Chapter 5 of the Class 11 Mathematics curriculum is titled “Complex Numbers,” which introduces students to the concept of numbers that extend beyond the real number system. A complex number is expressed in the form a+bi, where aaa and bbb are real numbers, and i. This chapter explores the basic operations with complex numbers, including addition, subtraction, multiplication, and division. Students also learn about the geometric representation of complex numbers on the complex plane, where the real part corresponds to the x-axis and the imaginary part to the y-axis. The chapter provides a foundation for understanding how complex numbers are applied in various mathematical fields, such as algebra, calculus, and engineering.
In addition to basic operations, the chapter delves into the concepts of the modulus and argument of a complex number, along with their significance in solving problems. The modulus represents the distance of a complex number from the origin, and the argument gives the angle formed with the positive real axis. Students also explore the polar form of complex numbers, which makes complex number multiplication and division simpler. The chapter lays the groundwork for more advanced topics like De Moivre’s Theorem, roots of complex numbers, and their applications in solving equations. Mastering complex numbers is essential for students as it forms the basis for understanding functions, transformations, and many advanced concepts in higher mathematics.
Exercise 5.1 (Complex Numbers)
Express each of the complex numbers given in Exercises 1 to 10 in the form a + ib.
1. (5i) (-3/5i)
Solution:
The complex number in the form a+ib is: −3+0i
Thus, the result is a=−3 and b=0.
2. i9 + i19
Solution:
Step 1: Simplify i9
We know that the powers of iii repeat every 4 terms. So, to find i9, we divide 9 by 4 and look at the remainder:
9÷4=2 remainder 1
Thus, i9=i1=i.
Step 2: Simplify i19
Similarly, divide 19 by 4:
19÷4=4 remainder 3
Thus, i19=i3=−i.
Step 3: Add the results
Now add i9+i19: i+(−i)=
Final Answer: The sum of i9+i19 is i0+0i, or simply 0.
3. I-39
Solution:
Simplify i−39
Since the powers of iii repeat every 4 terms, we first find the remainder when −39 is divided by 4:
−39÷4= 1
This means i−39 is equivalent to i1 (since the remainder is 1).
Evaluate i1
We know that i1=i.
Final Answer: Thus, i−39=i.
4. 3(7 + i7) + i(7 + i7)
Solution:
Step 1: Distribute the constants in each term
- For 3(7+i7): 3(7+i7)= 3×7+3×i7= 21+21i
- For i(7+i7): i(7+i7)= i×7+i×i7= 7i+i2×7i
We know that i2=−1, so: 7i+i2×7=7i+(−1)×7=7i−7
Step 2: Combine the terms
Now, combine the results from both terms:
(21+21i)+(7i−7))
First, combine the real parts:
21−7=14
Next, combine the imaginary parts:
21i+7i=28
Final Answer: The simplified expression is: 14+28i
5. (1 – i) – (–1 + i6)
Solution:
(1 – i) – (–1 + i6) = 1 – i + 1 – i6
= 2 – i7
Hence,
(1 – i) – (–1 + i6) = 2 – 7i
6.
Solution:
Given: (1/5 + 2i/5) – (4 + 5i/2)
Step 1: Separate real and imaginary parts:
- Real part: (1/5) – 4
- Imaginary part: (2/5)i – (5/2)i
Step 2: Simplify:
- Real part: -19/5
- Imaginary part: -21/10i
Step 3: Combine: -19/5 – 21/10i
Therefore, the answer is -19/5 – 21/10i.
7.
Solution:
Step 1: Simplify the innermost brackets:
Inside the first bracket, we have:
(1/3 + 7i/3) + (4 + i/3)
Combining the real and imaginary parts:
(1/3 + 4) + (7i/3 + i/3)
This simplifies to:
13/3 + 8i/3
Step 2: Simplify the outer brackets:
Now, we have:
(13/3 + 8i/3) – (-4/3 + i)
To subtract complex numbers, we subtract the real parts and the imaginary parts separately:
(13/3 – (-4/3)) + (8i/3 – i)
This simplifies to: 17/3 + 5i/3
Therefore, the solution to the given expression is: 17/3 + 5i/3
8. (1 – i)4
Solution:
Step 1: Square the binomial:
(1 – i)² = 1² – 2(1)(i) + i² = 1 – 2i – 1 = -2i
Step 2: Square the result from Step 1:
(-2i)² = (-2)² * i² = 4 * (-1) = -4
Therefore, (1 – i)⁴ = -4.
9. (1/3 + 3i)3
Solution:
To solve (1/3 + 3i)³, we can use the formula for the cube of a binomial:
(a + b)³ = a³ + 3a²b + 3ab² + b³
In this case, a = 1/3 and b = 3i.
