Chapter 1 Complex Numbers Ex 1.3 Solutions

Maharashtra State Board 11th Maths Solutions Chapter 1 Complex Numbers Ex 1.3

(ii) √3 + √2 i
Solution:
Let z = √3 + √2 i
a = √3, b = √2,
i.e. a > 0, b > 0

(iii) -8 + 15i
Solution:
Let z = -8 + 15i
a = -8, b = 15, i.e. a < 0, b > 0

(iv) -3(1 – i)
Solution:
Let z = -3(1 – i) = -3 + 3i
a = -3, b = 3, i.e. a < 0, b > 0
|z| = = 3√2
Here, (-3, 3) lies in 2nd quadrant.
amp(z) = π –

(v) -4 – 4i
Solution:
Let z = -4 – 4i
a = -4, b = -4, i.e. a < 0, b < 0

(vi) √3 – i
Solution:
Let z = √3 – i
a = √3, b = -1, i.e. a > 0, b < 0

(vii) 3
Solution:
Let z = 3 + 0i
a = 3, b = 0
z is a real number, it lies on the positive real axis.
|z|= = 3
and amp (z) = 0

(viii) 1 + i
Solution:
Let z = 1 + i
a = 1, b = 1, i.e. a > 0, b > 0
|z| =
Here, (1, 1) lies in 1st quadrant.
amp (z) =

(ix) 1 + i√3
Solution:
Let z = 1 + i√3
a = 1, b = √3, i.e. a > 0, b > 0
|z| =
Here, (1, √3) lies in 1st quadrant.
amp (z) =

(x) (1 + 2i)²(1 – i)
Solution:
Let z = (1 + 2i)²(1 – i)
= (1 + 4i + 4i²) (1 – i)
= [1 + 4i + 4(-1)] (1 – i) ….[∵ i²= -1]
= (-3 + 4i) (1 – i)
= -3 + 3i + 4i – 4i²
= -3 + 7i – 4(-1)
= -3 + 7i + 4
∴ z = 1 + 7i
∴ a = 1, b = 7, i. e. a > 0, b > 0
∴ |z| =
Here, (1, 7) lies in 1st quadrant.
∴ amp(z) =

(ii) -i
Solution:
Let z = -i = 0 – i
a = 0, b = -1
z lies on negative imaginary Y-axis.
|z| = r = = 1 and
θ = amp z = 270° =
The polar form of z = r (cos θ + i sin θ)
= 1 (cos 270° + i sin 270°)
= 1 (cos + i sin )
The exponential form of z =

(iii) -1
Solution:
Let z = -1 = -1 + 0.i
a = -1, b = 0
z lies on negative real X-axis.
|z| = r = = 1 and
θ = amp z = 180° = π
The polar form of z = r (cos θ + i sin θ)
= 1 (cos 180° + i sin 180°)
= 1 (cos π + i sin π)
The exponential form of z =

(iv)
Solution:

(v)
Solution:

(vi)
Solution:

(ii)
Solution:

(iii)
Solution:

(iv)
Solution:

(v)
Solution:

(vi) [/latex]
Solution:

(ii)
Solution:
z = (2 + 3i) (2 – 3i)
= 4 – 9i²
= 4 – 9(-1) …..[∵ i²= -1]
= 13
|z|²=
= 2²+ 3²
= 4 + 9
= 13
∴

(iii) (z + ) is real
Solution:
(z + ) = (2 + 3i) + (2 – 3i)
= 2 + 3i + 2 – 3i
= 4, which is a real number.
∴ z + is real.

(iv) z – = 6i
Solution:
z – = (2 + 3i) – (2 – 3i)
= 2 + 3i – 2 + 3i
= 6i

(ii)
Solution:

(iii)
Solution:

(iv)
Solution: