chapter 5 complex numbers and quadratic...
TRANSCRIPT
Class XI Chapter 5 – Complex Numbers and Quadratic Equations Maths
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Exercise 5.1
Question 1:
Express the given complex number in the form a + ib:
Answer
Question 2:
Express the given complex number in the form a + ib: i9 + i19
Answer
Question 3:
Express the given complex number in the form a + ib: i–39
Answer
Class XI Chapter 5 – Complex Numbers and Quadratic Equations Maths
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Question 4:
Express the given complex number in the form a + ib: 3(7 + i7) + i(7 + i7)
Answer
Question 5:
Express the given complex number in the form a + ib: (1 – i) – (–1 + i6)
Answer
Question 6:
Express the given complex number in the form a + ib:
Answer
Question 7:
Express the given complex number in the form a + ib:
Answer
Class XI Chapter 5 – Complex Numbers and Quadratic Equations Maths
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Question 8:
Express the given complex number in the form a + ib: (1 – i)4
Answer
Question 9:
Express the given complex number in the form a + ib:
Answer
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Question 10:
Express the given complex number in the form a + ib:
Answer
Question 11:
Find the multiplicative inverse of the complex number 4 – 3i
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Answer
Let z = 4 – 3i
Then, = 4 + 3i and
Therefore, the multiplicative inverse of 4 – 3i is given by
Question 12:
Find the multiplicative inverse of the complex number
Answer
Let z =
Therefore, the multiplicative inverse of is given by
Question 13:
Find the multiplicative inverse of the complex number –i
Answer
Let z = –i
Therefore, the multiplicative inverse of –i is given by
Question 14:
Class XI Chapter 5 – Complex Numbers and Quadratic Equations Maths
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Express the following expression in the form of a + ib.
Answer
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Exercise 5.2
Question 1:
Find the modulus and the argument of the complex number
Answer
On squaring and adding, we obtain
Since both the values of sin θ and cos θ are negative and sinθ and cosθ are negative in
III quadrant,
Thus, the modulus and argument of the complex number are 2 and
respectively.
Question 2:
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Find the modulus and the argument of the complex number
Answer
On squaring and adding, we obtain
Thus, the modulus and argument of the complex number are 2 and
respectively.
Question 3:
Convert the given complex number in polar form: 1 – i
Answer
1 – i
Let r cos θ = 1 and r sin θ = –1
On squaring and adding, we obtain
Class XI Chapter 5 – Complex Numbers and Quadratic Equations Maths
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This is
the required polar form.
Question 4:
Convert the given complex number in polar form: – 1 + i
Answer
– 1 + i
Let r cos θ = –1 and r sin θ = 1
On squaring and adding, we obtain
It can be written,
Class XI Chapter 5 – Complex Numbers and Quadratic Equations Maths
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This is the required polar form.
Question 5:
Convert the given complex number in polar form: – 1 – i
Answer
– 1 – i
Let r cos θ = –1 and r sin θ = –1
On squaring and adding, we obtain
This is the
required polar form.
Question 6:
Convert the given complex number in polar form: –3
Answer
–3
Let r cos θ = –3 and r sin θ = 0
On squaring and adding, we obtain
Class XI Chapter 5 – Complex Numbers and Quadratic Equations Maths
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This is the required polar form.
Question 7:
Convert the given complex number in polar form:
Answer
Let r cos θ = and r sin θ = 1
On squaring and adding, we obtain
Class XI Chapter 5 – Complex Numbers and Quadratic Equations Maths
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This is the required polar form.
Question 8:
Convert the given complex number in polar form: i
Answer
i
Let r cosθ = 0 and r sin θ = 1
On squaring and adding, we obtain
This is the required polar form.
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Exercise 5.3
Question 1:
Solve the equation x2 + 3 = 0
Answer
The given quadratic equation is x2 + 3 = 0
On comparing the given equation with 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
Question 2:
Solve the equation 2x2 + x + 1 = 0
Answer
The given quadratic equation is 2x2 + x + 1 = 0
On comparing the given equation with ax2 + bx + c = 0, we obtain
a = 2, b = 1, and c = 1
Therefore, the discriminant of the given equation is
D = b2 – 4ac = 12 – 4 × 2 × 1 = 1 – 8 = –7
Therefore, the required solutions are
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Question 3:
Solve the equation x2 + 3x + 9 = 0
Answer
The given quadratic equation is x2 + 3x + 9 = 0
On comparing the given equation with 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 = 9 – 36 = –27
Therefore, the required solutions are
Question 4:
Solve the equation –x2 + x – 2 = 0
Answer
The given quadratic equation is –x2 + x – 2 = 0
On comparing the given equation with ax2 + bx + c = 0, we obtain
a = –1, b = 1, and c = –2
Therefore, the discriminant of the given equation is
D = b2 – 4ac = 12 – 4 × (–1) × (–2) = 1 – 8 = –7
Therefore, the required solutions are
Question 5:
Solve the equation x2 + 3x + 5 = 0
Answer
The given quadratic equation is x2 + 3x + 5 = 0
On comparing the given equation with ax2 + bx + c = 0, we obtain
Class XI Chapter 5 – Complex Numbers and Quadratic Equations Maths
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a = 1, b = 3, and c = 5
Therefore, the discriminant of the given equation is
D = b2 – 4ac = 32 – 4 × 1 × 5 =9 – 20 = –11
Therefore, the required solutions are
Question 6:
Solve the equation x2 – x + 2 = 0
Answer
The given quadratic equation is x2 – x + 2 = 0
On comparing the given equation with ax2 + bx + c = 0, we obtain
a = 1, b = –1, and c = 2
Therefore, the discriminant of the given equation is
D = b2 – 4ac = (–1)2 – 4 × 1 × 2 = 1 – 8 = –7
Therefore, the required solutions are
Question 7:
Solve the equation
Answer
The given quadratic equation is
On comparing the given equation with ax2 + bx + c = 0, we obtain
a = , b = 1, and c =
Therefore, the discriminant of the given equation is
D = b2 – 4ac = 12 – = 1 – 8 = –7
Therefore, the required solutions are
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Question 8:
Solve the equation
Answer
The given quadratic equation is
On comparing the given equation with ax2 + bx + c = 0, we obtain
a = , b = , and c =
Therefore, the discriminant of the given equation is
D = b2 – 4ac =
Therefore, the required solutions are
Question 9:
Solve the equation
Answer
The given quadratic equation is
This equation can also be written as
On comparing this equation with ax2 + bx + c = 0, we obtain
a = , b = , and c = 1
Therefore, the required solutions are
Class XI Chapter 5 – Complex Numbers and Quadratic Equations Maths
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Question 10:
Solve the equation
Answer
The given quadratic equation is
This equation can also be written as
On comparing this equation with ax2 + bx + c = 0, we obtain
a = , b = 1, and c =
Therefore, the required solutions are
Class XI Chapter 5 – Complex Numbers and Quadratic Equations Maths
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NCERT Miscellaneous Solutions
Question 1:
Evaluate:
Answer
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Question 2:
For any two complex numbers z1 and z2, prove that
Re (z1z2) = Re z1 Re z2 – Im z1 Im z2
Answer
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Question 3:
Reduce to the standard form.
Answer
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Question 4:
If x – iy = prove that .
Answer
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Question 5:
Convert the following in the polar form:
(i) , (ii)
Answer
(i) Here,
Let r cos θ = –1 and r sin θ = 1
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On squaring and adding, we obtain
r2 (cos2 θ + sin2 θ) = 1 + 1
⇒ r2 (cos2 θ + sin2 θ) = 2
⇒ r2 = 2 [cos2 θ + sin2 θ = 1]
∴z = r cos θ + i r sin θ
This is the required polar form.
(ii) Here,
Let r cos θ = –1 and r sin θ = 1
On squaring and adding, we obtain
r2 (cos2 θ + sin2 θ) = 1 + 1
⇒r2 (cos2 θ + sin2 θ) = 2
⇒ r2 = 2 [cos2 θ + sin2 θ = 1]
Class XI Chapter 5 – Complex Numbers and Quadratic Equations Maths
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∴z = r cos θ + i r sin θ
This is the required polar form.
Question 6:
Solve the equation
Answer
The given quadratic equation is
This equation can also be written as
On comparing this equation with ax2 + bx + c = 0, we obtain
a = 9, b = –12, and c = 20
Therefore, the discriminant of the given equation is
D = b2 – 4ac = (–12)2 – 4 × 9 × 20 = 144 – 720 = –576
Therefore, the required solutions are
Question 7:
Solve the equation
Answer
Class XI Chapter 5 – Complex Numbers and Quadratic Equations Maths
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The given quadratic equation is
This equation can also be written as
On comparing this equation with ax2 + bx + c = 0, we obtain
a = 2, b = –4, and c = 3
Therefore, the discriminant of the given equation is
D = b2 – 4ac = (–4)2 – 4 × 2 × 3 = 16 – 24 = –8
Therefore, the required solutions are
Question 8:
Solve the equation 27x2 – 10x + 1 = 0
Answer
The given quadratic equation is 27x2 – 10x + 1 = 0
On comparing the given equation with ax2 + bx + c = 0, we obtain
a = 27, b = –10, and c = 1
Therefore, the discriminant of the given equation is
D = b2 – 4ac = (–10)2 – 4 × 27 × 1 = 100 – 108 = –8
Therefore, the required solutions are
Question 9:
Solve the equation 21x2 – 28x + 10 = 0
Answer
The given quadratic equation is 21x2 – 28x + 10 = 0
On comparing the given equation with ax2 + bx + c = 0, we obtain
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a = 21, b = –28, and c = 10
Therefore, the discriminant of the given equation is
D = b2 – 4ac = (–28)2 – 4 × 21 × 10 = 784 – 840 = –56
Therefore, the required solutions are
Question 10:
If find .