Step 1: Calculate a³ and b³:
- a³ = (1/3)³ = 1/27
- b³ = (3i)³ = 27i³ = 27(-i) = -27i (since i³ = -i)
Step 2: Calculate 3a²b and 3ab²:
- 3a²b = 3(1/3)²(3i) = i
- 3ab² = 3(1/3)(3i)² = -9
Step 3: Combine all the terms:
(1/3 + 3i)³ = 1/27 – 27i + i – 9
Step 4: Combine like terms:
(1/3 + 3i)³ = -242/27 – 26i
Therefore, the solution is -242/27 – 26i.
10. (-2 – 1/3i)3
Solution:
To solve this, we can use the formula for the cube of a binomial:
(a + b)³ = a³ + 3a²b + 3ab² + b³
In this case, a = -2 and b = -1/3i.
Step 1: Calculate a³ and b³:
- a³ = (-2)³ = -8
- b³ = (-1/3i)³ = -1/27i³ = 1/27i (since i³ = -i)
Step 2: Calculate 3a²b and 3ab²:
- 3a²b = 3(-2)²(-1/3i) = -4i
- 3ab² = 3(-2)(-1/3i)² = -2/3
Step 3: Combine all the terms:
(-2 – 1/3i)³ = -8 + 1/27i – 4i – 2/3
Step 4: Combine like terms:
(-2 – 1/3i)³ = -202/27 – 107/27i
Therefore, the solution is -202/27 – 107/27i.
Find the multiplicative inverse of each of the complex numbers given in Exercises 11 to 13.
11. 4 – 3i
Solution:
The expression 4 – 3i is already in its simplest form, which is the standard form of a complex number: a + bi.
Here,
- a = 4 (real part)
- b = -3 (imaginary part)
Therefore, 4 – 3i is the solution.
12. √5 + 3i
Solution:
The expression √5 + 3i is already in its simplest form. It’s a complex number in the standard form a + bi, where:
- a = √5 (real part)
- b = 3 (imaginary part)
Therefore, √5 + 3i is the solution.
13. – i
Solution:
-i = 0 – 1i, Therefore, the standard form is 0 – 1i.
14. Express the following expression in the form of a + ib:
Solution:
Step 1: Simplify the numerator:
We can use the difference of squares formula:
(a + b)(a – b) = a² – b²
Applying this to the numerator, we get:
(3 + i√5)(3 – i√5) = 3² – (i√5)² = 9 – 5i² = 9 + 5 = 14
Step 2: Simplify the denominator:
We can distribute the negative sign in the second bracket:
(√3 + √2i) – (√3 – i√2) = √3 + √2i – √3 + i√2
Combining like terms:
√3 – √3 + √2i + i√2 = 2√2i
Step 3: Divide the numerator by the denominator:
Now we have:
14 / (2√2i)
To simplify further, we can rationalize the denominator by multiplying both numerator and denominator by √2i:
(14 * √2i) / (2√2i * √2i) = (14√2i) / (4i²)
Since i² = -1, the denominator becomes:
(14√2i) / (-4)
Simplifying further:
-7√2i / 2
Therefore, the solution to the given expression is -7√2i / 2.
Exercise 5.2 (Complex Numbers)
Find the modulus and the arguments of each of the complex numbers in Exercises 1 to 2.
1. z = – 1 – i √3
Solution:
Given: z = -1 – i√3
Modulus (|z|)
The modulus of a complex number z = x + iy is given by:
|z| = √(x² + y²)
For z = -1 – i√3, x = -1 and y = -√3.
So, |z| = √((-1)² + (-√3)²) = √(1 + 3) = √4 = 2
Argument (θ)
The argument of a complex number z = x + iy is given by:
θ = arctan(y/x)
For z = -1 – i√3, x = -1 and y = -√3.
So, θ = arctan((-√3)/(-1)) = arctan(√3)
Since the point (-1, -√3) lies in the third quadrant, we need to add π to the principal value of arctan(√3).
Therefore, θ = π + π/3 = 4π/3
Hence, the modulus of the complex number is 2, and the argument is 4π/3.
2. z = -√3 + i
Solution:
Given: z = -√3 + i
Modulus (|z|)
|z| = √((-√3)² + 1²) = √(3 + 1) = 2
Argument (θ)
θ = arctan(1/(-√3)) = -π/6
Since the point (-√3, 1) lies in the second quadrant, we need to add π to the principal value of arctan(1/(-√3)).
Therefore, θ = π – π/6 = 5π/6
Hence, the modulus of the complex number is 2, and the argument is 5π/6.
