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All questions of Complex Numbers and Quadratic Equation for JEE Exam

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Imaginary part of −i(3i + 2) is
a)−2
b)2
c)3
d)−3
Correct answer is option 'A'. Can you explain this answer?

Geetika Shah answered
(-i)(3i) +2(-i) =-3(i^2)-2i =-3(-1)-2i =3-2i since i=√-1 =3+(-2)i comparing with a+bi,we get b=(-2)

If z1 and z2 are non real complex numbers such that z1+z2 and z1z2 are real numbers, then
  • a)
  • b)
  • c)
  • d)
    none of these
Correct answer is option 'A'. Can you explain this answer?

Siddharth answered
since 
Z1 + Z2 & Z1Z2 are the real numbers 

therefore

Z1 = conjugate of Z2 

 Z2 = conjugate of Z1

so  

option D is correct

according to me 

apke according kon sa option sahi h  


 

 Find the reciprocal (or multiplicative inverse) of -2 + 5i 
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'A'. Can you explain this answer?

Gaurav Kumar answered
-2 + 5i
multiplicative inverse of -2 + 5i is
1/(-2+5i)
= 1/(-2+5i) * ((-2-5i)/(-2-5i))
= -2-5i/(-2)^2 -(5i)^2
= -2-5i/4-(-25)
= -2-5i/4+25
= -2-5i/29
= -2/29 -5i/29

If the sum of n terms of an A.P. is 2n2 + 5n, then find the 4th term.
  • a)
    4n – 3
  • b)
    3n – 4       
  • c)
    4n + 3
  • d)
    3n + 4
Correct answer is option 'C'. Can you explain this answer?

Rohit Jain answered
Sum of first n terms of AP, Sn = 2n2 + 5n
 
Now choose n =1 and put in the above formula,
First term = 2+5 = 7
 
Now put n=2 to get the sum of first two terms = 2x4 + 5x 2= 8 + 10 = 18
 
This means 
 
first term + second term = 18
 
but first term =7 as calculated above
 
so, second term = 18-7 = 11
 
So common difference becomes, 11-7 = 4
 
So the AP becomes, 7, 11, 15, ....
nth term = a + (n - 1)d = 7 + (n-1)4 = 7 + 4n - 4 = 3 + 4n

Express the following in standard form : 
  • a)
    3+3i
  • b)
    2 + 2i
  • c)
    1 + 2i
  • d)
    0 + 2i
Correct answer is option 'D'. Can you explain this answer?

Gaurav Kumar answered
(8 - 4i) - (-2 - 3i) + (-10 + 3i)
=> 8 - 4i + 2 + 3i-10 + 3i
=> 8 + 2 - 10 -  4i + 3i + 3i  =>0 + 2i

The argument of the complex number -1 – √3
  • a)
    π/3
  • b)
    4π/3
  • c)
    -π/3
  • d)
    -2π/3
Correct answer is option 'D'. Can you explain this answer?

Om Desai answered
z = a + ib
a = -1, b = -√3
(-1, -√3) lies in third quadrant.
Arg(z) = -π + tan-1(b/a)
= - π + tan-1(√3)
= - π + tan-1(tan π/3)
= - π + π/3
= -2π/3

  • a)
  • b)
  • c)
  • d)
Correct answer is option 'C'. Can you explain this answer?

Riya Banerjee answered
(x+iy)(x−iy) = (a+ib)(a−ib)/(c+id)(c−id) 
⇒x2−iy2 = √[(a2−i2b2)/(c2−i2d2)]
⇒x2+y2 = √[(a2+b2)/(c2+d2)]  [1i2 = -1]
(x2+y2)2 = (a2+b2)/(c2+d2)

Find the real numbers x and y such that : (x + iy)(3 + 2i) = 1 + i
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'A'. Can you explain this answer?

Suresh Iyer answered
(x + iy)(3 + 2i) = (1 + i)
x + iy = (1 + i)/(3 + 2i)
x + iy = [(1 + i) * (3 - 2i)] / [(3 + 2i)*(3 - 2i)]
x + iy = (3 + 3i - 2i + 2) / [(3)2 + (2)2]
x + iy = (5 + i)/[ 9 + 4]
= (5 + i) / 13
=> 13x + 13iy = 5+i
13x = 5         13y = 1
x = 5/13         y = 1/13

Find the real numbers x and y such that :
a)
b)
c)
d)
Correct answer is option 'c'. Can you explain this answer?

