)算等判斷求解A，|z1z2|0z1z20z1z2eq o(z,sup10()1eq o(z,sup10()2，真命題；B，z1eq o(z,sup10()2eq o(z,sup10()1eq o(z,sup10()2z2，真命題；C，|z1|z2|z1|2|z2|2z1eq o(z,sup10()1z2eq o(z,sup10()2，真命題；D，當(dāng)|z1|z2|時，可取z11，z2i，顯然zeq oal(2,1)1，zeq oal(2,2)1，即zeq oal(2,1)zeq oal(2,2)，假命題4(2015湖北省八校聯(lián)考)已知i是

9、虛數(shù)單位，z1i，eq xto(z)為z的共軛復(fù)數(shù)，則復(fù)數(shù)eq f(z2,xto(z)在復(fù)平面上對應(yīng)的點(diǎn)的坐標(biāo)為_解析：z1i，則eq f(z2,xto(z)eq f(1i2,1i)eq f(2i,1i)eq f(2i1i,1i1i)1i，則復(fù)數(shù)eq f(z2,xto(z)在復(fù)平面上對應(yīng)的點(diǎn)的坐標(biāo)為(1,1)答案：(1,1)5下面是關(guān)于復(fù)數(shù)zeq f(2,1i)的四個命題：p1：|z|2, p2：z22i，p3：z的共軛復(fù)數(shù)為1i, p4：z的虛部為1.其中的真命題為_解析：|z|eq blc|rc|(avs4alco1(f(2,1i)eq f(2,r(2)eq r(2)，z2eq blc(rc)(avs4alco1(f(2,1i)22i，zeq f(2,1i)eq f(21i,2)1i，eq xto(z)1i.答案：p2，p46復(fù)數(shù)zeq blc(rc)(avs4alco1(f(1i,r(2)2 015在復(fù)平面內(nèi)對應(yīng)的點(diǎn)的坐標(biāo)為_解析：eq blc(rc)(avs4alco1(f(1i,r(2)2i，eq blc(rc)(avs4alco1(f(1i,r(2)2 015eq blc(rc)(avs4alco1(f(1i,r(2)2 014eq f(1i,r(2)eq blcrc(avs4alco1(blc(rc)(avs4alco