Convert each of the complex numbers given in Exercises 3 to 8 in the polar form:
3. 1 – i
Solution:
Given: z = 1 – i
Modulus (|z|)
|z| = √(1² + (-1)²) = √2
Argument (θ)
θ = arctan((-1)/1) = -π/4
Since the point (1, -1) lies in the fourth quadrant, the argument is -π/4.
Polar Form:
z = |z|(cosθ + i sinθ) = √2(cos(-π/4) + i sin(-π/4))
Therefore, the polar form of 1 – i is √2(cos(-π/4) + i sin(-π/4)).
4. – 1 + i
Solution:
Given: z = -1 + i
Modulus (|z|)
|z| = √((-1)² + 1²) = √2
Argument (θ)
θ = arctan(1/(-1)) = -π/4
Since the point (-1, 1) lies in the second quadrant, we need to add π to the principal value of arctan(1/(-1)).
Therefore, θ = π – π/4 = 3π/4
Polar Form:
z = |z|(cosθ + i sinθ) = √2(cos(3π/4) + i sin(3π/4))
Therefore, the polar form of -1 + i is √2(cos(3π/4) + i sin(3π/4)).
5. – 1 – i
Solution:
Given: z = -1 – i
Modulus (|z|)
|z| = √((-1)² + (-1)²) = √2
Argument (θ)
θ = arctan((-1)/(-1)) = π/4
Since the point (-1, -1) lies in the third quadrant, we need to add π to the principal value of arctan(1).
Therefore, θ = π + π/4 = 5π/4
Polar Form:
z = |z|(cosθ + i sinθ) = √2(cos(5π/4) + i sin(5π/4))
Therefore, the polar form of -1 – i is √2(cos(5π/4) + i sin(5π/4)).
6. – 3
Solution:
Given: z = -3
We can write this as z = -3 + 0i.
Modulus (|z|)
|z| = √((-3)² + 0²) = 3
Argument (θ)
Since -3 lies on the negative real axis, its argument is π.
Polar Form:
z = |z|(cosθ + i sinθ) = 3(cosπ + i sinπ)
Therefore, the polar form of -3 is 3(cosπ + i sinπ).
7. 3 + i
Solution:
Given: z = 3 + i
Modulus (|z|)
|z| = √(3² + 1²) = √10
Argument (θ)
θ = arctan(1/3)
Since the point (3, 1) lies in the first quadrant, the argument is arctan(1/3).
Polar Form:
z = |z|(cosθ + i sinθ) = √10(cos(arctan(1/3)) + i sin(arctan(1/3)))
Therefore, the polar form of 3 + i is √10(cos(arctan(1/3)) + i sin(arctan(1/3))).
8. i
Solution:
Given: z = i
We can write this as z = 0 + 1i.
Modulus (|z|)
|z| = √(0² + 1²) = 1
Argument (θ)
Since i lies on the positive imaginary axis, its argument is π/2.
Polar Form:
z = |z|(cosθ + i sinθ) = 1(cos(π/2) + i sin(π/2))
Therefore, the polar form of i is cos(π/2) + i sin(π/2).
Exercise 5.3 (Complex Numbers)
Solve each of the following equations:
1. x2 + 3 = 0
Solution:
Given equation: x² + 3 = 0
Subtracting 3 from both sides: x² = -3
Taking the square root of both sides: x = ±√(-3)
Since the square root of a negative number is not a real number, we introduce the imaginary unit “i,” where i² = -1.
Therefore, we can rewrite the equation as: x = ±√(3 * -1) x = ±√3 * √(-1) x = ±√3i
So, the solutions to the equation x² + 3 = 0 are: x = √3i and x = -√3i
2. 2x2 + x + 1 = 0
Solution:
To solve the quadratic equation 2x² + x + 1 = 0, we can use the quadratic formula:
x = (-b ± √(b² – 4ac)) / (2a)
For the given equation, a = 2, b = 1, and c = 1.
Substituting these values into the formula, we get:
x = (-1 ± √(1² – 4*2*1)) / (2*2)
= (-1 ± √(-7)) / 4
Since the discriminant (b² – 4ac) is negative, the roots of the equation are complex.
We can express the roots in the form a + bi, where a and b are real numbers and i is the imaginary unit (√-1).
x = (-1 ± √7i) / 4
Therefore, the solutions to the equation 2x² + x + 1 = 0 are:
x = (-1 + √7i) / 4
x = (-1 – √7i) / 4
3. x2 + 3x + 9 = 0
Solution:
To solve the quadratic equation x² + 3x + 9 = 0, we can use the quadratic formula:
x = (-b ± √(b² – 4ac)) / (2a)
For the given equation, a = 1, b = 3, and c = 9.