Answer
Question 11:
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If a + ib = , prove that a2 + b2 =
Answer
On comparing real and imaginary parts, we obtain
Hence, proved.
Question 12:
Let . Find
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(i) , (ii)
Answer
(i)
On multiplying numerator and denominator by (2 – i), we obtain
On comparing real parts, we obtain
(ii)
On comparing imaginary parts, we obtain
Question 13:
Find the modulus and argument of the complex number .
Answer
Let , then
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On squaring and adding, we obtain
Therefore, the modulus and argument of the given complex number are
respectively.
Question 14:
Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of –6 – 24i.
Answer
Let
It is given that,
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Equating real and imaginary parts, we obtain
Multiplying equation (i) by 3 and equation (ii) by 5 and then adding them, we obtain
Putting the value of x in equation (i), we obtain
Thus, the values of x and y are 3 and –3 respectively.
Question 15:
Find the modulus of .
Answer
Question 16:
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If (x + iy)3 = u + iv, then show that .
Answer
On equating real and imaginary parts, we obtain
Hence, proved.
Question 17:
If α and β are different complex numbers with = 1, then find .
Answer
Let α = a + ib and β = x + iy
It is given that,
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Question 18:
Find the number of non-zero integral solutions of the equation .
Answer
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Thus, 0 is the only integral solution of the given equation. Therefore, the number of non-
zero integral solutions of the given equation is 0.
Question 19:
If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that
(a2 + b2) (c2 + d2) (e2 + f2) (g2 + h2) = A2 + B2.
Answer
On squaring both sides, we obtain
(a2 + b2) (c2 + d2) (e2 + f2) (g2 + h2) = A2 + B2
Hence, proved.
Question 20:
If , then find the least positive integral value of m.
Answer
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Therefore, the least positive integer is 1.
Thus, the least positive integral value of m is 4 (= 4 × 1).
38 [XI – Mathematics]
CHAPTER - 5
COMPLEX NUMBERS ANDQUADRATIC EQUATIONS
KEY POINTS
The imaginary number –1= i, is called iota
For any integer k, i4k =1, i4k+1 = i, i4k+2 = –1, i4k+3 = –i
a b ab if both a and b are negative real numbers
A number of the form z = a + ib, where a, b R is called a complexnumber.
a is called the real part of z, denoted by Re(z) and b is called theimaginary part of z, denoted by Im(z)
a + ib = c + id if a = c, and b = d
z1 = a + ib, z2 = c + id.
In general, we cannot compare and say that z1 > z2 or z1 < z2
but if b, d = 0 and a > c then z1 > z2
i.e. we can compare two complex numbers only if they are purely real.
0 + i 0 is additive identity of a complex number.
–z = –a + i(–b) is called the Additive Inverse or negative of z = a + ib
1 + i 0 is multiplicative identity of complex number.
z = a – ib is called the conjugate of z = a + ib
39 [XI – Mathematics]
z–1 = 2 2 2
1 a ib zz za b
is called the multiplicative Inverse of
z = a + ib (a 0, b 0)
The coordinate plane that represents the complex numbers is called thecomplex plane or the Argand plane
Polar form of z = a + ib is,
z = r (cos + i sin) where r = 2 2a b = |z| is called the modulus of z,
is called the argument or amplitude of z.
The value of such that, –< is called the principle argument of z.
|z1 + z2| |z1| + |z2|
|z1z2| = |z1|. |z2|
n n 211
2 2
zz, z z , z z z z , z z z
z z
|z1 – z2| |z1| + |z2|
|z1 – z2| 1 2z z
If z1 = r1 (cos 1 + i sin 1)
z2 = r2 (cos 2 + i sin 2)
then z1z2 = r1 r2 [cos (1 + 2) + i sin (1 + 2)]
1 11 2 1 2
2 2cos sin
z ri
z r
For the quadratic equation ax2 + bx + c = 0, a, b, c R, a 0,
if b2 – 4ac < 0 then it will have complex roots given by,
2b i 4ac bx
2a
40 [XI – Mathematics]
a ib is called square root of z = a + ib
a ib x iy
squaring both sides we get
a + ib = x2 – y2 + 2i(xy)
x2 – y2 = a, 2xy = b. Solving these we get x and y.
