Q1: Express the given complex number in the form a + ib:
Ans:
Q2: Express the given complex number in the form a + ib: i^{9}+ i^{19}
Ans:
Q3: Express the given complex number in the form a + ib: i^{–39}
Ans:
Q4: Express the given complex number in the form a + ib: 3(7 + i7) + i(7 + i7)
Ans:
Q5: Express the given complex number in the form a + ib: (1 – i) – (–1 i6)
Ans:
Q6: Express the given complex number in the form a + ib:
Ans:
Q7: Express the given complex number in the form a + ib:
Ans:
Q8: Express the given complex number in the form a + ib: (1 – i)^{4}
Ans:
Q9: Express the given complex number in the form a + ib:
Ans:
Q10: Express the given complex number in the form a + ib:
Ans:
Q11: Find the multiplicative inverse of the complex number 4 – 3i
Ans:
Let z = 4 – 3i
Then, = 4 +3i and
Therefore, the multiplicative inverse of 4 – 3i is given by
Q12: Find the multiplicative inverse of the complex number
Ans:
Let z =
Therefore, the multiplicative inverse of is given by
Q13: Find the multiplicative inverse of the complex number –i
Ans: Let z = –i
Therefore, the multiplicative inverse of –i is given by
Q14: Express the following expression in the form of a + ib.
Ans:
Question 1: Evaluate:
Answer:
Question 2: For any two complex numbers z_{1} and z_{2}, prove that
Re (z_{1}z_{2}) = Re z_{1 }Re z_{2} – Im z_{1} Im z_{2}
Answer:
Question 3: Reduce to the standard form.
Answer:
[ On Multiplying numerator and denomunator by (14 + 5i)
Question 4: If x – iy = prove that .
Answer:
[ On Multiplying numerator and denomunator by (c + id)]
Question 5: Convert the following in the polar form:
(i) ,
(ii)
Answer:
(i) Here,
Let r cos θ = –1 and r sin θ = 1
On squaring and adding, we obtain
r^{2} (cos^{2} θ + sin^{2} θ) = 1+ 1
⇒ r^{2} (cos^{2} θ + sin^{2} θ) = 2
⇒ r^{2} = 2 [cos^{2} θ + sin^{2} 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
r^{2} (cos^{2} θ + sin^{2} θ) = 1+ 1
⇒r^{2} (cos^{2} θ +sin^{2} θ) = 2
⇒ r^{2} = 2 [cos^{2} θ + sin^{2} θ = 1]
∴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 ax^{2} + bx + c = 0, we obtain
a = 9, b = –12, and c = 20
Therefore, the discriminant of the given equation is
D = b^{2} – 4ac = (–12)^{2} – 4 × 9 × 20 = 144 – 720 = –576
Therefore, the required solutions are
Question 7: Solve the equation
Answer:
The given quadratic equation is
This equation can also be written as
On comparing this equation with ax^{2} + bx + c = 0, we obtain
a = 2, b = –4, and c = 3
Therefore, the discriminant of the given equation is
D = b^{2} – 4ac = (–4)^{2} – 4 × 2 × 3 = 16 – 24 = –8
Therefore, the required solutions are
Question 8: Solve the equation 27x^{2} – 10x + 1 = 0
Answer:
The given quadratic equation is 27x^{2} – 10x + 1 = 0
On comparing the given equation with ax^{2} + bx + c = 0, we obtain
a = 27, b = –10, and c = 1
Therefore, the discriminant of the given equation is
D = b^{2} – 4ac = (–10)^{2} – 4 × 27 × 1 = 100 – 108 = –8
Therefore, the required solutions are
Question 9: Solve the equation 21x^{2} – 28x + 10 = 0
Answer:
The given quadratic equation is 21x^{2} – 28x + 10 = 0
On comparing the given equation with ax^{2} + bx + c = 0, we obtain
a = 21, b = –28, and c = 10
Therefore, the discriminant of the given equation is
D = b^{2} – 4ac = (–28)^{2} – 4 × 21 × 10 = 784 – 840 = –56
Therefore, the required solutions are
Question 9: If find .
Answer:
Question 10: If a + ib = , prove that a^{2} + b^{2} =
Answer:
On comparing real and imaginary parts, we obtain
Hence, proved.
Question 10: Let . Find
(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 11: Find the modulus and argument of the complex number .
Answer:
Let , then
On squaring and adding, we obtain
Therefore, the modulus and argument of the given complex number are respectively.
Question 12: Find the real numbers x and y if (x – iy) (3+5i) is the conjugate of –6 – 24i.
Answer:
Let
It is given that,
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 13: Find the modulus of .
Answer:
Question 14: If (x + iy)^{3} = u + iv, then show that .
Answer:
On equating real and imaginary parts, we obtain
Hence, proved.
Question 15: 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,
Question 16: Find the number of nonzero integral solutions of the equation .
Answer:
Thus, 0 is the only integral solution of the given equation. Therefore, the number of nonzero integral solutions of the given equation is 0.
Question 17: If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that
(a^{2} + b^{2}) (c^{2}+ d^{2}) (e^{2} + f^{2}) (g^{2} + h^{2}) = A^{2} +B^{2}.
Answer:
On squaring both sides, we obtain
(a^{2} + b^{2}) (c^{2} + d^{2}) (e^{2} + f^{2}) (g^{2} + h^{2}) = A^{2} + B^{2}
Hence, proved.
Question 18: If , then find the least positive integral value of m.
Answer:
Therefore, the least positive integer is 1.
Thus, the least positive integral value of m is 4 (= 4 × 1).
209 videos443 docs143 tests

1. What are complex numbers? 
2. How do you add complex numbers? 
3. How do you multiply complex numbers? 
4. How do you find the conjugate of a complex number? 
5. How do you divide complex numbers? 
209 videos443 docs143 tests


Explore Courses for JEE exam