Substituting these values into the formula, we get:
x = (-3 ± √(3² – 4*1*9)) / (2*1)
= (-3 ± √(-27)) / 2
Since the discriminant (b² – 4ac) is negative, the roots of the equation are complex.
We can simplify the expression further:
x = (-3 ± √(27 * -1)) / 2
= (-3 ± 3√3i) / 2
Therefore, the solutions to the equation x² + 3x + 9 = 0 are:
x = (-3 + 3√3i) / 2
x = (-3 – 3√3i) / 2
4. –x2 + x – 2 = 0
Solution:
-x2 + x – 2 = 0
Multiplying both sides by -1 to make the leading coefficient positive:
x2 – x + 2 = 0
Now, we can use the quadratic formula:
x = (-b ± √(b^2 – 4ac)) / (2a)
Here, a = 1, b = -1, and c = 2.
Substituting these values:
x = (1 ± √((-1)^2 – 4*1*2)) / (2*1)
= (1 ± √(-7)) / 2
Since the discriminant is negative, the roots are complex:
x = (1 ± √7i) / 2
So, the solutions are:
x = (1 + √7i) / 2
x = (1 – √7i) / 2
5. X2 + 3x + 5 = 0
Solution:
To solve the quadratic equation 5x² + 3x + 5 = 0, we can use the quadratic formula:
x = (-b ± √(b² – 4ac)) / (2a)
Here, a = 5, b = 3, and c = 5.
Substituting these values into the formula, we get:
x = (-3 ± √(3² – 455)) / (2*5)
Simplifying the expression under the radical:
x = (-3 ± √(-91)) / 10
Since the value under the square root is negative, the equation has no real solutions. The solutions are complex numbers involving the imaginary unit “i”:
x = (-3 ± √91i) / 10
6. x2 – x + 2 = 0
Solution:
To solve the quadratic equation x² – x + 2 = 0, we can use the quadratic formula:
x = (-b ± √(b² – 4ac)) / (2a)
Here, a = 1, b = -1, and c = 2.
Substituting these values into the formula, we get:
x = (1 ± √((-1)² – 412)) / (2*1)
Simplifying the expression under the radical:
x = (1 ± √(-7)) / 2
Since the value under the square root is negative, the equation has no real solutions. The solutions are complex numbers involving the imaginary unit “i”:
x = (1 ± √7i) / 2
7. √2x2 + x + √2 = 0
Solution:
To solve the quadratic equation √2x² + x + √2 = 0, we can use the quadratic formula:
x = (-b ± √(b² – 4ac)) / (2a)
Here, a = √2, b = 1, and c = √2.
Substituting these values into the formula, we get:
x = (-1 ± √(1² – 4√2√2)) / (2*√2)
Simplifying the expression under the radical:
x = (-1 ± √(-7)) / (2√2)
Since the value under the square root is negative, the equation has no real solutions. The solutions are complex numbers involving the imaginary unit “i”:
x = (-1 ± √7i) / (2√2)
We can rationalize the denominator by multiplying both the numerator and denominator by √2:
x = (-√2 ± √14i) / 4
8. √3x2 – √2x + 3√3 = 0
Solution:
To solve the quadratic equation √3x² – √2x + 3√3 = 0, we can use the quadratic formula:
x = (-b ± √(b² – 4ac)) / (2a)
Here, a = √3, b = -√2, and c = 3√3.
Substituting these values into the formula, we get:
x = (√2 ± √((-√2)² – 4√33√3)) / (2*√3)
Simplifying the expression under the radical:
x = (√2 ± √(-34)) / (2√3)
Since the value under the square root is negative, the equation has no real solutions. The solutions are complex numbers involving the imaginary unit “i”:
x = (√2 ± √34i) / (2√3)
We can rationalize the denominator by multiplying both the numerator and denominator by √3:
x = (√6 ± √102i) / 6
9. x2+ x + 1/√2 = 0
Solution:
To solve the quadratic equation x² + x + 1/√2 = 0, we can use the quadratic formula:
x = (-b ± √(b² – 4ac)) / (2a)
Here, a = 1, b = 1, and c = 1/√2.
Substituting these values into the formula, we get:
x = (-1 ± √(1² – 41(1/√2))) / (2*1)
Simplifying the expression under the radical:
x = (-1 ± √(1 – 2√2)) / 2
Since the value under the square root is negative, the equation has no real solutions. The solutions are complex numbers involving the imaginary unit “i”:
x = (-1 ± i√(2√2 – 1)) / 2
10. x2+ x/√2 + 1 = 0
Solution:
To solve the quadratic equation x² + x/√2 + 1 = 0, we can use the quadratic formula:
x = (-b ± √(b² – 4ac)) / (2a)
Here, a = 1, b = 1/√2, and c = 1.
