Math topic中文

Four operations

The four operations of complex numbers: addition, subtraction, multiplication and division of complex numbers expressed in cartesian form.

Practice questions

  1. 1Solve z4=−4−4√3iz^4 = -4 - 4√3\mathrm{i}, expressing your answers in the form reiθr\mathrm{e}^{\mathrm{i}\theta}, where r>0r > 0 an…
  2. 2已知复数z1=1−3iz_{1}=1-3i,z2=3+iz_{2}=3+i,则()(\quad) A. ∣z1+z2∣=6  |z_{1}+z_{2}|=6\:\: B. z‾1−z2=−2+2i{\overline{z}}_{1}-{z}_{2}=-2+2i C. z1z2=6−8iz_{1}z_{2}=6-8i …
  3. 3Solve the equation iz5=−32i z^5 = -32, giving your roots in the form reiθre^{i\theta}, where r>0r>0 and −π<θ≤π-\pi<\theta\le\pi. Ske…
  4. 4Given that z=2−iz = 2 - \text{i} is a root of the equation 4z4−12z3+17z2+pz+q=04z^4 - 12z^3 + 17z^2 + pz + q = 0, where pp and qq are real,…
  5. 5Show that for any complex number z=eiθz = e^{i\theta}, zn+1zn=2cos⁡nθ.z^{n} + \dfrac{1}{z^{n}} = 2\cos n\theta . By taking n=1n = 1, sh…
  6. 6已知复数z=x+yi(x,y∈R)z=x+yi(x,y\in R),则()(\quad) A. z2⩾0z^{2}\geqslant 0 B. zz的虚部是yiyi C. 若z=1+2iz=1+2i,则x=1x=1,y=2y=2 D. ∣z∣=x2+y2|z|=\sqrt{x^{2}+y^{2}}…
  7. 7已知a∈Ra∈R,若1+ai2+i \dfrac {1+ai}{2+i}为实数,则a=a= ___ .
  8. 8Solve z4=−4−43 iz^{4} = -4 - 4\sqrt{3}\,i, expressing your answers in the form reiθre^{i\theta}, where r>0r > 0 and −π<θ≤π-\pi < \theta \le \pi…
  9. 9One of the roots of the equation 2z4−14z3+33z2−26z+p=0, where p is a constant2z^4 - 14z^3 + 33z^2 - 26z + p = 0\text{, where } p \text{ is a constant} is 3+i3 + \mathrm{i}…
  10. 10已知A(1 , 2)A(1\:,\:2),B(a ,1)B(a\:,1),C(2 , 3)C(2\:,\:3),D(−1,b)(D(-1,b)(其中aa,b∈R)b\in R)是复平面上的四点,且向量AB→,CD→,\overset{\rightarrow}{{AB}}{,}{\overset{\rightarrow}{{CD}}{,}}…
  11. 11已知ii为虚数单位,则i1+i\dfrac{i}{1+i}的虚部等于()() A. 12i\dfrac{1}{2}i B. 12\dfrac{1}{2} C. 14\dfrac{1}{4} D. −14i-\dfrac{1}{4}i…
  12. 12The complex number zz satisfies the equation f(z)=0\mathrm{f}(z) = 0, where f(z)=(1−ai)z2−2iz−10−20i\mathrm{f}(z) = (1 - a\mathrm{i})z^2 - 2\mathrm{i}z - 10 - 20\mathrm{i}…
  13. 13Do not use a calculator in answering this question. Given that x=1+3ix = 1 + 3\mathrm{i} is a root of the equation x3+ax2+18x+b=0,x^3 + ax^2 + 18x + b = 0,…
  14. 14One of the roots of the equation 2z4−14z3+33z2−26z+p=0, where p is a constant2z^{4} - 14z^{3} + 33z^{2} - 26z + p = 0\text{, where } p \text{ is a constant} is…
  15. 15Find, in a simplified form, the cartesian equation of the locus of the point P representing the complex number z where ∣z+3∣=2 ℜ(z)|z+3|=2\,\Re(z)…