Hansa Sharma answered
(x + iy) (3 + 2i)
= 3x + 2xi + 3iy + 3i*y = 1+i
= 3x-2y + i(2x+3y) = 1+i
= 3x-2y-1 = 0 ; 2x + 3y -1 = 0  
on equating real and imaginary parts on both sides
on solving two equations
x= 5/13 ; y = 1/13  

Let  Suppose α1 and B1are the roots of the equation x2 – 2x sec a + 1 = 0 and α2 and β2 are the roots of the equation x2 + 2x tanθ – 1 = 0. If α1 > b1 and α2 > β2, then α1 + β2 equals (JEE Adv. 2016)
  • a)
    2 (secθ – tanθ)
  • b)
    2 secθ
  • c)
    –2 tanθ
  • d)
    0
Correct answer is option 'C'. Can you explain this answer?

Krishna Iyer answered
x2 – 2x secθ + 1 = 0 ⇒ x = secθ ± tanθ and x2 + 2x tanθ – 1 = 0 ⇒ x = –tanθ ± secθ
α1,β1 are roots of x2 – 2x secθ + 1 = 0 and α1> b1
∴ α1 = secθ – tanθ and b1 = secθ + tanθ α2,
β2 are roots of x2 + 2x tanθ – 1 = 0 and α2 > β2
∴ α2 = -tanθ + secθ, β2 = – tanθ – secθ
∴ α1 + β2 = secθ – tanθ – tanθ – secθ = – 2tanθ

Express the following in standard form :
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'B'. Can you explain this answer?

Pooja Shah answered
first write above equation in complex number format , ie using iota
(3-4i) / (2-3i)*(2+3i) / (2+3i) = (6+9i-8i+12) / 13=(18/13)+(i/13)

  • a)
    10
  • b)
    4
  • c)
    8
  • d)
    6
Correct answer is option 'D'. Can you explain this answer?

Geetika Shah answered
Let √(5 – 12i) = x + iy
Squaring both sides, we get
5 – 12i = x2 + 2ixy +(iy)2 = x2 – y2 + 2xyi.
Comparing real and imaginary parts , we get
5 = x2 – y2 ———– (1) and xy = – 6 ———— (2)
Squaring (1), we get
25 = (x2 – y2)2 = (x2 + y2)2 – 4x2y2
⇒ 25 = (x2 + y2)2 – 4(– 6)2
⇒ (x2 + y2)2 = 169
⇒ x2 + y2 = 13 ———- (3)
Adding (1) and (3) we get
2x2 = 18
⇒ x = ± 3.
Subtracting (1) from (3) we get
2y2 = 8
⇒ y = ± 2.
Hence, square root of √(5 – 12i) is (3 – 2i)
Similarly, √(5 + 12i) is (3 + 2i)
√(5 + 12i) + √(5 – 12i)
⇒ (3 + 2i) + (3 - 2i)
⇒ 6

  • a)
  • b)
  • c)
  • d)
Correct answer is option 'C'. Can you explain this answer?

Hansa Sharma answered
(a + ib)1/2 = (x + iy)
Squaring both sides,
a + ib = (x + iy)2
a + ib = x2 - y2 + 2ixy
Equating real and imaginary 
a = x2 - y2    b = 2xy............(1)
Using (x2 + y2)2 = (x2 - y2)2 + 4xy 
(x2 + y2)2 = a2 + b2
(x2 + y2) = (a2 + b2)1/2.......(2)
Adding (1) and (2)
2x2 = (a2 + b2)1/2 + a
x = +-{1/2(a2 + b2)1/2 + a}1/2
Subtract (2) from (1)
2y2 = (a2 + b2)1/2 - a
y = x = +-{1/2(a2 + b2)1/2 - a}1/2
Therefore, (a+ib)1/2 = x+iy
=> +-{1/2(a2 + b2)1/2 + a}1/2 + i+-{1/2(a2 + b2)1/2 - a}1/2

Multiplicative inverse of the non zero complex number x + iy (x,y ∈ R,)
  • a)
  • b)
  • c)
    - 
  • d)
    none of these
Correct answer is option 'C'. Can you explain this answer?

Impact Learning answered
Let u be multiplicative inverse
zu = 1
u = 1/z
u = 1/(x+iy)
Rationalise it 
[1/(x+iy)]*[(x-iy)/(x-iy)]
= (x-iy)/(x2+y2)
u = x/(x2+y2) +i(-y)/(x2+y2)

For all complex numbers z1z2 such that |z1| = 12 and |z2 - 3 - 4i| = 5 the  minimum value of |z1 -z2| is
  • a)
    0
  • b)
    2
  • c)
    7
  • d)
    10
Correct answer is option 'B'. Can you explain this answer?

Ram Mohith answered
The equation |z| = 12 represents a circle with origin as center and radius equal 12. Similarly, |z - 3 - 4i| = 5 represents a circle with (3,4) as center and radius equal to 5. Note that is circle passes through origin. |z1 - z2| represents the distance between any two points one each on these two circles.

