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 Argand 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 BYJUâ€™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
- Eulerâ€™s Formula
- De Moivreâ€™s Theorem

**Also, Check: **

- 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.

**Question 1: **

Write the given complex number (1 â€“ i) â€“ ( â€“1 + i6) in the form a + ib

**Solution:**

Given Complex number: (1 â€“ i) â€“ ( â€“1 + i6)

Multiply (-) by the term inside the second bracket ( â€“1 + i6)

= 1 â€“ i +1 – i6

= 2 – 7i, which is of the form a + ib.

**Question 2: **

Express the given complex number (-3) in the polar form.

**Solution: **

Given, complex number is -3.

Let r cos Î¸ = -3 â€¦(1)

and r sin Î¸ = 0 â€¦(2)

Squaring and adding (1) and (2), we get

r^{2}cos^{2}Î¸ + r^{2}sin^{2}Î¸ = (-3)^{2}

Take r^{2 }outside from L.H.S, we get

r^{2}(cos^{2}Î¸ + sin^{2}Î¸) = 9

We know that, cos^{2}Î¸ + sin^{2}Î¸ = 1, then the above equation becomes,

r^{2} = 9

r = 3 (Conventionally, r > 0)

Now, subsbtitute the value of r in (1) and (2)

3 cos Î¸ = -3 and 3 sin Î¸ = 0

cos Î¸ = -1 and sin Î¸ = 0

Therefore, Î¸ = Ï€

Hence, the polar representation is,

-3 = r cos Î¸ + i r sin Î¸

3 cos Ï€ + 3 sin Ï€ = 3(cos Ï€ + i sin Ï€)

Thus, the required polar form is 3 cos Ï€+ 3i sin Ï€ = 3(cos Ï€+i sin Ï€)

**Question 3:**

Solve the given quadratic equation 2x^{2Â }+ x + 1 = 0.

**Solution:**

Given quadratic equation: 2x^{2Â }+ x + 1 = 0

Now, compare the given quadratic equation with the general form ax^{2Â }+ bx + c = 0

On comparing, we get

a = 2, b = 1 and c = 1

Therefore, the discriminant of the equation is:

D = b^{2}– 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 Â± âˆšD]/2a

x = [-1 Â± âˆš-7]/2(2)

We know that, âˆš-1 = i

x = [-1 Â± âˆš7i] / 4

Hence, the solution for the given quadratic equation is (-1 Â± âˆš7i) / 4.

**Question 4:**

For any two complex numbers z_{1} and z_{2}, show that Re(z_{1}z_{2}) = Rez_{1} Rez_{2}– Imz_{1}Imz_{2}

**Solution:**

Given: z_{1} and z_{2} are the two complex numbers

To prove: Re(z_{1}z_{2}) = Rez_{1} Rez_{2}– Imz_{1}Imz_{2}

Let z_{1} = x_{1}+iy_{1} and z_{2} = x_{2}+iy_{2}

Now, z_{1} z_{2} =(x_{1}+iy_{1})(x_{2}+iy_{2})

Now, split the real part and the imaginary part from the above equation:

â‡’ x_{1}(x_{2}+iy_{2}) +iy_{1}(x_{2}+iy_{2})

Now, multiply the terms:

= x_{1}x_{2}+ix_{1}y_{2}+ix_{2}y_{1}+i^{2}y_{1}y_{2}

We know that, i^{2} = -1, then we get

= x_{1}x_{2}+ix_{1}y_{2}+ix_{2}y_{1}-y_{1}y_{2}

Now, again seperate the real and the imaginary part:

= (x_{1}x_{2} -y_{1}y_{2}) +i (x_{1}y_{2}+x_{2}y_{1})

From the above equation, take only the real part:

â‡’ Re (z_{1}z_{2}) =(x_{1}x_{2} -y_{1}y_{2})

It means that,

â‡’ Re(z_{1}z_{2}) = Rez_{1} Rez_{2}– Imz_{1}Imz_{2}

Hence, the given statement is proved.

**Question 5:**

Find the modulus of [(1+i)/(1-i)] – [(1-i)/(1+i)]

**Solution:**

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+i^{2}+2i-1-i^{2}+2i)) / (1^{2}+1^{2})

Now, cancel out the terms,

= 4i/2

= 2i

Now, take the modulus,

| [(1+i)/(1-i)] – [(1-i)/(1+i)]| =|2i| = âˆš2^{2} = 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 |z
^{2}-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 x
^{2}+px+8=0 is 2. - Express each of the following complex numbers in the form a+ib
- 3(7 + i7) + i (7 + i7)
- i
^{9}+i^{19} - [(â…“)+3i]
^{3}

- Solve the following quadratic equations:

Â Â Â Â Â Â Â (a) x^{2}+3x+5 = 0

Â Â Â Â Â Â Â (b) x^{2}+x+(1/âˆš2)= 0

Â Â Â 5. Determine the real numbers x and y if (x-iy)(3+5i) is the conjugate of -6-24i.

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