Important Properties of Conjugate, Modulus & Argument | Mathematics (Maths) Class 11 - Commerce PDF Download

D. Important Properties Of Conjugate / Modulus / Argument

If z , z₁ , z₂C then ;
(a) 

(b) 

|z| ≥ 0  ;  |z| ≥  Re (z)  ;    |z| ≥ Im (z) ; 

(c) 

(i) amp (z₁ . z₂) =  amp  z₁ + amp z₂ + 2 kπ  . k ∈I

(ii) amp  = amp z₁ - amp z₂ + 2 kπ   ;     k∈ I\

(iii) amp(zⁿ) = n amp(z)  +  2kπ .
where proper value of  k  must be chosen  so  that  RHS  lies  in  (- π , π ].

Ex.4 The maximum & minimum values of  are

Sol.  denotes set of points on or inside a circle with centre (- 3, 0) and radius 3.
 denotes the distance of P from A

Ex.5 Let z₁ , z₂ be two complex numbers represented by points on the circle  respectively , then

Sol.

Maximum value of  are collinear.

Ex.6 Prove that if z₁ and z₂ are two complex numbers and c > 0, then

Sol.

Incorporating the number c > 0, the last term on the RHS can be written

Ex.7 If    are the points A, B, C in the Argand Plane such that,
prove that ABC is an equilateral triangle .

Sol . Let z₂ - z₃ = p ; z₃ - z1 = q ; z₁ - z₂ = r ⇒ p + q + r = 0

Given condition, pq + qr + rp = 0   ⇒ p (q + r) + qr = 0   ⇒  p (- p) + qr = 0

⇒ p² = qr 

Ex.8 If z & w are two complex numbers simultaneously satisfying the equations,  then

Sol.

z³ = - w⁵ ⇒ |z| ³ =|w|⁵ ⇒ |z| ⁶ = |w|¹⁰  .....(1)

Ex.9 The complex numbers whose real and imaginary parts are integers and satisfy the relation  forms a rectangle on the Argand plane, the length of whose diagonal is

Sol.

 = 2 (x² + y²) (x² - y²) = 350 ⇒ (x² - y²) (x² + y²) = 175 = 35.5 = 25.7

= x² + y² = 25 & x² - y² = 7 ⇒ x = ± 4 & y = ± 3

Ex.10 Find the area bounded by the curve, arg z   in the complex plane .

Sol.

required area, the equilateral triangle OPQ with side 4

Ex.11 

Find the complex number where the curves arg   intersect.

Sol.

Ex.12 if  then prove that is purely real.

Sol. The given relation can be written as

Ex.13 For every real number a ≥ 0 find all the complex numbers z that satisfy the equation 2|z| – 4 az + 1 + ia = 0.

Sol. We have 2|z| – 4 az + 1 + ia = 0
Put z = x + i y,

We get,   = 4 ax – 1 + 4aiy – ia  or 4(x² + y²) = (4 ax – 1)²    .....(1)

and a = 4 ay (by separating imaginary and real parts)

⇒ y =1/2 and 4x² + 1/4– 16 a² x² – 1 + 8 ax = 0  ⇒ x² (16 - 64 a²) + 32 ax - 3 = 0