新高考復(fù)數(shù)題目及答案

一、單項(xiàng)選擇題(每題2分,共10題)1.已知復(fù)數(shù)\(z=1+2i\),則\(\overline{z}\)（\(\overline{z}\)為\(z\)的共軛復(fù)數(shù)）為（）A.\(1-2i\)B.\(-1+2i\)C.\(2+i\)D.\(2-i\)答案：A2.復(fù)數(shù)\(z=\frac{2i}{1-i}\),則\(|z|=\)()A.\(1\)B.\(\sqrt{2}\)C.\(\sqrt{3}\)D.\(2\)答案：B3.若\((1+i)z=2\),則\(z\)的虛部為()A.\(-1\)B.\(1\)C.\(-i\)D.\(i\)答案：A4.復(fù)數(shù)\(z=\frac{3-i}{1+i}\)的實(shí)部為()A.\(1\)B.\(2\)C.\(-1\)D.\(-2\)答案：A5.已知\(i\)為虛數(shù)單位,復(fù)數(shù)\(z=1+i+i^2+i^3\),則\(|z|=\)()A.\(0\)B.\(1\)C.\(\sqrt{2}\)D.\(2\)答案：C6.復(fù)數(shù)\(z=\frac{1+i}{1-i}\)的共軛復(fù)數(shù)是()A.\(i\)B.\(-i\)C.\(1+i\)D.\(1-i\)答案：B7.若復(fù)數(shù)\(z=a+bi\)(\(a,b\inR\))滿足\(z(1+i)=2i\),則\(a+b=\)()A.\(0\)B.\(1\)C.\(2\)D.\(3\)答案：B8.復(fù)數(shù)\(z=\frac{2}{1-i}\),則\(z^2=\)()A.\(2i\)B.\(-2i\)C.\(2\)D.\(-2\)答案：A9.已知復(fù)數(shù)\(z\)滿足\(z(1-i)=1+i\),則\(z^2=\)()A.\(-1\)B.\(1\)C.\(-i\)D.\(i\)答案：A10.復(fù)數(shù)\(z=\frac{1-3i}{1+i}\)在復(fù)平面內(nèi)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).第四象限答案：C二、多項(xiàng)選擇題(每題2分,共10題)1.已知復(fù)數(shù)\(z=a+bi\)(\(a,b\inR\)),則下列說法正確的是()A.若\(z=0\),則\(a=0\)且\(b=0\)B.若\(z\)為純虛數(shù),則\(a=0\)且\(b\neq0\)C.若\(z\)為實(shí)數(shù),則\(b=0\)D.若\(z\)為虛數(shù),則\(b\neq0\)答案:ABCD2.對于復(fù)數(shù)\(z=\frac{1+i}{1-i}\),以下結(jié)論正確的是（）A.\(z=i\)B.\(|z|=1\)C.\(z^2=-1\)D.\(z\)的共軛復(fù)數(shù)為\(-i\)答案:ABC3.已知復(fù)數(shù)\(z=1+i\),則（）A.\(z+\overline{z}=2\)B.\(z-\overline{z}=2i\)C.\(z\cdot\overline{z}=2\)D.\(\frac{z}{\overline{z}}=i\)答案:ABC4.下列復(fù)數(shù)中,是純虛數(shù)的是（）A.\(z=\frac{1+i}{1-i}\)B.\(z=(1+i)^2\)C.\(z=\frac{1}{i}\)D.\(z=i(1-i)\)答案:BC5.已知復(fù)數(shù)\(z=a+bi\)（\(a,b\inR\)）,若\(z+2\overline{z}=3+4i\),則（）A.\(a=1\)B.\(b=4\)C.\(a=-1\)D.\(b=-4\)答案:AD6.復(fù)數(shù)\(z=\frac{2-i}{i}\)（\(i\)為虛數(shù)單位）,則（）A.\(z=-1-2i\)B.\(|z|=\sqrt{5}\)C.\(z\)的實(shí)部為\(-1\)D.\(z\)的虛部為\(-2\)答案:ABCD7.已知復(fù)數(shù)\(z\)滿足\((1+2i)z=4+3i\),則（）A.\(z=2-i\)B.