Substituting these values into the formula, we get:
x = (-(1/√2) ± √((1/√2)² – 411)) / (2*1)
Simplifying the expression under the radical:
x = (-1/√2 ± √(-7/2)) / 2
Since the value under the square root is negative, the equation has no real solutions. The solutions are complex numbers involving the imaginary unit “i”:
x = (-1 ± i√7) / (2√2)
We can rationalize the denominator by multiplying both the numerator and denominator by √2:
x = (-√2 ± i√14) / 4
Miscellaneous Exercise (Complex Numbers)
1.
Solution:
[i18 + (1/i)25]3
Step 1: Simplify the powers of i:
- i18: We can break this down into (i4)4 * i2. Since i4 = 1, we get 1 * (-1) = -1.
- (1/i)25: This is the same as i(-25). Now, i25= (i4)6 * i = 1 * i = i. So, 1/i25 = 1/i.
Step 2: Substitute the simplified values:
The expression becomes:
(-1 + 1/i)3
Step 3: Find a common denominator:
The common denominator is i. So, we get:
((-i + 1) / i)3
Step 4: Cube the expression:
((-i + 1)3) / i3
Step 5: Expand the numerator:
Using the binomial expansion formula (a + b)3 = a3 + 3a2b + 3ab2 + b3, we get:
(-i3 + 3i2 – 3i + 1) / i3
Simplifying the numerator:
(-(-i) + 3(-1) – 3i + 1) / (-i)
(i – 3 – 3i + 1) / (-i)
(-2i – 2) / (-i)
Step 6: Rationalize the denominator:
To get rid of the complex number in the denominator, we can multiply both the numerator and denominator by i:
((-2i – 2) * i) / (-i * i)
(2i2 + 2i) / 1
Since i2 = -1, we have:
(-2 + 2i) / 1
So, the final answer is: -2 + 2i
2. For any two complex numbers z1 and z2, prove that
Re (z1z2) = Re z1 Re z2 – Im z1 Im z2
Solution:
Let’s consider two complex numbers:
z1 = a + bi, z2 = c + di
where a, b, c, and d are real numbers.
Now, let’s calculate their product:
z1 * z2 = (a + bi)(c + di) = ac + adi + bci + bdi2 = (ac – bd) + (ad + bc)i (since i2 = -1)
So, the real part of z1 * z2 is (ac – bd).
Now, let’s calculate the right-hand side of the equation:
Re(z1) * Re(z2) – Im(z1) * Im(z2) = a * c – b * d
As you can see, both expressions are equal to (ac – bd).
Therefore, we have proven that:
Re(z1 * z2) = Re(z1) * Re(z2) – Im(z1) * Im(z2)
3. Reduce to the standard form.
Solution:
Step 1: Simplify the fractions
To simplify the fractions, we need to find a common denominator. The common denominator for 1-4i and 1+i is their product: (1-4i)(1+i) = 1 – 16i2 = 17. (i2= -1)
So, the expression becomes: (((1+i) – 2(1-4i))/17) * (3-4i)/(5+i)
Simplifying the numerator: ((-1+9i)/17) * (3-4i)/(5+i)
Step 2: Multiply the fractions
To multiply the fractions, we multiply the numerators and denominators separately: ((-1+9i)(3-4i))/((17)(5+i))
Multiplying the numerators: (-3 + 4i + 27i – 36i^2)
Simplifying the numerator: (33 + 31i)
Step 3: Rationalize the denominator
To rationalize the denominator, we multiply both the numerator and denominator by the conjugate of the denominator: ((33 + 31i)(5-i))/((17)(5+i)(5-i))
Multiplying the numerators and denominators: (165 – 33i + 155i – 31i^2) / (17 * 26)
Simplifying: (196 + 122i) / 442
Step 4: Separate the real and imaginary parts: 196/442 + (122/442)i
Therefore, the simplified expression is: 196/442 + (122/442)i
4.
