2024-2025學(xué)年高中數(shù)學(xué)第3章數(shù)系的擴(kuò)充與復(fù)數(shù)的引入3.3復(fù)數(shù)的幾何意義課時(shí)素養(yǎng)評(píng)價(jià)含解析蘇教版選修2-2

PAGE課時(shí)素養(yǎng)評(píng)價(jià)十三復(fù)數(shù)的幾何意義(25分鐘·60分)一、選擇題(每小題5分,共25分)1.(2024·全國(guó)卷Ⅰ)設(shè)復(fù)數(shù)z滿意|z-i|=1,z在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)為(x,y),則()A.(x+1)2+y2=1B.(x-1)2+y2=1C.x2+(y-1)2=1D.x2+(y+1)2=1【解析】選C.z=x+yi,z-i=x+(y-1)i,|z-i|=QUOTE=1,則x2+(y-1)2=1.2.如圖所示,向量,所對(duì)應(yīng)的復(fù)數(shù)分別為z1,z2,則z1·z2= ()A.4+2i B.2+iC.2+2i D.3+i【解析】選A.由題圖可知,z1=1+i,z2=3-i,則z1·z2=(1+i)(3-i)=4+2i.3.已知復(fù)數(shù)z=QUOTE-i5(i為虛數(shù)單位),則復(fù)數(shù)QUOTE= ()A.2-i B.-2-iC.2+i D.-2+i【解析】選C.z=QUOTE-i5=2-i,所以QUOTE=2+i.4.向量對(duì)應(yīng)的復(fù)數(shù)z1=-3+2i,對(duì)應(yīng)的復(fù)數(shù)z2=1-i,則|+|= ()A.QUOTE B.QUOTE C.QUOTE D.5【解題指南】解答本題的關(guān)鍵是把復(fù)數(shù)運(yùn)算轉(zhuǎn)化為向量的運(yùn)算.【解析】選C.因?yàn)橄蛄繉?duì)應(yīng)的復(fù)數(shù)z1=-3+2i,對(duì)應(yīng)的復(fù)數(shù)z2=1-i,所以=(-3,2),=(1,-1),則+=(-2,1),所以|+|=QUOTE.5.已知復(fù)數(shù)z滿意|z|2-3|z|+2=0,則復(fù)數(shù)z對(duì)應(yīng)點(diǎn)的軌跡是 ()A.一個(gè)圓 B.兩個(gè)圓C.兩點(diǎn) D.線段【解析】選B.由|z|2-3|z|+2=0,得(|z|-1)·(|z|-2)=0,所以|z|=1或|z|=2.由復(fù)數(shù)模的幾何意義知,z對(duì)應(yīng)點(diǎn)的軌跡是兩個(gè)圓.二、填空題(每小題5分,共15分)6.若復(fù)數(shù)z滿意z-1=cosθ+isinθ,則當(dāng)θ=____________時(shí)|z|有最大值,最大值為_(kāi)___________.

【解析】因?yàn)閦-1=cosθ+isinθ,所以z=(1+cosθ)+isinθ,所以|z|=QUOTE=QUOTE,當(dāng)cosθ=1時(shí),即θ=2kπ(k∈Z)時(shí),|z|有最大值2.答案:2kπ(k∈Z)27.設(shè)復(fù)數(shù)z=(x-1)+(y-QUOTE)i(x,y∈R),若|z|≤2,則y≤QUOTEx的概率為_(kāi)___________.

【解析】復(fù)數(shù)z=(x-1)+(y-QUOTE)i(x,y∈R),由|z|≤2得(x-1)2+(y-QUOTE)2≤4,表示圓心在C(1,QUOTE),半徑為2的圓及其內(nèi)部,總區(qū)域的面積為4π,如圖,求得y≤QUOTEx對(duì)應(yīng)的弓形的面積為QUOTE-QUOTE,由幾何概型的概率公式,得事務(wù)A=QUOTE發(fā)生的概率為P(A)=QUOTE=QUOTE-QUOTE.答案:QUOTE-QUOTE8.已知實(shí)數(shù)x,y,a滿意a2+2a+2xy+(a+x-y)i=0,則點(diǎn)P(x,y)的軌跡方程為_(kāi)______________.