If z2/z1 is imaginary, then   is
  • a)
     2/3
  • b)
    1
  • c)
     3/2
  • d)
    5
Correct answer is option 'B'. Can you explain this answer?

Tejas Verma answered
Let   
Using Componendo and Dividenode, we get

an imaginary number


= an imaginary number

= an imaginary number

Let (x0, y0) be the solution of the following equations
(2x)ℓn2 = (3y)ℓn3
3ℓnx =2ℓny
Then x0 is (2011)
  • a)
  • b)
  • c)
  • d)
    6
Correct answer is option 'C'. Can you explain this answer?

Manish Aggarwal answered
We have (2x)ℓn2 = (3y)ℓn3
⇒ ℓn2. ℓn2x = ℓn3. ℓn3y
⇒ ℓn2. ℓn2x = ℓn3. (ℓn3 + ℓny) ...(1)
Also given 3ℓnx = 2lny
⇒ ℓnx. ℓn3 = ℓny. ℓn2 ⇒ ℓny =
Substituting this value of ℓny in equation (1), we get
ℓn2. ℓn2x = ℓn3 
⇒ (ℓn2)2 ℓn2x = (ℓn3)2 ℓn2 + (ℓn3)2 ℓnx
⇒ (ℓn2)2 ℓn2x = (ℓn3)2 (ℓn2 + ℓnx)
⇒ (ℓn2)2 ℓn2x – (ℓn3)2 ℓn2x = 0
⇒ [(ℓn2)2 – (ℓn3)2] ℓn2x = 0  ⇒ ℓn2x = 0
⇒ 2x = 1   or x = 

If the ratio of the roots of ax2 + 2bx + c = 0 is same as the ratio of the roots of px2 + 2qx + r = 0 then
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'C'. Can you explain this answer?

Aryan Khanna answered
 Let α,β be roots of ax3+bx+c=0
γ,δ be roots of px2+2qx+r=0
α/β = γ/δ …………(1) and  
β/α = δ/γ ………….(2)
(1) + (2)
⇒ α/β + β/α = γ/δ + δ/γ
= [(22+β2)/αβ + 2]= [γ2+δ2]/γδ + 2
⇒ [(α)2+(β)2+2αβ]/αβ ​= [γ2+δ2+2γδ]/γδ
​= [(α+β)2]/αβ = [(γ+δ)2]/γδ
⇒ (4b2/a2)/(c/a) = (4q2/p2)/(r/p)
⇒ b2/ac = q2/pr.

The complex numbers sin x + i cos 2x and cos x – i sin 2x are conjugate to each to other, for
  • a)
    x = nπ
  • b)
    x = 0
  • c)
    x = (n + 1/2)π
  • d)
    No value of x
Correct answer is option 'D'. Can you explain this answer?

Rocky Gupta answered
Option (a) :- for x = nπ
let n=1
so, x= 1 x π =π
therefore , sinx + icos2x
sinπ + icos 2π
0 + i= i
Now,
cosx - isin2x
cosπ - isin2π
or, -1 -0 = -1
hence no conjugate
Option (b) :- x = 0
sin0 + icos0
0+i= i
now ,
cos0 - isin0
= 1
hence no conjugate.
option (c) :- x = (n+1/2)π
if n= 1
x=3π/2
sin3π/2 + icos3π
-1 - i
Now,
cos3π/2 - isin3π
0-0= 0
hence no conjugate
Thus option D is correct

  • a)
    - i
  • b)
    1
  • c)
    i
  • d)
    -1
Correct answer is option 'C'. Can you explain this answer?

Raghav Bansal answered
x = (√3+i)/2 
x3 = 1/8(√3+i)3
Apply formula (a+b)3 = a3 + b3 + 3a2b + 3ab2
= (3√3 + i3 + 3*3*i + 3*√3*i2)/8 
= (3√3 - i + 9i - 3√3)/8 
= 8i/8
= i

Number of rational roots of the equation 
  • a)
    1
  • b)
    2
  • c)
    3
  • d)
    4
Correct answer is option 'A'. Can you explain this answer?