VERY SHORT ANSWER TYPE QUESTIONS (1 MARK)
1. Evaluate :
(i) –16 3 25 36 625
(ii) 16 25 49 49 14i i i
(iii) 31
– –1
(iv)35
351
ii
2. Find x and y if 3x + (2y – 3) i is equal to 2 + 4i
3. Find additive inverse of 6 – –49i i
4. Find multiplicative inverse of 2 + i
5. Find modulus of –25 7
6. If 1 22 4 , 3 – 5z i z i the find
(i) Re (z1z2) (ii) Im (z1z2)
(iii) Re (z1–z2) (iv) Im (z1 + z2)
7. Express 1
1 i in the form of a + ib.
8. If modulus of z is 2 and argument of z is 56
then write z in form of
a + ib.
41 [XI – Mathematics]
9. If z1 = 2(cos 30° + i sin 30°), z2 = 3(cos 60 + i sin 30°)
Find Re (z1z2)
10. Find the value of 4
1n
n is any positive integer.
11. Find conjugate of i 7
12. Find the solution of equation x2 + 3 = 0 in complex numbers.
13. Represent sin cos3 3
i in polar form
14. Represent in (a + ib) 3 3
3 cos sin2 2
i
15. If z1 = cos 30° + i sin 30°
z2 = cos 60° + i sin 60°
then find (i) 1
2
zz ; (ii) 1 2z z
SHORT ANSWER TYPE QUESTIONS (4 MARKS)
16. For Complex numbers z1 = –1 + i, z2 = 3 – 2i
show that,
Im (z1z2) = Re (z1) Im(z2) + Im (z1) Re (z2)
17. If x + iy 1 i1 i
, prove that x2 + y2 = 1
18. Find real value of such that,
1 i cos1 2i cos
is a real number
19. If z 5i
1z 5i
, show that z is a real number.
20. Find the value of x and y if
x2 – 7x + 9yi = y2 i + 20i – 12
42 [XI – Mathematics]
21. E xpress in the form of a + ib
314 3
ii
22. Convert the following in polar form :
(i) –3 2 3 2 i (ii) 3 – 1 – 3 1
2 2i
(iii) i (1 + i ) (iv)5 –2 – 3
ii
23. If 13x iy a ib , prove that,
x ya b
= 4(a2 – b2)
24. For complex numbers z1 = 6 + 3i, z2 = 3 – i find 1
2
zz
25. If n2 2i
12 2i
,find the least positive integral value of n.
26. Solve
(i) ix2 + 4x – 4i = 0 (ii) x2 – (2 + i )x – (1 – 7i ) = 0
(iii) 2x2 + 3ix + 2 = 0 (iv) 2 32 0
2x x
27. Find square root of the following complex number
(i) –7 + 24i (ii) –3 – 4i
(iii) –15 – 8i (iv) 7 – 30 –2
LONG ANSWER TYPE QUESTIONS (6 MARKS)
28. If z1, z2 are complex numbers such that, 1 2
21
z 3z1
3 z z
and |z2| 1
then find |z1|
29. If x = –1 + i then find the value of x4 + 4x3 + 4x2 + 2
43 [XI – Mathematics]
*30. If z = x + iy and 1– iz
wz i
show that if |w| = 1 then z is purely real.
*31. Prove by Principle of Mathematical Induction
(cos + i sin )n = cos n + i sin n
if n is any natural number.
ANSWERS
1. (i) 0; (ii) 19; (iii) i ; (iv) 0
2.2 7
,3 2
x y 3. –7 – 6i
4.2
–3 3
i5. 74
6. (i) 26; (ii) 2; (iii) –1 ; (iv) –1
7.1
–2 2
i8. – 3 1i
9. 0 10. 1
11. i 12. 3 i
13. cos sin6 6
i 14. –3i
15. (i) 1; (ii) 1;
18. 2 12
n
20. x = 4, 3; y = 5, 4
21.–2 1425 25
i
22. (i)3 3
6 cos sin4 4
i
; (ii)–5 –5
1 cos sin12 12
i
;
44 [XI – Mathematics]
(iii) 3 3
2 cos sin4 4
i
; (iv) 2 cos sin4 4
i
24. 1
2
3 12
z iz
25. n = 4
26. (i) 2i, 2i (ii) 3 – i, –1 + 2i
(iii) 1
, – 22
i i (iv)2 2
2i
27. (i) ± (3 + 4i ) (ii) ±(1 – 2i )
(iii) ± (1 – 4i ) (iv) ± (5 – 32i )
28. |z1| = 3
29. 6