【解析】由題意知,QUOTE消去a化簡(jiǎn)得(x-1)2+(y+1)2=2.答案:(x-1)2+(y+1)2=2三、解答題(每小題10分,共20分)9.求當(dāng)實(shí)數(shù)m為何值時(shí),復(fù)數(shù)z=(m2-8m+15)+(m2+3m-28)i在復(fù)平面內(nèi)的對(duì)應(yīng)點(diǎn)分別滿意下列條件:(1)位于第四象限.(2)位于x軸的負(fù)半軸上.【解析】(1)由題意,得QUOTE解得QUOTE即-7<m<3.故當(dāng)-7<m<3時(shí),復(fù)數(shù)z的對(duì)應(yīng)點(diǎn)位于第四象限.(2)由題意知QUOTE由②得m=-7或m=4.因m=-7不符合不等式①,舍去,m=4符合.所以當(dāng)m=4時(shí),復(fù)數(shù)z位于x軸負(fù)半軸上.10.已知關(guān)于t的一元二次方程t2+(2+i)t+2xy+(x-y)i=0(x,y∈R).(1)當(dāng)方程有實(shí)根時(shí),求點(diǎn)(x,y)的軌跡方程;(2)若方程有實(shí)根,求方程的實(shí)根的取值范圍.【解析】(1)設(shè)實(shí)根為m,則m2+(2+i)m+2xy+(x-y)i=0,即(m2+2m+2xy)+(m+x-y)i=0.依據(jù)復(fù)數(shù)相等的充要條件得QUOTE由②得m=y-x代入①得(y-x)2+2(y-x)+2xy=0,即(x-1)2+(y+1)2=2.故點(diǎn)(x,y)的軌跡方程為(x-1)2+(y+1)2=2.(2)由(1)知點(diǎn)(x,y)的軌跡是一個(gè)圓,圓心為(1,-1),半徑r=QUOTE,設(shè)方程的實(shí)根為m,則直線m+x-y=0與圓(x-1)2+(y+1)2=2有公共點(diǎn),所以QUOTE≤QUOTE,即|m+2|≤2,即-4≤m≤0.故方程的實(shí)根的取值范圍為[-4,0].(20分鐘·40分)1.(5分)已知復(fù)數(shù)z滿意z·i=3-4i(i為虛數(shù)單位),則|z|= ()A.3 B.4 C.5 D.QUOTE【解析】選C.方法一:因?yàn)閦·i=3-4i,所以z=QUOTE=QUOTE=-3i+4i2=-4-3i,所以|z|=QUOTE=5.方法二:z=QUOTE,所以|z|=QUOTE=QUOTE=5.2.(5分)(多選題)以下選項(xiàng)能滿意復(fù)數(shù)z=QUOTE在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)在第四象限的是 ()A.m=-2 B.m=-QUOTE C.m=0 D.m=1【解析】選BC.因?yàn)閦=QUOTE=QUOTE=QUOTE+QUOTEi在復(fù)平面內(nèi)對(duì)應(yīng)的點(diǎn)為QUOTE,且在第四象限,所以QUOTE解得-1<m<1.3.(5分)已知復(fù)平面內(nèi)三點(diǎn)A,B,C,點(diǎn)A對(duì)應(yīng)的復(fù)數(shù)為3+i,對(duì)應(yīng)的復(fù)數(shù)為2-i,對(duì)應(yīng)的復(fù)數(shù)為5+2i,則點(diǎn)C對(duì)應(yīng)的復(fù)數(shù)為_(kāi)___________.

【解析】設(shè)C(x,y),=-=(5,2)-(2,-1)=(3,3),所以(x,y)-(3,1)=(3,3),所以x=6,y=4,所以點(diǎn)C對(duì)應(yīng)的復(fù)數(shù)為6+4i.答案:6+4i【一題多解】=-,所以對(duì)應(yīng)的復(fù)數(shù)為5+2i-(2-i)=3+3i,=+,且對(duì)應(yīng)的復(fù)數(shù)即為A點(diǎn)對(duì)應(yīng)的復(fù)數(shù)3+i.所以對(duì)應(yīng)的復(fù)數(shù)為3+i+(3+3i)=6+4i.也即C點(diǎn)對(duì)應(yīng)的復(fù)數(shù)為6+4i.答案:6+4i4.(5分)復(fù)數(shù)z=x+yi(x,y∈R)滿意條件|z-4i|=|z+2|,則2x+4y的最小值為_(kāi)___________.

【解析】由|z-4i|=|z+2|得|x+(y-4)i|=|x+2+yi|,所以x2+(y-4)2=(x+2)2+y2,即x+2y=3,所以2x+4y=2x+QUOTE≥2QUOTE=2QUOTE=4QUOTE,當(dāng)且僅當(dāng)x=2y=QUOTE時(shí),2x+4y取得最小值4QUOTE.答案:4QUOTE5.(10分)已知a∈R,z=(a2-2a+4)-(a2-2a+2)i所對(duì)應(yīng)的點(diǎn)在第幾象限?復(fù)數(shù)z對(duì)應(yīng)的點(diǎn)的軌跡是什么?【解析】由a2-2a+4=(a-1)2+3≥3,-(a2-2a+2)=-(a-1)2-1≤-1,所以復(fù)數(shù)z的實(shí)部為正數(shù),復(fù)數(shù)z的虛部為負(fù)數(shù),因此,復(fù)數(shù)z的對(duì)應(yīng)點(diǎn)在第四象限.設(shè)z=x+yi(x,y∈R),則QUOTE消去a2-2a得:y=-x+2(x≥3).所以復(fù)數(shù)z的對(duì)應(yīng)點(diǎn)的軌跡是一條射線,方程為y=-x+2(x≥3).6.(10分)設(shè)復(fù)數(shù)z1,z2滿意QUOTE=QU