Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Complex Numbers question

2003 · Shift 0 · Q99
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Complex Numbers
  5. /2003 · Shift 0 · Q99

Complex Numbers question

2003 · Shift 0 · Q99

JEE MainMathematicsComplex NumbersMCQ+4 / −1
If (1+i1−i)x=1{\left( {{{1 + i} \over {1 - i}}} \right)^x} = 1(1−i1+i​)x=1 then :
  1. A
    x = 2n + 1, where n is any positive integer
  2. B
    x = 4n , where n is any positive integer
  3. C
    x = 2n, where n is any positive integer
  4. D
    x = 4n + 1, where n is any positive integer.
View written solutionFree

Correct answer: B

  1. Simplify the complex number

We need to solve:

(1+i1−i)x=1\left(\frac{1+i}{1-i}\right)^x = 1(1−i1+i​)x=1

First, simplify the base:

1+i1−i⋅1+i1+i=(1+i)21−i2\frac{1+i}{1-i} \cdot \frac{1+i}{1+i} = \frac{(1+i)^2}{1-i^2}1−i1+i​⋅1+i1+i​=1−i2(1+i)2​

Since i2=−1i^2=-1i2=−1,

1−i2=1−(−1)=21-i^2 = 1-(-1)=21−i2=1−(−1)=2

and

(1+i)2=1+2i+i2=1+2i−1=2i(1+i)^2 = 1+2i+i^2 = 1+2i-1=2i(1+i)2=1+2i+i2=1+2i−1=2i

So,

1+i1−i=2i2=i\frac{1+i}{1-i} = \frac{2i}{2}=i1−i1+i​=22i​=i

Thus the equation becomes:

ix=1i^x = 1ix=1

  1. Use powers of iii

The powers of iii repeat with period 444:

i1=i,i2=−1,i3=−i,i4=1i^1=i,\quad i^2=-1,\quad i^3=-i,\quad i^4=1i1=i,i2=−1,i3=−i,i4=1

Therefore,

ix=1  ⟺  x is a multiple of 4i^x=1 \iff x \text{ is a multiple of } 4ix=1⟺x is a multiple of 4

So,

x=4nx=4nx=4n

where nnn is a positive integer.

  1. Check options
  • A: x=2n+1x=2n+1x=2n+1 gives odd numbers, not always multiples of 444 ❌
  • B: x=4nx=4nx=4n gives multiples of 444 ✅
  • C: x=2nx=2nx=2n gives even numbers, but not all are multiples of 444 ❌
  • D: x=4n+1x=4n+1x=4n+1 gives remainder 111 modulo 444 ❌
  1. Final answer

The correct option is:

B\boxed{\text{B}}B​

PreviousNext

More from Complex Numbers

  • z and w are two nonzero complex numbers such that ∣z∣=∣w∣ and Arg z + Arg w =π then z equals2002 · MCQ
  • If ∣z−4∣<∣z−2∣, its solution is given by :2002 · MCQ
  • The locus of the centre of a circle which touches the circle ∣z−z1​∣=a and ∣z−z2​∣=b externally (z,z1​&z2​ are complex numbers) will be :2002 · MCQ
  • Let z be a complex number such that ∣z∣=1. If k+zˉ2+k2z​=kz,k∈R, then the maximum distance of k+ik2 from the circle ∣z−(1+2i)∣=1 is :2025 · MCQ
  • Let z∈C be such that z−2+iz2+3i​=2+3i. Then the sum of all possible values of z2 is :2025 · MCQ
  • If z1​,z2​,z3​∈C are the vertices of an equilateral triangle, whose centroid is z0​, then k=1∑3​(zk​−z0​)2 is equal to2025 · MCQ
  • Let A={z∈C:∣z−2−i∣=3},B={z∈C:Re(z−iz)=2} and S=A∩B. Then ∑z∈S​∣z∣2 is equal to ​.2025 · Numerical
  • Let the product of ω1​=(8+i)sinθ+(7+4i)cosθ and ω2​=(1+8i)sinθ+(4+7i)cosθ be α+iβ, i=−1​. Let p and q be the maximum and the minimum values of α+β…2025 · MCQ