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

2011 · Shift 0 · Q33
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. /2011 · Shift 0 · Q33

Complex Numbers question

2011 · Shift 0 · Q33

JEE MainMathematicsComplex NumbersMCQ+4 / −1
Let α ,β\alpha \,,\betaα,β be real and z be a complex number. If z2+αz+β=0{z^2} + \alpha z + \beta = 0z2+αz+β=0 has two distinct roots on the line Re z = 1, then it is necessary that :
  1. A
    β ∈(−1,0)\beta \, \in ( - 1,0)β∈(−1,0)
  2. B
    ∣β ∣=1\left| {\beta \,} \right| = 1∣β∣=1
  3. C
    β ∈(1,∞)\beta \, \in (1,\infty )β∈(1,∞)
  4. D
    β ∈(0,1)\beta \, \in (0,1)β∈(0,1)
View written solutionFree

Correct answer: C

  1. Roots lie on the line Re⁡(z)=1\operatorname{Re}(z)=1Re(z)=1

Since the quadratic z2+αz+β=0z^2+\alpha z+\beta=0z2+αz+β=0 has real coefficients (α,β∈R\alpha,\beta\in\mathbb Rα,β∈R), any non-real roots occur in conjugate pairs.

If both distinct roots lie on the line Re⁡(z)=1\operatorname{Re}(z)=1Re(z)=1, they must be of the form z1=1+it,z2=1−itz_1=1+it,\qquad z_2=1-itz1​=1+it,z2​=1−it for some real t≠0t\neq 0t=0.

The condition "two distinct roots" gives t≠0.t\neq 0.t=0.

  1. Use Vieta's formulas

For the quadratic z2+αz+β=0,z^2+\alpha z+\beta=0,z2+αz+β=0, we have z1+z2=−α,z1z2=β.z_1+z_2=-\alpha,\qquad z_1z_2=\beta.z1​+z2​=−α,z1​z2​=β.

Now, z1+z2=(1+it)+(1−it)=2.z_1+z_2=(1+it)+(1-it)=2.z1​+z2​=(1+it)+(1−it)=2. So, −α=2⇒α=−2.-\alpha=2\quad\Rightarrow\quad \alpha=-2.−α=2⇒α=−2.

Also, z1z2=(1+it)(1−it)=1+t2.z_1z_2=(1+it)(1-it)=1+t^2.z1​z2​=(1+it)(1−it)=1+t2. Thus, β=1+t2.\beta=1+t^2.β=1+t2.

Since t≠0t\neq 0t=0, we get t2>0⇒β=1+t2>1.t^2>0\quad\Rightarrow\quad \beta=1+t^2>1.t2>0⇒β=1+t2>1.

Hence, β∈(1,∞).\beta\in(1,\infty).β∈(1,∞).

  1. Check the options
  • A: β∈(−1,0)\beta\in(-1,0)β∈(−1,0) — false
  • B: ∣β∣=1|\beta|=1∣β∣=1 — false, since β>1\beta>1β>1
  • C: β∈(1,∞)\beta\in(1,\infty)β∈(1,∞) — true
  • D: β∈(0,1)\beta\in(0,1)β∈(0,1) — false
  1. Conclusion

The necessary condition is β∈(1,∞).\boxed{\beta\in(1,\infty)}.β∈(1,∞)​.

PreviousNext

More from Complex Numbers

  • If ω(e1) is a cube root of unity, and (1+ω)7=A+Bω. Then (A,B) equals :2011 · MCQ
  • The number of complex numbers z such that ∣z−1∣=∣z+1∣=∣z−i∣ equals :2010 · MCQ
  • If ​z−z4​​=2, then the maximum value of ∣z∣ is equal to :2009 · MCQ
  • The conjugate of a complex number is i−11​ then that complex number is :2008 · MCQ
  • If ∣z+4∣≤3, then the maximum value of ∣z+1∣ is :2007 · MCQ
  • If z2+z+1=0, where z is complex number, then value of (z+z1​)2+(z2+z21​)2+(z3+z31​)2+..........+(z6+z61​)2…2006 · MCQ
  • The value of k=1∑10​(sin112kπ​+icos112kπ​) is :2006 · MCQ
  • If the cube roots of unity are 1, ω,ω2 then the roots of the equation (x−1)3 + 8 = 0, are :2005 · MCQ