## Rajasthan Board RBSE Class 11 Maths Chapter 5 Complex Numbers Miscellaneous Exercise

Question 1.

Real and imaginary parts of complex number \(\frac { 1+i }{ 1-i }\) respectively are:

(A) 1, 1

(B) 0, 0

(C) 0, 1

(D) 1, 0

Solution:

Real part = 0

Imaginary part = 1

Hence, option (C) is correct.

Question 2.

If 2 + (2a + 5ib) = 8 + 10i, then:

(A) a = 2, b = 3

(B) a = 2, b = -3

(C) a = 3, b = 2

(D) a = 3, b = -2

Solution:

2 + (2a + 5ib) = 8 + 10i

⇒ (2 + 2a) + 5bi = 8 + 10i

On comparing

2 + 2a = 8

⇒ 2a =6

⇒ a = 3

and 5b = 10

⇒ b = 2

So, a = 3 and b = 2

Hence, option (C) is correct.

Question 3.

The multiplicative inverse of 3 – i is:

Solution:

Let z = 3 – i

Let multiplicative inverse of z

Question 4.

Solution:

Question 5.

If z_{1}, z_{2} ∈ C then which statement is true:

(A) |z_{1} – z_{2}| ≥ |z_{1}| + |z_{2}|

(B) |z_{1} – z_{2}| ≤ |z_{1}| + |z_{2}|

(C) |z_{1} + z_{2}| ≥ |z_{1} – z_{2}|

(D) |z_{1} + z_{2}| ≤ |z_{1} – z_{2}|

Solution:

statement |z_{1} – z_{2}| ≤ |z_{1}| + |z_{2}| is true.

Hence, option (B) is correct.

Question 6.

If |z – 3| = |z + 3|, then z lies in

(A) at x-axis

(B) at y-axis

(C) on line x = y

(D) on line x = -y

Solution:

Let z = x + iy

It is situated on the axis.

Hence, option (B) is correct.

Question 7.

Write the principal argument of -2.

Solution:

-2 = -2 + 0i

Let -2 + 0i = r(cos θ + i sin θ)

⇒ r cos θ = -2 ….(i)

r sin θ = 0 ………(ii)

⇒ tan θ = \(\frac { 0 }{ -2 }\) = 0

Hence, principal arument of -z is π.

Question 8.

Write polar form of \(\frac { 5\surd 3 }{ 2 } +\frac { 5 }{ 2 } i\)

Solution:

Question 9.

What is the value of 4 + 5ω^{4} + 3ω^{5}?

Solution:

Question 10.

Write \(\frac { 1 }{ 1-cos\theta +isin\theta }\) in the form of a + ib.

Solution:

Question 11.

What is the number of non-zero integer roots of |1 – i|^{x} = 2^{x}?

Solution:

Comparing power of both sides

\(\frac { x }{ 2 }\) = x, which is not possible (because x is not zero)

Thus, the root of equation |1 – i|^{x} = 2^{x} is not possible.

Hence, the number of roots is zero.

Question 12.

If z_{1}, z_{2} ∈ C prove that

(i) |z_{1} – z_{2}| ≤ |z_{1}| + |z_{2}|

(ii) |z_{1} + z_{2}| ≥ |z_{1}| – |z_{2}|

(iii) |z_{1} + z_{2} + ….. + z_{n}| ≤ |z_{1}| + |z_{2}| + ……. + |z_{n}|

Solution:

Question 13.

If |z_{1}| = 1 = |z_{2}| then prove that

Solution:

According to question,

Question 14.

Solution:

Question 15.

If |z_{1}| = |z_{2}| and arg z_{1} + arg z_{2} = 0 then prove that z_{1} = \(\bar { { z }_{ 2 } }\).

Solution:

Question 16.

If θ_{1} and θ_{2} are arguments of complex numbers z_{1} and z_{2} respectively then prove that:

z_{1}\(\bar { { z }_{ 2 } }\) + \(\bar { { z }_{ 1 } }\)z_{2} = 2|z_{1}||z_{2}| cos(θ_{1} – θ_{2})

Solution:

Question 17.

Prove that:

(i) (a + bω + cω^{2}) (a + bω^{2} + cω) = a^{2} + b^{2} + c^{2} – ab – bc – ca

(ii) (a + b + c) (a + bω + cω^{2}) (a + bω^{2} + cω) = a^{3} + b^{3} + c^{3} – 3abc

Solution:

Question 18.

If α, β are two complex numbers and |β| = 1 then prove that \(\left| \frac { \beta -\alpha }{ 1-\bar { \alpha } \beta } \right| =1\)

Solution:

Question 19.

If α and β roots of equation px^{2} – qx – r = 0, then find the equation which has roots 1/α and 1/β

Solution:

α, β are roots of equation px^{2} – qx – r = 0

⇒ rx^{2} + qx – p = 0

Hence, required equation is rx^{2} + qx – p = 0

Question 20.

Find conditions for which equation lx^{2} – 2mx + n = 0 has roots such that one root is P times the other roots.

Solution:

Given equation lx^{2} – 2mx + n = 0

Let it’s one root is a and another root is Pα.

Then, the sum of roots

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