\(|z|=\sqrt{5}\)C.\(z\)的共軛復(fù)數(shù)為\(2+i\)D.\(z\)在復(fù)平面內(nèi)對應(yīng)的點(diǎn)位于第四象限答案:ABCD8.若復(fù)數(shù)\(z=\frac{1}{1+i}\),則（）A.\(z=\frac{1-i}{2}\)B.\(|z|=\frac{\sqrt{2}}{2}\)C.\(z\)的實(shí)部為\(\frac{1}{2}\)D.\(z\)的虛部為\(-\frac{1}{2}\)答案:ABD9.已知復(fù)數(shù)\(z\)滿足\(z^2=-4\),則（）A.\(z=2i\)B.\(z=-2i\)C.\(|z|=2\)D.\(z\)的共軛復(fù)數(shù)為\(\pm2i\)答案:BCD10.復(fù)數(shù)\(z=\frac{3+i}{1-i}\),則（）A.\(z=1+2i\)B.\(|z|=\sqrt{5}\)C.\(z\)的實(shí)部為\(1\)D.\(z\)的虛部為\(2\)答案:ABCD三、判斷題(每題2分,共10題)1.復(fù)數(shù)\(z=a+bi\)(\(a,b\inR\)）的虛部是\(bi\)。(）答案：×2.若\(z\)是復(fù)數(shù),則\(z^2\geq0\)。()答案：×3.純虛數(shù)的共軛復(fù)數(shù)是它本身。()答案：×4.兩個(gè)復(fù)數(shù)相等當(dāng)且僅當(dāng)它們的實(shí)部和虛部分別相等。()答案：√5.復(fù)數(shù)\(z=\frac{1}{i}\)化簡后為\(i\)。()答案：×6.復(fù)數(shù)\(z=2+3i\)的模\(|z|=\sqrt{13}\)。()答案：√7.若\(z_1,z_2\)為復(fù)數(shù),且\(z_1+z_2\)為實(shí)數(shù),則\(z_1,z_2\)的虛部互為相反數(shù)。()答案：√8.復(fù)數(shù)\(z=\frac{2-2i}{1+i}\)化簡后為\(-2i\)。()答案：√9.復(fù)數(shù)\(z=3-4i\)在復(fù)平面內(nèi)對應(yīng)的點(diǎn)位于第四象限。()答案：√10.若\(z\)為復(fù)數(shù),則\(z\overline{z}=|z|^2\)。()答案：√四、簡答題(每題5分,共4題)1.已知復(fù)數(shù)\(z=2+3i\),求\(z\)的共軛復(fù)數(shù)\(\overline{z}\)以及\(z\overline{z}\)的值。答案：\(\overline{z}=2-3i\),\(z\overline{z}=(2+3i)(2-3i)=4+9=13\)。2.化簡復(fù)數(shù)\(z=\frac{3+2i}{1-i}\)。答案:\(z=\frac{(3+2i)(1+i)}{(1-i)(1+i)}=\frac{3+3i+2i+2i^2}{2}=\frac{1+5i}{2}=\frac{1}{2}+\frac{5}{2}i\)。3.已知\(z=1-2i\),求\(|z|\)。答案:\(|z|=\sqrt{1^2+(-2)^2}=\sqrt{5}\)。4.若復(fù)數(shù)\(z\)滿足\(z(1+i)=2\),求\(z\)。答案:\(z=\frac{2}{1+i}=\frac{2(1-i)}{(1+i)(1-i)}=\frac{2(1-i)}{2}=1-i\)。五、討論題(每題5分,共4題)1.討論復(fù)數(shù)在生活中的應(yīng)用。答案:如在電子技術(shù)中用于電路分析,在導(dǎo)航定位中用于信號處理等,能更精確地描述和處理相關(guān)問題。2.談?wù)剰?fù)數(shù)與實(shí)數(shù)的聯(lián)系與區(qū)別。答案:聯(lián)系是實(shí)數(shù)是復(fù)數(shù)的特殊情況(虛部為0)。區(qū)別在于復(fù)數(shù)包含虛數(shù)部分,運(yùn)算規(guī)則有所不同,復(fù)數(shù)能解決一些實(shí)數(shù)無法解決的問題,如方程\(x^2+1=0\)在復(fù)數(shù)域有