CLASS 11 MATHS CHAPTER-5 COMPLEX NUMBER AND QUADRATIC EQUATIONS

Exercise 5.1

Express each of the complex numbers given in the exercises 1 to 10 in the form a + ib:

Question1. Express each of the complex number in the form a + ib.

(5i)(−3/5i)

Solution :
 

Question2. Express each of the complex number in the form a + ib.

i⁹+i i¹⁹

Solution :
 

Question3. Express each of the complex number in the form a + ib.

i⁻³⁹

Solution :
Evaluate the complex number

i⁻³⁹ = i^4×9−3

i⁻³⁹=(i⁴)⁻⁹.i⁻³

i⁻³⁹=i⋯[i=−1]

 i⁻³⁹=i

 We get the final answer

Question4. Express each of the complex number in the form a + ib.

3(7+i7)+i(7+i7)

Solution :
Here Evaluate the complex number

3(7+i7)+i(7+i7)=21+21i+7i+7i²

3(7+i7)+i(7+i7)=21+28i+7i²⋯[i²=−1]

3(7+i7)+i(7+i7)=14+28i

We get the final answer

Question5. Express each of the complex number in the form a + ib.

(1−i)−(−1+i6)

Solution :
Here Evaluate the complex number

(1−i) − (−1+6i) = 1−i+1 − i6

(1−i)−(−1+6i) = 2−7i

We get the final answer

Question6.

Solution :

Question7.

Solution :
 

we get the answear

Question8. Express each of the complex number in the form a + ib.

(1−i)⁴

Ans. Here Evaluate the complex number

(1−i)⁴=[1+i²−2i]²

(1−i)⁴=[1−1−2i]2

(1−i)⁴ =(−2i)×(−2i)
(1−i)⁴ =−4

Question 9.Express each of the complex number in the form a + ib.

(1/3+3i)³

Solution :
 

Question10. Express each of the complex number in the form a + ib.

(−2−1/3 i)³

Solution :
 

Find the multiplicative inverse of each of the complex numbers given in the exercises 11 to 13.

Question11. Find the multiplicative inverse of each of the complex numbers given

4−3i

Solution :
 

Question12. Find the multiplicative inverse of the complex number √5 + 3i

Question13. Find the multiplicative inverse of each of the complex numbers given

−i

Solution :
Multiplicative Inverse of Let  z = −i
Then,
z¯=i  and   | z | ² = 1² =1
Therefore, the multiplicative inverse of  −i is given by z⁻¹ = z¯ | z| ²=i/1=i
Here we got final answer
 

Question14. Express the following expression in the form of a + ib

Solution :

Exercise 5.2

Find the modulus and the argument of each of the complex numbers in exercises 1 to 2.

Question1. z = – 1 – i√3

Solution :
Given:

The complex number is

z = – 1 – i√3

Let  rcosθ  = -1   and   rsin θ = –√3
Squaring and adding

(rcosθ )²+(rsinθ )²=(−1)² + (–√3)²

r²(cos²θ +sin²θ )=1+3r² = 4[cos²θ +sin²θ =1]

r= √4 = 2[   Conventionally,   r⁰ ]
Modulus  =2

2cosθ =-1   and   2sin θ = –√3cosθ =−12  and   sin θ =–√3/2
Since both the values of  sinθ   and   cos θ

Question2.

z = -√3 + i

Solution :
Given: Let  rcosθ =–√3 and   rsin θ =1

 

squaring and adding

(rcosθ )²+(rsinθ )²=(–√3)²+(−1)²

 

r²=3+1=4 [cos²θ +sin²θ =1]r = √ 4=2LLL[   Conventionally,   r⁰ ]

 

Modulus  =22cosθ = –√ 3and   2sin θ =1

 

cosθ =−√3/2  and   sin θ =12θ = π -π /6=5π/ 6LL[ As   θ   lies in the II quadrant ]

 

Thus, the modulus and argument of the complex number  −3–√+i  are   2   and   5π 6

 

Respectively,

Question3. Convert each of the complex numbers given in the polar form:

1−i

Solution :
Given: The complex number is
1−i

Let  rcosθ = 1   and   rsinθ = −1

squaring and adding

r²cos²θ+r²sin²θ=12+(−1)²⇒r²(cos²θ+sin²θ)=1+1

r²=2/r= √2[   Conventionally,   r>0]

√2cosθ  = 1   and   √2sinθ =-1cosθ  = 1/√2  and   sin θ =-1/√2 

θ =-π/ 4[   As   θ   liesin the IV quadrant   ]

1-i=rcos θ +irsin θ = √2 cos(−π /4) + i√2sin(−π /4 ) = √2 [cos(−π/4)+isin(−π /4)]

Required polar form

Question4. Convert each of the complex numbers given in the polar form:

−1+i

Solution :The complex number is

−1+i

 

Let  rcosθ =-1   and   rsin θ =1

 

Squaring and adding

r²cos²θ +r²sin²θ =(-1)²+12r²(cos²θ +sin²θ )=1+1r²=2r=  √2

 

2–√cosθ =-1   and   2–√sinθ =1

 

cosθ = -√12  and   √2sinθ = 1 /θ = π -π/ 4=3π /4L[As  θ   lies in the II quadrant]

 

It can be written,

 -1+i=rcos θ +irsin θ =2–√cos3π 4+i  √2sin3π 4 =  √2(cos3π 4+isin3π / 4)

 

Required polar form

Question5. Convert each of the complex numbers given in the polar form:

−1−i

Solution :The complex number is

−1−i

 

Let  rcosθ =-1   and   rsin θ =-1

 

Squaring and adding

r²cos²θ +r²sin²θ =(-1)²+(−1)²r²(cos²θ +sin²θ )=1+1

r²=2

r=  √2

 

