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

2023 · 24 Jan · Shift 1 · Q24
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. /2023 · 24 Jan · Shift 1 · Q24

Complex Numbers question

2023 · 24 Jan · Shift 1 · Q24

JEE MainMathematicsComplex NumbersMCQ+4 / −1
Let p,q∈R\mathrm{p,q\in\mathbb{R}}p,q∈R and (1−3i)200=2199(p+iq),i=−1{\left( {1 - \sqrt 3 i} \right)^{200}} = {2^{199}}(p + iq),i = \sqrt { - 1}(1−3​i)200=2199(p+iq),i=−1​ then p+q+q2\mathrm{p+q+q^2}p+q+q2 and p−q+q2\mathrm{p-q+q^2}p−q+q2 are roots of the equation.
  1. A
    x2+4x−1=0{x^2} + 4x - 1 = 0x2+4x−1=0
  2. B
    x2−4x+1=0{x^2} - 4x + 1 = 0x2−4x+1=0
  3. C
    x2+4x+1=0{x^2} + 4x + 1 = 0x2+4x+1=0
  4. D
    x2−4x−1=0{x^2} - 4x - 1 = 0x2−4x−1=0
View written solutionFree

Correct answer: B

  1. Write the complex number in polar form

Given 1−3i1-\sqrt{3}i1−3​i its modulus is r=12+(3)2=4=2.r=\sqrt{1^2+(\sqrt{3})^2}=\sqrt{4}=2.r=12+(3​)2​=4​=2.

Its argument is in the fourth quadrant, with tan⁡θ=−31=−3,\tan\theta=\frac{-\sqrt{3}}{1}=-\sqrt{3},tanθ=1−3​​=−3​, so θ=−π3.\theta=-\frac{\pi}{3}.θ=−3π​.

Thus, 1−3i=2(cos⁡(−π3)+isin⁡(−π3)).1-\sqrt{3}i=2\left(\cos\left(-\frac{\pi}{3}\right)+i\sin\left(-\frac{\pi}{3}\right)\right).1−3​i=2(cos(−3π​)+isin(−3π​)).

  1. Raise to the power 200

Using De Moivre's theorem, (1−3i)200=2200(cos⁡(−200π3)+isin⁡(−200π3)).\left(1-\sqrt{3}i\right)^{200}=2^{200}\left(\cos\left(-\frac{200\pi}{3}\right)+i\sin\left(-\frac{200\pi}{3}\right)\right).(1−3​i)200=2200(cos(−3200π​)+isin(−3200π​)).

Given (1−3i)200=2199(p+iq),\left(1-\sqrt{3}i\right)^{200}=2^{199}(p+iq),(1−3​i)200=2199(p+iq), so p+iq=2(cos⁡(−200π3)+isin⁡(−200π3)).p+iq=2\left(\cos\left(-\frac{200\pi}{3}\right)+i\sin\left(-\frac{200\pi}{3}\right)\right).p+iq=2(cos(−3200π​)+isin(−3200π​)).

  1. Simplify the angle

Reduce −200π3-\frac{200\pi}{3}−3200π​ modulo 2π2\pi2π.

Since 200≡2(mod6),200\equiv 2 \pmod{6},200≡2(mod6), we get −200π3≡−2π3(mod2π).-\frac{200\pi}{3}\equiv -\frac{2\pi}{3} \pmod{2\pi}.−3200π​≡−32π​(mod2π).

Hence, p+iq=2(cos⁡(−2π3)+isin⁡(−2π3)).p+iq=2\left(\cos\left(-\frac{2\pi}{3}\right)+i\sin\left(-\frac{2\pi}{3}\right)\right).p+iq=2(cos(−32π​)+isin(−32π​)).

Now, cos⁡(−2π3)=cos⁡(2π3)=−12,\cos\left(-\frac{2\pi}{3}\right)=\cos\left(\frac{2\pi}{3}\right)=-\frac12,cos(−32π​)=cos(32π​)=−21​, sin⁡(−2π3)=−sin⁡(2π3)=−32.\sin\left(-\frac{2\pi}{3}\right)=-\sin\left(\frac{2\pi}{3}\right)=-\frac{\sqrt3}{2}.sin(−32π​)=−sin(32π​)=−23​​.

Therefore, p+iq=2(−12−i32)=−1−3i.p+iq=2\left(-\frac12-i\frac{\sqrt3}{2}\right)=-1-\sqrt3 i.p+iq=2(−21​−i23​​)=−1−3​i.

So, p=−1,q=−3.p=-1,\qquad q=-\sqrt3.p=−1,q=−3​.

  1. Find the two roots mentioned

First root: p+q+q2=−1−3+(−3)2=−1−3+3=2−3.p+q+q^2=-1-\sqrt3+(-\sqrt3)^2=-1-\sqrt3+3=2-\sqrt3.p+q+q2=−1−3​+(−3​)2=−1−3​+3=2−3​.

Second root: p−q+q2=−1+3+3=2+3.p-q+q^2=-1+\sqrt3+3=2+\sqrt3.p−q+q2=−1+3​+3=2+3​.

Thus the roots are 2−3and2+3.2-\sqrt3 \quad \text{and} \quad 2+\sqrt3.2−3​and2+3​.

  1. Form the quadratic equation

For roots α=2−3\alpha=2-\sqrt3α=2−3​ and β=2+3\beta=2+\sqrt3β=2+3​,

Sum: α+β=(2−3)+(2+3)=4.\alpha+\beta=(2-\sqrt3)+(2+\sqrt3)=4.α+β=(2−3​)+(2+3​)=4.

Product: αβ=(2−3)(2+3)=4−3=1.\alpha\beta=(2-\sqrt3)(2+\sqrt3)=4-3=1.αβ=(2−3​)(2+3​)=4−3=1.

Hence the quadratic is x2−(α+β)x+αβ=0,x^2-(\alpha+\beta)x+\alpha\beta=0,x2−(α+β)x+αβ=0, that is, x2−4x+1=0.x^2-4x+1=0.x2−4x+1=0.

  1. Match with the options

This is Option B.

PreviousNext

More from Complex Numbers

  • The value of (1+sin92π​−icos92π​1+sin92π​+icos92π​​)3 is2023 · MCQ
  • Let z1​=2+3i and z2​=3+4i. The set S={z∈C:∣z−z1​∣2−∣z−z2​∣2=∣z1​−z2​∣2} represents a2023 · MCQ
  • Let z be a complex number such that ​z+iz−2i​​=2,ze−i. Then z lies on the circle of radius 2 and centre :2023 · MCQ
  • For two non-zero complex numbers z1​ and z2​, if Re(z1​z2​)=0 and Re(z1​+z2​)=0, then which of the following are possible? A. Im(z1​)>0…2023 · MCQ
  • Let α=8−14i,A={z∈c:z2−(z)2−112iαz−αz​=1} and B={z∈c:∣z+3i∣=4}. Then z∈A∩B∑​(Reolimitsz−Imolimitsz)…2023 · Numerical
  • Let z=1+i and z1​=zˉ(1−z)+z1​1+izˉ​. Then π12​arg(z1​) is equal to ​.2023 · Numerical
  • For all z∈C on the curve C1​:∣z∣=4, let the locus of the point z+z1​ be the curve C2​. Then :2023 · MCQ
  • The complex number z=cos3π​+isin3π​i−1​ is equal to :2023 · MCQ