Solution:
Given: x – iy = √(a-ib) / √(c-id)
Step 1: Rationalize the denominator
To simplify the right-hand side, we need to rationalize the denominator. We can do this by multiplying both the numerator and denominator by the complex conjugate of the denominator:
x – iy = √(a-ib) * √(c+id) / √(c-id) * √(c+id)
Simplifying: x – iy = √((a-ib)(c+id)) / √(c2 + d2)
Step 2: Expand the numerator
Multiplying the terms in the numerator:
x – iy = √(ac + adi – bci – bdi2) / √(c2 + d2)
Since i2 = -1:
x – iy = √(ac + adi – bci + bd) / √(c2 + d2)
Step 3: Equate real and imaginary parts
Now, we can equate the real and imaginary parts on both sides of the equation:
- Real part: x = √(ac + bd) / √(c2 + d2)
- Imaginary part: -y = √(ad – bc) / √(c2 + d2)
Step 4: Square both sides of each equation
Squaring both sides of the real part equation:
x2 = (ac + bd) / (c2 + d2)
Squaring both sides of the imaginary part equation:
y2 = (ad – bc)2 / (c2 + d2)2
Step 5: Add the squared equations
Adding the two squared equations:
x2 + y2 = (ac + bd) / (c2 + d2) + (ad – bc)2 / (c2 + d2)2
Step 6: Combine the fractions
To combine the fractions, we need a common denominator:
x2 + y2 = [(ac + bd)(c2 + d2) + (ad – bc)2] / (c2 + d2)2
Step 7: Expand and simplify the numerator
Expanding the numerator:
x2 + y2 = (a2c2 + 2abcd + b2d2 + a2d2 – 2abcd + b2c2) / (c2 + d2)2
Simplifying:
x2 + y2 = (a2c2 + a2d2 + b2c2 + b2d2) / (c2 + d2)2
Step 8: Factor the numerator
Factoring the numerator:
x2 + y2 = [(a2 + b2)(c2 + d2)] / (c2 + d2)2
Step 9: Cancel common factors
Canceling the common factor (c2 + d2):
x2 + y2 = (a2 + b2) / (c2 + d2)
Step 10: Square both sides
Squaring both sides to get the desired form:
(x2 + y2)2 = (a2 + b2)2 / (c2 + d2)2
Therefore, we have proven that: (x2 + y2)2 = (a2 + b2)2 / (c2 + d2)2
5. Convert the following into the polar form:
(i) , (ii)
Solution:
A complex number z = x + iy can be represented in polar form as:
z = r(cosθ + i sinθ)
Where: r: Modulus of z = √(x² + y²) and θ: Argument of z = arctan(y/x)
Let’s convert the given complex numbers:
i) (1 + 7i)/(2 – i)²
Step 1: Simplify the denominator:
(2 – i)² = 4 – 4i + i² = 3 – 4i
Step 2: Multiply numerator and denominator by the conjugate of the denominator:
(1 + 7i)/(3 – 4i) = (1 + 7i)(3 + 4i) / (3 – 4i)(3 + 4i)
= (3 + 4i + 21i + 28i²) / (9 – 16i²)
= (-25 + 25i) / 25
= -1 + i
Step 3: Convert to polar form:
r = √((-1)² + 1²) = √2
θ = arctan(-1/1) = -π/4 (since the point (-1, 1) is in the second quadrant)
Using trigonometric form:
z = √2(cos(-π/4) + i sin(-π/4))
ii) (1 + 3i)/(1 – 2i)
Step 1: Multiply numerator and denominator by the conjugate of the denominator:
(1 + 3i)/(1 – 2i) = (1 + 3i)(1 + 2i) / (1 – 2i)(1 + 2i)
= (1 + 2i + 3i + 6i²) / (1 – 4i²)
= (-5 + 5i) / 5
= -1 + i
Step 2: Convert to polar form:
(Same as part (i))
r = √((-1)² + 1²) = √2
θ = arctan(-1/1) = -π/4
Using trigonometric form:
z = √2(cos(-π/4) + i sin(-π/4))
Therefore, the polar forms using trigonometric functions are:
(i) √2(cos(-π/4) + i sin(-π/4))
(ii) √2(cos(-π/4) + i sin(-π/4))
Solve each of the equations in Exercises 6 to 9.
6. 3x2 – 4x + 20/3 = 0
Solution:
7. x2 – 2x + 3/2 = 0
Solution:
Let’s solve the equation: x2−2x+2/3=0
Steps to solve:
1. Identify the coefficients of the quadratic equation:
- a = 1
- b = -2
- c = 3/2
2. Apply the quadratic formula:
The quadratic formula is given by:
8. 27x2 – 10x + 1 = 0
Solution:
Let’s solve the quadratic equation:
27x2−10x+1=0
We can either factor this equation or use the quadratic formula. Let’s try factoring:
We need to find two numbers that multiply to 27 and add up to -10. After some trial and error, we find that -9 and -1 satisfy these conditions.
So, we can factor the equation as follows:
(27x−1)(x−1)=0
Now, we can set each factor equal to zero and solve for x:
- 27x – 1 = 0
- 27x = 1
- x = 1/27
- x – 1 = 0
- x = 1
Therefore, the solutions to the equation 27x2 – 10x + 1 = 0 are x = 1/27 and x = 1.