 √2cosθ =-1   and   √2sinθ =-1

cosθ = -1/√2  and   sin θ =-1/√2 

θ -(π – π /4)−3π/ 4  [As  0   lies in the III quadrant]

 

 -1-i = rcos θ +irsin θ = √2cos−3π / 4 + i √2 sin − 3π / 4 = –√2(cos− 3π/ 4 + isin−3π/ 4)

 

Required polar form

Question6. Convert each of the complex numbers given in the polar form:

−3

Solution :

Question7. Convert the given complex number in polar form √3 + i3–√+i 

Solution :
 

Question8. Convert each of the complex numbers given in the polar form:

i

Solution :
 

Exercise 5.3

 

Solve each of the following equations:

Question1. Solve the equation x² + 3 = 0

Solution :
Given: Quadratic equation  x²+3=0

General form  ax²+bx+c=0

We obtain  a=1,b=0,  and   c=3

Therefore, the discriminant of the given equation is

D=b²−4ac=0²−4× 1 × 3=-12

 Therefore, the required solutions are

Question2. Solve the equation 2x² + x + 1 = 0

Solution :
Given: 2x² + x + 1 = 0

Quadratic equation  2x²+x+1=0

 

General form  ax²+bx+c=0

 

We obtain  a=2,b=1,  and   c=1

 

Therefore, the discriminant of the given equation is

, 

Question3. Solve the equation x² + 3x + 9 = 0

Solution :
Given: x² + 3x + 9 = 0

Quadratic equation  x²+3x+9=0

 

General form  ax²+bx+c=0

 

We obtain  a=1,b=3,  and   c=9

 

Therefore, the discriminant of the given equation is

D=b²−4ac=32−4× 1 × 9=-27

 

Therefore, the required solutions are

=−b ± √D / 2a=−3± −√27 / 2 × 1=−3± 3√3i / 2

Question4. Solve the equation –x² + x – 2 = 0

Solution :
 

Question5. Solve the equation x² + 3x + 5 = 0

Solution :

Question6. Solve the equation x² – x + 2 = 0

Solution :
 

Question7. √2x² + x + √2 =0

Solution :
Given: √2x² + x + √2 =0

Question8. √3x² − √2x + 3√3 =0

Solution :
Given: √3x² − √2x + 3√3 =0

Question9. x² + x +1/√2 = 0

Solution :
Given: x² + x +1/√2 = 0

 ,

Question10. x² + x/√2 + 1 =0

Solution :

Miscellaneous Exercise

Question1. Evaluate: [i¹⁸ + (1 / i)²⁵ ]³

Solution :
Given: [i¹⁸ + (1 / i)²⁵ ]³
 

Question2. For any two complex numbers z₁ and z₂, prove that
Re (z₁z₂)  =  Re z₁  Re z₂  –  Im z₁ Im z₂

Solution :
Let z₁ = x₁+  iy₁  and z₂ = x₂ + iy₂
∴z₁z₂ = (x₁+iy₁) (x₂ + iy₂)
=x1(x₂ + iy₂)  +  iy₁(x₂ + ty₂)
=x₁x₂  +  ix₁y₂  +  iy₁x₂  +  i2y₁y₂
=x₁x₂ + ix₁y₂ + iy₁x₂ − y₁y₂
=(x₁x₂  −  y₁y₂) +  i(x₂y₂  +  y₁x₂)
⇒Re(z₁ z₂)=  x₁x₂  −  y₁y₂
⇒Re (z₁ z₂)  =  Rez₁ Rez₂  −  lmz₁ Imz₂
Hence, proved.
Question3. Reduceto the standard form.

Solution :


Question4. If prove that

Solution :
 


Question5. Convert the following in the polar form:
(i) 
(ii) 

Solution :
 

This is the required polar form.

(ii) Here

Solve each of the equations in exercises 6 to 9:
Question6. 3x² -4x + 20/3 = 0

Solution :
 

Question7.  x² -2x + 3/2 = 0

Solution. Given:

Question8. Solve the equation 27x² – 10x + 1 = 0

Solution :
Given:

Question9. Solve the equation 21x² – 28x + 10 = 0

Solution :
Given:

Question10. If find 

Solution :
 


Question11. If prove that 

Solution :



Question12. Let z₁=2−i,z₂ =−2+i. Find  
(i) Re (z₁z₂/z₁),
(ii) Im (1z₁/ z₁)

Solution :
 


Question13. Find the modulus and argument of the complex number 

Solution :
 


Question14. Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of –6 – 24i.

Solution :
Let = (x−ty) (3+5i)z
=3x + 5xi − 3yi − 5yi²
=3x + 5xi − 3yi + 5y
=(3x+5y) + i(5x−3y)
∴z =(3x+5y) −i (5x−3y)
It is given that,z=−6−24i
∴(3x+5y) −i(5x−3y) =−6−24i
Equating real and imaginary parts, we obtain
3x+5y=−6
5x−3y=24
Multiplying equation (i) by 3 and equation (ii) by 5 and then adding them, we obtain
Putting the value ofxin equation (i),
we obtain 3(3)+ 5y=−6⇒ 5y=−6−15 ⇒y=−3
Thus, the values of x and y are 3 and − 3 respectlvely.
Question15. Find the modulus of 

Solution :
 

Question16. If then show that 
Ans. Given:

Question17. If  α and  β are different complex numbers with | β | = 1  then find 

Solution :
 

Question 18. Find the number of non-zero integral solutions of the equation |1−i|^x = 2^x

Solution :


Question19. If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that
(a² + b²) (c² + d²) (e² + f²) (g² + h²) = A² + B².

Solution :
 

Question20. If then find the least positive integral value of m.

Solution :