JEE Mathematics Complex Numbers MCQs Set 25

Answer: (a) \(-\dfrac32i\)

Explanation:
1. \(5+12i=(3+2i)^2\) and \(5-12i=(3-2i)^2\), so the principal roots are \(3+2i\) and \(3-2i\).
2. Numerator \(=6\), denominator \(=4i\), and \(\dfrac6{4i}=-\dfrac{3}{2}i\).

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Answer: (b) less than 1

Explanation:
1. By the triangle inequality, \(\left|\sum\lambda_ia_i\right|\le\sum\lambda_i|a_i|\).
2. Since \(|a_i| \lt 1\) and \(\sum\lambda_i=1\), \(\sum\lambda_i|a_i| \lt 1\). So the value is less than \(1\).

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Answer: (b) \(\tan^{-1}\left(\dfrac BA\right)\)

Explanation:
1. The argument of a product is the sum of the arguments: \(\sum\tan^{-1}\dfrac{b_k}{a_k}=\arg(A+iB)\).
2. \(\arg(A+iB)=\tan^{-1}\dfrac BA\) (up to a multiple of \(\pi\)).

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Answer: (a) \(\sin u\cos v\left[\cos(x+y-u-v)+i\sin(x+y-u-v)\right]\)

Explanation:
1. \(\cot u+i=\dfrac{\cos u+i\sin u}{\sin u}\) and \(1+i\tan v=\dfrac{\cos v+i\sin v}{\cos v}\).
2. The quotient is \(\sin u\cos v\cdot\dfrac{e^{i(x+y)}}{e^{iu}e^{iv}}=\sin u\cos v\,e^{i(x+y-u-v)}\).

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Answer: (c) \(1+c\cos\theta\)

Explanation:
1. Squaring \(\sqrt{1-c^2}=nc-1\): \(1-c^2=n^2c^2-2nc+1\Rightarrow c(1+n^2)=2n\), i.e. \(1+n^2=\dfrac{2n}c\).
2. \((1+nz)\left(1+\dfrac nz\right)=1+n^2+n\left(z+\dfrac1z\right)=\dfrac{2n}c+2n\cos\theta\).
3. So the expression is \(\dfrac c{2n}\left(\dfrac{2n}c+2n\cos\theta\right)=1+c\cos\theta\).

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Answer: (c) \(\dfrac32-2i\)

Explanation:
1. Let \(z=x+iy\): \(\sqrt{x^2+y^2}-x-iy=1+2i\). Comparing imaginary parts: \(y=-2\).
2. Real parts: \(\sqrt{x^2+4}=1+x\Rightarrow x^2+4=1+2x+x^2\Rightarrow x=\tfrac32\) (valid since \(1+x>0\)).
3. \(z=\dfrac32-2i\).

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Answer: (b) purely imaginary

Explanation:
1. \(|z_1+z_2|^2=|z_1|^2+|z_2|^2+2\text{Re}(z_1\bar z_2)\), so the given relation means \(\text{Re}(z_1\bar z_2)=0\).
2. Since \(z_1,z_2\ne0\), \(z_1\bar z_2\ne0\), so it is purely imaginary.

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Answer: (c) 0

Explanation:
1. \(\dfrac{z_1}{z_2}=ik\) with \(k\) a non-zero real number.
2. Then \(i\cdot\dfrac{z_1}{z_2}=i\cdot ik=-k\), a non-zero real number, whose argument is \(0\) (if \(k \lt 0\)) or \(\pi\) (if \(k>0\)). So \(0\) is a possible argument.

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Answer: (d) \(\dfrac52\sqrt2\)

Explanation:
1. \(|z-1|\le|z+i|\) means \(z\) is at least as close to \((1,0)\) as to \((0,-1)\): \((x-1)^2+y^2\le x^2+(y+1)^2\Rightarrow x+y\ge0\).
2. The disc has centre \((4,1)\) and must lie inside the half-plane \(x+y\ge0\): \(r\le\) distance from the centre to \(x+y=0\), which is \(\dfrac{5}{\sqrt2}=\dfrac52\sqrt2\).