9. 21x2 – 28x + 10 = 0
Solution:
we can use the quadratic formula: x = (-b ± √(b2 – 4ac)) / (2a)
In this equation, a = 21, b = -28, and c = 10.
Substituting these values into the formula, we get:
x = (28 ± √((-28)2 – 4 * 21 * 10)) / (2 * 21)
Simplifying:
x = (28 ± √(784 – 840)) / 42
x = (28 ± √(-56)) / 42
Since the discriminant (b2 – 4ac) is negative, the roots of the equation are complex numbers.
x = (28 ± √(56)i) / 42
Simplifying further:
x = (2 ± √(14)i) / 3
Therefore, the solutions to the equation 21x2 – 28x + 10 = 0 are:
x = (2 + √(14)i) / 3
x = (2 – √(14)i) / 3
10. If z1 = 2 – i, z2 = 1 + i, find
Solution:
Step 1: Calculate the sum and difference of z1 and z2
z1 + z2 = (2 – i) + (1 + i) = 3
z1 – z2 = (2 – i) – (1 + i) = 1 – 2i
Step 2: Calculate the absolute values of the sum and difference
|z1 + z2| = |3| = 3
|z1 – z2| = |1 – 2i| = √(12 + (-2)2) = √5
Step 3: Calculate the required expression
|z1 + z2 + 1| / |z1 – z2 + 1| = (3 + 1) / (√5 + 1)
= 4 / (√5 + 1)
= (4(√5 – 1)) / ((√5 + 1)(√5 – 1))
= (4√5 – 4) / (5 – 1)
= (√5 – 1)
Therefore, the value of the expression is (√5 – 1).
11.
Solution:
Given:
a + ib = (x + i)2 / (2x2 + 1)
Step 1: Expand the numerator
a + ib = (x2 + 2ix – 1) / (2x2 + 1)
Step 2: Separate real and imaginary parts
a = (x2 – 1) / (2x2 + 1)
b = 2x / (2x2 + 1)
Step 3: Calculate a2 + b2
a2 + b2 = [(x2 – 1)2 + (2x)2] / (2x2 + 1)2
Step 4: Simplify the numerator
a2 + b2 = (x4 – 2x2 + 1 + 4x2) / (2x2 + 1)2
a2 + b2 = (x4 + 2x2 + 1) / (2x2 + 1)2
Step 5: Factor the numerator
a2 + b2 = [(x2 + 1)2] / (2x2 + 1)2
Hence, we have proved that: a2 + b2 = (x2 + 1)2 / (2x2 + 1)2
13. Find the modulus and argument of the complex number.
Solution:
Step 1: Simplify the complex number
We’ll first simplify the complex number by multiplying both the numerator and denominator by the conjugate of the denominator:
(1 + 2i) / (1 – 3i) = [(1 + 2i)(1 + 3i)] / [(1 – 3i)(1 + 3i)]
= (1 + 5i + 6i2) / (1 – 9i2)
= (1 + 5i – 6) / (1 + 9)
= (-5 + 5i) / 10
= -1/2 + (1/2)i
Step 2: Find the modulus (|z|)
The modulus of a complex number z = a + bi is given by |z| = √(a² + b²).
Here, a = -1/2 and b = 1/2.
So, |z| = √((-1/2)² + (1/2)²) = √(1/4 + 1/4) = √(1/2) = √2/2
Step 3: Find the argument (arg z)
The argument of a complex number z = a + bi is the angle θ it makes with the positive real axis in the complex plane. It can be found using the formula:
θ = arctan(b/a)
Here, a = -1/2 and b = 1/2.
So, θ = arctan(1/2 / -1/2) = arctan(-1) = -π/4
However, since the complex number is in the second quadrant, we need to add π to the angle:
θ = -π/4 + π = 3π/4
Therefore, the modulus of the complex number is √2/2 and the argument is 3π/4.
14. Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of – 6 – 24i.
Solution:
Let’s denote the complex number (x – iy)(3 + 5i) as z.
We are given that z is the conjugate of -6 – 24i.
Therefore, z = -6 + 24i.
Now, let’s expand the product (x – iy)(3 + 5i):
(x – iy)(3 + 5i) = 3x + 5xi – 3yi – 5yi^2
Since i^2 = -1, we can simplify this to:
3x + 5xi – 3yi + 5y
Now, we can equate the real and imaginary parts of this expression to the real and imaginary parts of -6 + 24i:
Real part: 3x + 5y = -6 Imaginary part: 5x – 3y = 24
We now have a system of two linear equations with two unknowns. We can solve this system using any method, such as substitution or elimination.