  16. 16若z=1+2iz=1+2i,则∣z2∣=()|z^{2}|=(\quad) A. 5+4i5+4i B. −3+4i-3+4i C. 2525 D. 55
  17. 17设复数zz满足(1−i)z=1+i(1-i)z=1+i,则∣z∣=()|z|=() A. 00 B. 11 C. 2 \sqrt {2} D. 22
  18. 18已知i{i}为虚数单位,复数z=(a−i)2z={{\left( a-{i} \right)}^{2}},a∈Ra\in \mathrm{R},若复数zz是纯虚数,则∣z∣=\left| z \right|= A. 11 B.…
  19. 19在复平面内,复数z=13+iz= \dfrac {1}{3+i}对应的点位于()() A. 第一象限 B. 第二象限 C. 第三象限 D. 第四象限
  20. 20Do not use a calculator in answering this question. Three complex numbers are z1=1+iz_1 = 1 + i, z2=3−iz_2 = \sqrt{3} - i and z3=cos⁡π3+isin⁡π3z_3 = \cos \dfrac{\pi}{3} + i\sin \dfrac{\pi}{3}…
  21. 21设z=1−i1+i+2iz=\dfrac{1-{i}}{1+{i}}+2{i},则∣z∣=|z|= A. 00 B. 12\dfrac{1}{2} C. 11 D. 2\sqrt{2}
  22. 22The equation z3−2z+k=0z^{3}-2z+k=0, where kk is a real constant, has a root z=1+aiz = 1 + ai, where aa is a positive real constan…
  23. 23Do not use a calculator in answering this question. It is given that complex numbers z1z_1 and z2z_2 are the roots of th…
  24. 24Sketch on a single Argand diagram, the loci defined by −π4<arg⁡(z+1+2i)≤π4-\frac{\pi}{4} < \arg(z + 1 + 2\text{i}) \le \frac{\pi}{4} and…
  25. 25设复数z=1+2iz=1+ \sqrt {2}i,则z2−2zz^{2}-2z等于()() A. 3i3 i B. −3i-3 i C. 33 D. −3-3
  26. 26The complex number zz has modulus 2 and argument π8\frac{\pi}{8}. It is also given that w=1+iw = 1 + \mathrm{i}. Given tha…
  27. 27Solve the equation iz5=−32iz^5=-32, giving your roots in the form reiθre^{i\theta}, where r>0r>0 and −π<θ≤π-\pi<\theta\le\pi. Sketch…
  28. 28已知复数 z=3+4iz = 3 + 4i,求 ∣z∣|z|、zˉ\bar{z} 和 z2z^2。
  29. 29设a,b,c∈R,则复数(a+bi)(c+di)为实数的充要条件是( ) A. ad-bc=0 B. ac-bd=0 C. ac+bd=0 D. ad+bc=0
  30. 30The complex number zz satisfies the equation 2 z3=1+22a−1 i,\sqrt{2}\,z^{3} = 1 + \dfrac{2}{2a - 1}\,i , where aa is a real numbe…
  31. 31Find the set of values of α\alpha for which the expression 2−isin⁡2α1+2isin⁡2α\dfrac{2 - i\sin 2\alpha}{1 + 2i\sin 2\alpha} is purely re…
  32. 32Do not use a calculator in answering this question. It is given that α=−3+i\alpha = -\sqrt{3} + \mathrm{i} and β=cos⁡π3+isin⁡π3\beta = \cos\frac{\pi}{3} + \mathrm{i}\sin\frac{\pi}{3}…
  33. 33Given that z1=−i2z_1 = -\frac{\text{i}}{2} is a root of the equation 2z3+(i−8)z2+az+13i=02z^3 + (\text{i}-8)z^2 + az + 13\text{i} = 0, find th…
  34. 34设ii为虚数单位,则复数i−2i \dfrac {i-2}{i}的共轭复数是 ___ .
  35. 35Do not use a calculator in answering this question. Showing your working, find the complex numbers zz and ww which…
  36. 36已知zz是复数,若z+2iz+2i为实数(i(i为虚数单位)),且z−4z-4为纯虚数. (1)求复数zz; (2)若复数(z+mi)2(z+mi)^{2}在复平面上对应的点在第四象限,求实数mm的取值范围.