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Answer: (d) 1

Explanation:
1. Let \(\dfrac{Z_2}{Z_1}=ik\) with \(k\) real. Then the ratio is \(\left|\dfrac{2+3ik}{2-3ik}\right|\).
2. The numerator and denominator are conjugates, so they have equal modulus and the ratio is \(1\).

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Answer: (c) \(\sqrt5\)

Explanation:
1. With \(z=x+iy\): the imaginary part gives \(y+1=0\Rightarrow y=-1\). The real part: \(x+\sqrt2\sqrt{(x+1)^2+1}=0\).
2. So \(x\le0\) and \(x^2=2(x^2+2x+2)\Rightarrow x^2+4x+4=0\Rightarrow x=-2\).
3. \(z=-2-i\), \(|z|=\sqrt5\).

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Answer: (d) \(\{z:|z|=1,\ z\ne1\}\)

Explanation:
1. Rearranging: \(w-k=(\bar w-k)z\), so \(z=\dfrac{w-k}{\bar w-k}\).
2. For real \(k\), the numerator and denominator are conjugates, so \(|z|=1\). Also \(w-k\ne\bar w-k\) as \(\text{Im}\,w\ne0\), so \(z\ne1\).
3. Every such \(z\) (\(|z|=1\), \(z\ne1\)) arises for a suitable \(k\).

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Answer: (a) 0

Explanation:
1. \(z_2=\bar z_1\) and \(z_4=\bar z_3\). The sum of arguments is the argument of the product \(\dfrac{z_1z_2}{z_3z_4}=\dfrac{|z_1|^2}{|z_3|^2}\).
2. This is a positive real number, so its argument is \(0\).

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Answer: (a) \((-1,1)\)

Explanation:
1. For \(|z|=1\): \(\dfrac{1+z^2}z=\dfrac1z+z=2\cos\theta\), so the expression is \(\dfrac{2\cos\theta}{2i}=-i\cos\theta\) and \(a=-\cos\theta\).
2. \(z\ne\pm1\) excludes \(\cos\theta=\pm1\), so \(a\in(-1,1)\).

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Answer: (b) 7

Explanation:
1. The locus is an ellipse with foci \(\pm i\): \(2a=8\Rightarrow a=4\), \(c=1\), \(b^2=15\).
2. \(|z|\) ranges from the semi-minor axis \(b=\sqrt{15}\) to the semi-major axis \(a=4\). \(m+M=4+\sqrt{15}\approx7.87\).
3. \([m+M]=7\).

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Answer: (b) 8

Explanation:
1. \(\prod_{r=1}^{8}\tan\dfrac{r\pi}{17}=\sqrt{17}\), so \(b=17\).
2. \(\sum_{r=1}^{n}\tan^2\dfrac{r\pi}{2n+1}=n(2n+1)\), so with \(n=8\): \(a=8\cdot17=136\).
3. \(\dfrac ab=\dfrac{136}{17}=8\).

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Answer: (c) 8

Explanation:
1. \(z=e^{\pm i\pi/6}\), so \(z^r+z^{-r}=2\cos\dfrac{r\pi}6\).
2. \(\sum_{r=1}^5\left(2\cos\dfrac{r\pi}6\right)^2=\sum(2+2\cos\dfrac{r\pi}3)=10+2\sum_{r=1}^5\cos\dfrac{r\pi}3\).
3. \(\sum_{r=1}^5\cos\dfrac{r\pi}3=\tfrac12-\tfrac12-1-\tfrac12+\tfrac12=-1\), so the total is \(10-2=8\).