Nabanita Choudhury answered
Understanding the Equation
The equation we need to solve is
|x^2 - 2x - 3| + 4x = 0.
To find the number of rational roots, we must analyze the expression within the absolute value.
Step 1: Break Down the Absolute Value
The expression x^2 - 2x - 3 can be factored as:
(x - 3)(x + 1).
Thus, we identify the points where the expression changes sign:
- The roots are x = 3 and x = -1.
This gives us three intervals to consider:
1. x < />
2. -1 ≤ x < />
3. x ≥ 3
Step 2: Analyze Each Interval
- For x < />:
The expression is negative, so we have:
- (-(x^2 - 2x - 3)) + 4x = 0
- -x^2 + 2x + 3 + 4x = 0
- -x^2 + 6x + 3 = 0
- x^2 - 6x - 3 = 0 (after multiplying by -1).
The discriminant (D = b^2 - 4ac) is positive, indicating 2 rational roots.
- For -1 ≤ x < />:
Here, the expression is positive, so:
- (x^2 - 2x - 3) + 4x = 0
- x^2 + 2x - 3 = 0.
The discriminant is also positive, yielding 2 rational roots.
- For x ≥ 3:
The expression remains positive:
- (x^2 - 2x - 3) + 4x = 0, leading to the same equation as above.
Step 3: Conclusion
Combining all intervals, we find that the equation has:
1. 2 rational roots from x < />
2. 2 rational roots from -1 ≤ x < />
3. No new roots from x ≥ 3.
However, since the roots in the first and second intervals do not overlap and considering the behavior at the transition points, we conclude that there is only 1 unique rational root across all intervals.
Final Answer
The number of rational roots of the equation is 1 (Option A).

  • a)
    9p/10
  • b)
    5p/6
  • c)
    6p/5
  • d)
    none of these
Correct answer is option 'A'. Can you explain this answer?

Neha Joshi answered
sin(6π/5) + i(1+cos(6π/5))  or −sin(π/5) + i(1−cos(π/5)) lies in the second quadrant of complex plane hence its argument is given as
arg(x+iy) = π − tan-1 |y/x| (∀ x<0,y≥0)
= π−tan-1 |1−cos(π/5)/sin(π/5)|
= π−tan-1  |2sin2(π/10)/2sin(π/10)cos(π/10)|
= π−tan-1 |sin(π/10)/cos(π/10)|
= π − tan-1 |tan(π/10)|
= π−tan-1 (tan(π/10))  (∵ tan(π/10)>0)
= π−π/10(∵−π/2≤tan−1(x)≤π/2)
= 9π/10
 

If points corresponding to the complex numbers z1, z2, z3 and z4 are the vertices of a rhombus, taken in order, then for a non-zero real number k
  • a)
    z1 – z3 = i k( z2 –z4)
  • b)
    z1 – z2 = i k( z3 –z4)
  • c)
    z1 + z3 = k( z2 +z4)
  • d)
    z1 + z2 = k( z3 +z4)
Correct answer is option 'A'. Can you explain this answer?

Riya Banerjee answered
AC = z3 = z1 eiπ
= z1 (cosπ + i sinπ)
= z3 = z1(-1 + i(0))
= z3 = -z1
AC = z1 - z3
BC = z2 - z4
(z1 - z3)/(z2 - z4) = k
(z1 - z3) = eiπ/2(z2 - z4)
(z1 - z3) k(cosπ/2 + sinπ/2) (z2 - z4)
z1 - z3 = ki(z2 - z4)
z1 - z3 = ik(z2 - z4)
 

The sum of all real values of x satisfying the equation (x2 - 5 x+ 5) x2 +4x- 60 = 1 is : [JEE M 2016]
  • a)
    6
  • b)
    5
  • c)
    3
  • d)
    – 4
Correct answer is option 'C'. Can you explain this answer?

Ananya Das answered
(x2 - 5 x + 5) x2 + 4x-60=1
Case I x2 – 5x + 5 = 1 and
x2 + 4x – 60 can be any real number
⇒ x = 1, 4
Case II x2 – 5x + 5 = –1 and
x2 + 4x – 60 has to be an even number
⇒ x = 2, 3 where 3 is rejected because for x = 3,
x2 + 4x – 60 is odd.
Case III x2 – 5x + 5 can be any real number and
x2 + 4x – 60 = 0
⇒ x = –10, 6
⇒ Sum of all values of x = –10 + 6 + 2 + 1 + 4 = 3
 

If | z + 4 | ≤ 3, then the maximum value of | z + 1 | is[2007]
  • a)
    6
  • b)
    0
  • c)
    4
  • d)
    10
Correct answer is option 'A'. Can you explain this answer?

Ujjwal Mishra answered
There are two cases of |z+4| one is negative and the other one is positive
if we take negative term the value becomes -(z+4)≤3
z=-7
then if we put the value of z in second equation we get |-7+1|
=|-6|
=6

Chapter doubts & questions for Complex Numbers and Quadratic Equation - Chapter-wise Tests for JEE Main & Advanced 2025 is part of JEE exam preparation. The chapters have been prepared according to the JEE exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for JEE 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.

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