Question 11. If a + ib = , prove that a2 + b2 =
a + ib =
On comparing the real and imaginary parts, we have
a = and b =
a2 + b2 =
a2 + b2 =
Question 12. Let z1 = 2 – i, z2 = -2 + i. Find
z1 = 2 – i, z2 = -2 + i
(i) z1z2 = (2 – i)(-2 + i) = -4 + 2i + 2i – i2 = -4 + 4i – (-1) = -3 + 4i
= 2 + i
On multiplying numerator and denominator by (2 – i), we get
On comparing the real parts, we have
On comparing the imaginary part, we get
Question 13. Find the modulus and argument of the complex number
Let, z = , then
Let z = r cosθ + ir sinθ
r cosθ = and r sinθ =
On squaring and adding, we get
r2(cos2θ + sin2θ) =
cosθ = and sinθ =
= cosθ = and sinθ =
θ = [As θ lies in the II quadrant]
Question 14. Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of – 6 – 24i.
Let us assume z = (x – iy) (3 + 5i)
z = 3x + xi – 3yi – 5yi2 = 3x + 5xi – 3yi + 5y = (3x + 5y) + i(5x – 3y)
=(3x + 5y) – i(5x – 3y)
Also given, = -6 – 24i
(3x + 5y) – i(5x – 3y) = -6 -24i
After equating real and imaginary parts, we get
3x + 5y = -6 …… (i)
5x – 3y = 24 …… (ii)
After doing (i) x 3 + (ii) x 5, we have
(9x + 15y) + (25x – 15y) = -18 + 120
34x = 102
x = 102/34 = 3
Putting the value of x in equation (i), we get
3(3) + 5y = -6
5y = -6 – 9 = -15
y = -3
Therefore, the values of x and y are 3 and –3 respectively.
Question 15. Find the modulus of
Question 16. If (x + iy)3 = u + iv, then show that = 4(x2 – y2)
(x + iy)3 = u + iv
x3 + (iy)3 + 3 × x × iy(x + iy) = u + iv
x3 + i3y3 + 3x2yi + 3xy2 = u + iv
x3 – iy3 + 3x2yi – 3xy2 = u + iv
(x3 – 3xy2) + i(3x2y – y3) = u + iv
On equating real and imaginary parts, we get
u = x3 – 3xy2, v = 3x2y – y3
= x2 – 3y2 + 3x2 – y2
= 4x2 – 4y2
= 4(x2 – y2)
Question 17. If α and β are different complex numbers with |β| = 1, then find
Assume α = a + ib and β = x + iy
Given: |β| = 1
= x2 + y2 = 1 ….(1)
Question 18. Find the number of non-zero integral solutions of the equation |1 – i|x = 2x
|1 – i|x = 2t
x = 2x
2x – x = 0
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.
(a + ib)(c + id)(e + if)(g + ih) = A + iB
|(a + ib)(c + id)(e + if)(g + ih)| = |A + iB|
= |(a + ib)| × |(c + id)| × |(e + if)| × |(g + ih)| = |A + iB|
On squaring both sides, we get
(a2 + b2) (c2 + d2) (e2 + f2) (g2 + h2) = A2 + B2
Question 20. If, then find the least positive integral value of m.
im = 1
Hence, m = 4k, where k is some integer.
Hence, the least positive integer is 1.
Thus, the least positive integral value of m is 4 (= 4 × 1).
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