Let’s use the elimination method:
Multiply the first equation by 3 and the second equation by 5:
9x + 15y = -18 25x – 15y = 120
Add the two equations:
34x = 102
Divide both sides by 34:
x = 3
Substitute x = 3 into the first equation:
3(3) + 5y = -6 9 + 5y = -6 5y = -15 y = -3
Therefore, the values of x and y are: x = 3 and y = -3
15. Find the modulus of
Solution:
Step 1: Simplify the expression:
Let’s first simplify the expression: (1 + i)/(1 – i) – (1 – i)/(1 + i) = [(1 + i)2 – (1 – i)2] / [(1 – i)(1 + i)]
Expanding the numerator and denominator: = [(1 + 2i + i^2) – (1 – 2i + i2)] / (1 – i2)
Simplifying further: = (4i) / 2 = 2i
Step 2: Find the modulus:
The modulus of a complex number z = a + bi is given by |z| = √(a² + b²).
Here, a = 0 and b = 2.
So, |2i| = √(0² + 2²) = √4 = 2
Therefore, the modulus of the given expression is 2.
16. If (x + iy)3 = u + iv, then show that
Solution:
Step 1: Expand the cube of the complex number:
Given: (x + iy)³ = u + iv
Expanding the cube using the binomial theorem:
(x + iy)³ = x³ + 3x²iy + 3xi²y² + i³y³
Simplifying, remembering that i² = -1 and i³ = -i:
(x + iy)³ = x³ + 3ix²y – 3xy² – iy³
Step 2: Equate real and imaginary parts:
Now, we can equate the real and imaginary parts of both sides of the equation:
Real part: x³ – 3xy² = u Imaginary part: 3x²y – y³ = v
Step 3: Prove the given relation:
We need to prove:
(u/x) + (v/y) = 4(x² – y²)
Substituting the values of u and v from step 2:
(x³ – 3xy²)/x + (3x²y – y³)/y = 4(x² – y²)
Simplifying:
x² – 3y² + 3x² – y² = 4(x² – y²)
Combining like terms:
4x² – 4y² = 4(x² – y²)
Dividing both sides by 4:
x² – y² = x² – y²
This is an identity, which is always true.
17. If α and β are different complex numbers with |β| = 1, then find
Solution:
Step 1: Expand the numerator
a + ib = (x² + 2ix – 1) / (2x² + 1)
Step 2: Separate real and imaginary parts
a = (x² – 1) / (2x² + 1)
b = 2x / (2x² + 1)
Step 3: Calculate a² + b²
a² + b² = [(x² – 1)² + (2x)²] / (2x² + 1)²
Step 4: Simplify the numerator
a² + b² = (x⁴ – 2x² + 1 + 4x²) / (2x² + 1)²
a² + b² = (x⁴ + 2x² + 1) / (2x² + 1)²
Step 5: Factor the numerator
a² + b² = [(x² + 1)²] / (2x² + 1)²
Hence, we have proved that: a² + b² = (x² + 1)² / (2x² + 1)²
18. Find the number of non-zero integral solutions of the equation |1 – i|x = 2x.
Solution:
We know that |1-i| = √(1² + (-1)²) = √2.
So, the equation becomes:
√2 * x = 2x
Dividing both sides by x (assuming x ≠ 0):
√2 = 2
This equation has no solution as √2 ≠ 2.
Therefore, the number of non-zero integral solutions is 0.
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
Solution:
We are given: (a + ib)(c + id)(e + if)(g + ih) = A + iB
Taking the modulus of both sides:
|a + ib||c + id||e + if||g + ih| = |A + iB|
Squaring both sides:
|a + ib|² |c + id|² |e + if|² |g + ih|² = |A + iB|²
Using the property |a + ib|² = a² + b², we get:
(a² + b²)(c² + d²)(e² + f²)(g² + h²) = A² + B²
Hence, the given relation is proved.
20. If, then find the least positive integral value of m.
Solution:
We are given:
((1+i)/(1-i))m = 1
Step 1: Simplify the base:
Let’s simplify the base of the expression:
(1+i)/(1-i) = [(1+i)(1+i)] / [(1-i)(1+i)]
Multiplying the numerators and denominators:
= (1 + 2i + i²) / (1 – i²)
Since i² = -1, we get:
= (2i) / 2
= i
So, the equation becomes:
im = 1
Step 2: Find the smallest positive integer value of m:
We know that i⁴ = 1.
Therefore, the smallest positive integer value of m for which im = 1 is m = 4.
so, the least positive integral value of m is 4.
Hope you find these solutions helpful in understanding the concepts of Class 11 Chapter 5 Maths. Click here for solutions to Class 12 Maths Chapter.