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Answer: (b) 2

Explanation:
1. The last term is \(\dfrac{z^3}{1+z^3}\), so the sum is \(\sum_{k=1}^4\dfrac{z^k}{1+z^k}\).
2. Pair \(k\) with \(5-k\): \(\dfrac{z^k}{1+z^k}+\dfrac{z^{-k}}{1+z^{-k}}=\dfrac{z^k}{1+z^k}+\dfrac1{z^k+1}=1\).
3. Two pairs give \(2\).

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Answer: (c) 259

Explanation:
1. Equidistance: \(|f(z)-z|=|f(z)|\) for all \(z\), so \(|a+bi-1|=|a+bi|=8\).
2. \((a-1)^2+b^2=a^2+b^2=64\Rightarrow-2a+1=0\Rightarrow a=\tfrac12\), \(b^2=64-\tfrac14=\tfrac{255}4\).
3. \(m+n=255+4=259\).

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Answer: (b) 53

Explanation:
1. Let \(\zeta=\dfrac{1+\sqrt3i}2=e^{i\pi/3}\), so \(\zeta^6=1\): \(\zeta^{2016}=1,\ \zeta^{2017}=\zeta,\ \zeta^{2018}=\zeta^2=\dfrac{-1+\sqrt3i}2\).
2. \(f(\zeta)=a\zeta^2+b\zeta+c\): imaginary part \(\dfrac{\sqrt3}2(a+b)=2019\sqrt3\Rightarrow a+b=4038\). With \(a,b\le2019\): \(a=b=2019\). Real part: \(\dfrac{-a+b}2+c=2015\Rightarrow c=2015\).
3. \(f(1)=2019+2019+2015=6053\), remainder \(53\) on division by \(1000\).

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Answer: (c) 19

Explanation:
1. \(z=0\) is a solution. Otherwise, taking moduli: \(|z|^{2016}=|p|\), so \(|z|\) is fixed.
2. Arguments: \(2017\theta=\arg p-\theta+2\pi k\Rightarrow2018\theta=\arg p+2\pi k\), giving \(2018\) distinct values of \(\theta\).
3. So \(n=2018+1=2019\) and \(n-2000=19\).

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Answer: (c) 4

Explanation:
1. All four vertices are equidistant from the point \(1\), so \(1\) is the circumcentre of the square.
2. The circumcentre of a square is its centre, which is the average of the vertices: \(z_1+z_2+z_3+z_4=4\cdot1=4\).

Teacher's Note:
1. T
2. h
3. e
4.
5. c
6. e
7. n
8. t
9. r
10. e
11.
12. o
13. f
14.
15. a
16.
17. s
18. q
19. u
20. a
21. r
22. e
23.
24. i
25. s
26.
27. t
28. h
29. e
30.
31. m
32. e
33. a
34. n
35.
36. o
37. f
38.
39. i
40. t
41. s
42.
43. v
44. e
45. r
46. t
47. i
48. c
49. e
50. s
51. .

Answer: (c) 10

Explanation:
1. \(|iz+z_0|=|i|\,|z-iz_0|=|z-i(5+3i)|=|z+3-5i|\).
2. This is the distance from \(z\) to \(-3+5i\), where \(z\) lies in the disc of centre \(i\) and radius \(2\). The distance from \(-3+5i\) to \(i\) is \(|{-3}+4i|=5\).
3. Maximum \(=5+2=7\), minimum \(=5-2=3\), sum \(=10\).

Teacher's Note:
1. R
2. e
3. w
4. r
5. i
6. t
7. e
8.
9. t
10. h
11. e
12.
13. e
14. x
15. p
16. r
17. e
18. s
19. s
20. i
21. o
22. n
23.
24. a
25. s
26.
27. a
28.
29. d
30. i
31. s
32. t
33. a
34. n
35. c
36. e
37.
38. f
39. r
40. o
41. m
42.
43. a
44.
45. f
46. i
47. x
48. e
49. d
50.
51. p
52. o
53. i
54. n
55. t
56. .