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Complex Numbers and Quadratic Equations · Question 22

JEE Main 2026 (Online) 24th January Morning ShiftMCQMedium📈 Topic in 45% papers+4−1
1

Let S = \z C : | z-6iz-2i | = 1 and | z-8+2iz+2i | = 35 \. Then z S |z|2 is equal to :

  1. 413
  2. 398
  3. 385
  4. 423

Answer

(3) 385

Solution

| z-6iz-2i | = 1 |z-6i| = |z-2i|

z lies on the perpendicular bisector of the points 6i and 2i, i.e. on the line y = 4

Let z = x + 4i.

z-8+2i = (x-8) + 6i and z+2i = x + 6i

5|z-8+2i| = 3|z+2i| 25 [(x-8)2 + 36 ] = 9 (x2 + 36 )

16x2 - 400x + 2176 = 0

x2 - 25x + 136 = 0

(x-8)(x-17) = 0

z = 8 + 4i or z = 17 + 4i

z S |z|2 = (64+16) + (289+16)

= 80 + 305 = 385

Answer: (3)

JEE Main 2026 (Online) 24th January Morning Shift

Keep practising — Complex Numbers and Quadratic Equations

2

The smallest positive integral value of a, for which all the roots of x4 - ax2 + 9 = 0 are real and distinct, is equal to

  1. 7
  2. 3
  3. 4
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3

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4

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5

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6

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  1. 36 days
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  4. 42 days
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7

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  3. -1
  4. 256
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8

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  1. 2
  2. 3
  3. 0
  4. 1
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9

Let = -1 + i 32 and = -1 - i 32, i = -1. If (7 - 7 + 9 )20 + (9 + 7 - 7 )20 + (-7 + 9 + 7 )20 + (14 + 7 + 7 )20 = m10, then m is

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10

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  1. 3
  2. 6
  3. 1
  4. 4
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11

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