  37. 37已知z−\overset{-}{z}是复数zz的共轭复数,若z+2z−z+2\overset{-}{z}在复平面上的对应点位于第一象限,则zz的对应点位于()(\quad) A. 第一象限 B. 第二象限 C. 第三象限…
  38. 38A complex number zz varies with tt such that z=2cos⁡t+i(3sin⁡t),where 0≤t<2π.z = 2\cos t + \mathrm{i}(3\sin t), \quad \text{where } 0 \le t < 2\pi.…
  39. 39设复数i−21+i=a+bi(a,b∈R) \dfrac {i-2}{1+i}=a+bi(a,b∈R) 则a+b=()a+b=() A. 11 B. 22 C. −1-1 D. −2-2
  40. 40若复数a+ib−i=2−i \dfrac {a+i}{b-i}=2-i其中aa,bb是实数,则复数a+bia+bi在复平面内所对应的点位于()() A. 第一象限 B. 第二象限 C. 第三象限 D. 第四象限
  41. 41Solve the equation z5+16=163 iz^5 + 16 = 16\sqrt{3}\, i, giving the roots in the form reiθr e^{i\theta}, where r>0r>0 and −π<θ≤π-\pi < \theta \le \pi…
  42. 42The complex numbers aa and bb are given by a=−(1+3i)a = -(1 + \sqrt{3}\text{i}) and b=12(1−i)b = \frac{1}{2}(1 - \text{i}). Without…
  43. 43Do not use a graphing calculator in answering this question. The complex number ww is a root to the equation z4+az3+46z2+bz+125=0,z^4 + a z^3 + 46 z^2 + b z + 125 = 0 ,…
  44. 44Given that z1=r1(cos⁡θ1+isin⁡θ1)z_1 = r_1(\cos\theta_1 + \mathrm{i}\sin\theta_1) and z2=r2(cos⁡θ2+isin⁡θ2)z_2 = r_2(\cos\theta_2 + \mathrm{i}\sin\theta_2), pr…
  45. 45设复数zz的共轭复数为z‾\overline{z},且满足2z+z‾=3+i2z+\overline{z}=3+i,ω=sin⁡θ−icos⁡θ\omega=\sin\theta-i\cos\theta,复数z−ωz-\omega对应复平面的向量OM→.(1)\overrightarrow{OM}.(1)…
  46. 46若复数zz满足(2−i)z=4+3i(i(2-i)z=4+3i(i为虚数单位)),则∣z∣=|z|=
  47. 47Show that eiθ1−eiθ\frac{\text{e}^{\text{i}\theta}}{1 - \text{e}^{\text{i}\theta}} can be expressed as k(icot⁡θ2−1)k\left(\text{i}\cot\frac{\theta}{2} - 1\right)…
  48. 48Given that z=λ−4i1−λiz = \frac{\lambda - 4i}{1 - \lambda i}, λ∈R\lambda \in \mathbb{R} and arg⁡(z)=π\arg(z) = \pi, find the value of zz…
  49. 49Do not use a calculator in answering this question. The complex number zz has modulus 1 and argument θ\theta, where π2<θ<π\dfrac{\pi}{2}<\theta<\pi…
  50. 50Solve the equation z5+32=0z^{5}+32=0, expressing your answers in the form reiθre^{i\theta}, where r>0r>0 and −π<θ≤π-\pi<\theta\le\pi…
  51. 51若ii为虚数单位,则(3−6i)(1+9i)=()(3-6i)(1+9i)=(\quad) A. 51+33i51+33i B. 51−33i51-33i C. 57+21i57+21i D. 57−21i57-21i
  52. 52Let zz be the complex number −1+i3-1+i\sqrt{3}. Find arg⁡(z),\arg(z), the real number aa such that arg⁡(z(z+a))=5π6\arg\big(z(z+a)\big)=\frac{5\pi}{6}…
  53. 53已知ii是虚数单位,则(1−i1+i)3=()(\dfrac{1-i}{1+i})^{3}=(\quad) A. 11 B. ii C. −i-i D. −1-1
  54. 54ii为虚数单位,5i−2=()\dfrac{5}{i-2}=(\quad) A. i+2i\text{+}2 B. i-2i\text{-}2 C. -2-i\text{-}2\text{-}i D. 2-i2\text{-}i
  55. 55An arithmetic series has first term aa and common difference dd, where a>0a > 0 and d≠0d \ne 0. The first, sixth and nin…
  56. 56下列四个选项中,正确的为()(\quad) A. 复平面内实轴上的点都表示实数,虚轴上的点都表示纯虚数 B. 已知复数z=x+yi(x,y∈R)z=x+yi(x,y\in R),且∣z−2∣=3|z-2|=\sqrt{3},则yx\dfrac{y}{x}的…
  57. 57The points P and Q are represented by the complex numbers p and q respectively, where p=2+2i , arg⁡(q)=23π and ∣q∣=2.p=\sqrt{2}+\sqrt{2}i\,,\ \arg(q)=\dfrac{2}{3}\pi\text{ and }|q|=2.…
  58. 58(1+i)(2−i)=()(1+i)(2-i)=() A. −3−i-3-i B. −3+i-3+i C. 3−i3-i D. 3+i3+i
  59. 59已知ii是虚数单位,复数z=1+2iz=1+2i,则iz‾=()i \overline {z}=() A. 2−i2-i B. 2+i2+i C. −2−i-2-i D. −2+i-2+i
  60. 60已知复数z1=2+ai(a∈R)z_{1}=2+ai(a∈R),z2=1−2iz_{2}=1-2i,若z1z2 \dfrac {z_{1}}{z_{2}}为纯虚数,则∣z1∣=()|z_{1}|=() A. 2 \sqrt {2} B. 3 \sqrt {3}…