Get Important questions for class 11 Maths Chapter 5 – Complex Numbers and Quadratic Equations here. Students can get different types of questions covered in this chapter. This chapter provides detailed information about the complex numbers and how to represent complex numbers in the Arg and plane. These important questions are taken from the previous year question papers. They also cover the NCERT problems, as the majority of the questions are asked from the NCERT textbook. Practice and solve all the important Maths questions from class 11 chapters at BYJU’S.
Class 11 Maths Chapter 5 – Complex Numbers and Quadratic Equations cover the following important concepts, such as:
- Quadratic Equations
- Algebra of Complex Numbers
- Modulus and Conjugate of Complex Numbers
- Argand Plane and Polar Representation
- Euler’s Formula
- De Moivre’s Theorem
- Important 1 Mark Questions for CBSE Class 11 Maths
- Important 4 Marks Questions for CBSE Class 11 Maths
- Important 6 Marks Questions for CBSE Class 11 Maths
Class 11 Chapter 5 – Complex Numbers and Quadratic Equations Important Questions with Solutions
Practice the important problems from class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations here.
Write the given complex number (1 – i) – ( –1 + i6) in the form a + ib
Given Complex number: (1 – i) – ( –1 + i6)
Multiply (-) by the term inside the second bracket ( –1 + i6)
= 1 – i +1 – i6
= 2 – 7i, which is of the form a + ib.
Express the given complex number (-3) in the polar form.
Given, complex number is -3.
Let r cos θ = -3 …(1)
and r sin θ = 0 …(2)
Squaring and adding (1) and (2), we get
r2cos2θ + r2sin2θ = (-3)2
Take r2 outside from L.H.S, we get
r2(cos2θ + sin2θ) = 9
We know that, cos2θ + sin2θ = 1, then the above equation becomes,
r2 = 9
r = 3 (Conventionally, r > 0)
Now, subsbtitute the value of r in (1) and (2)
3 cos θ = -3 and 3 sin θ = 0
cos θ = -1 and sin θ = 0
Therefore, θ = π
Hence, the polar representation is,
-3 = r cos θ + i r sin θ
3 cos π + 3 sin π = 3(cos π + i sin π)
Thus, the required polar form is 3 cos π+ 3i sin π = 3(cos π+i sin π)
Solve the given quadratic equation 2x2 + x + 1 = 0.
Given quadratic equation: 2x2 + x + 1 = 0
Now, compare the given quadratic equation with the general form ax2 + bx + c = 0
On comparing, we get
a = 2, b = 1 and c = 1
Therefore, the discriminant of the equation is:
D = b2– 4ac
Now, substitute the values in the above formula
D = (1)2 – 4(2)(1)
D = 1- 8
D = -7
Therefore, the required solution for the given quadratic equation is
x =[-b ± √D]/2a
x = [-1 ± √-7]/2(2)
We know that, √-1 = i
x = [-1 ± √7i] / 4
Hence, the solution for the given quadratic equation is (-1 ± √7i) / 4.
For any two complex numbers z1 and z2, show that Re(z1z2) = Rez1 Rez2– Imz1Imz2
Given: z1 and z2 are the two complex numbers
To prove: Re(z1z2) = Rez1 Rez2– Imz1Imz2
Let z1 = x1+iy1 and z2 = x2+iy2
Now, z1 z2 =(x1+iy1)(x2+iy2)
Now, split the real part and the imaginary part from the above equation:
⇒ x1(x2+iy2) +iy1(x2+iy2)
Now, multiply the terms:
We know that, i2 = -1, then we get
Now, again seperate the real and the imaginary part:
= (x1x2 -y1y2) +i (x1y2+x2y1)
From the above equation, take only the real part:
⇒ Re (z1z2) =(x1x2 -y1y2)
It means that,
⇒ Re(z1z2) = Rez1 Rez2– Imz1Imz2
Hence, the given statement is proved.
Find the modulus of [(1+i)/(1-i)] – [(1-i)/(1+i)]
Given: [(1+i)/(1-i)] – [(1-i)/(1+i)]
Simplify the given expression, we get:[(1+i)/(1-i)] – [(1-i)/(1+i)] = [(1+i)2– (1-i)2]/ [(1+i)(1-i)]
= (1+i2+2i-1-i2+2i)) / (12+12)
Now, cancel out the terms,
Now, take the modulus,
| [(1+i)/(1-i)] – [(1-i)/(1+i)]| =|2i| = √22 = 2
Therefore, the modulus of [(1+i)/(1-i)] – [(1-i)/(1+i)] is 2.
Practice Problems for Class 11 Maths Chapter 5
Solve the below-given problems from class 11 Maths Chapter 5:
- If |z2-1|= |z|2+1, prove that z lies on the imaginary axis.
- Compute the value of p, such that the difference of the roots of the equation is x2+px+8=0 is 2.
- Express each of the following complex numbers in the form a+ib
(i) 3(7 + i7) + i (7 + i7)
(ii) i9 +i19
- Solve the following quadratic equations:
(a) x2+3x+5 = 0
(b) x2+x+(1/√2)= 0
- Determine the real numbers x and y if (x-iy)(3+5i) is the conjugate of -6-24i.
- If arg (z – 1) = arg (z + 3i), then find (x – 1) : y, where z = x + iy.
- Find the complex number satisfying the equation z + √2 |(z + 1)| + i = 0.
- Solve the system of equations Re (z2) = 0, |z| = 2.
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