    # Properties of Complex Numbers • If x, y are real and x + iy = 0 then x = 0, y = 0.

• Proof:

Since, x + iy = 0 = 0 + i0, hence by the definition of equality of two complex numbers it follows that, x = 0 and y = 0.

• If x, y, p, q are real and x + iy = p + iq then x = p and y = q.

• Proof:

Since x + iy = p + iq

Hence x − p = -i(y − q) ⇒ (x − p)2 = i2 (y − q)2

⇒ (x − p)2 + (y − q)2 = 0 (Since i2 = -1)          (1)

Since x, y, p, q are real, (x − p)2 and (y − q)2are both non-negative. Hence the equation (1) issatisfied if each square is separately zero. Hence,

(x − p)2 = 0 or x = p and (y − q)2 = 0 or y = q.

• Like real numbers, the set of complex numbers also satisfies the commutative, associative and distributive laws i.e., if z1, z2 and z3 be three complex numbers then,

• z1 + z2 = z2 + z1 (commutative law for addition) and z1. z2 = z2. z1 (commutative law for multiplication).

• (z1 + z2+ z3 = z1 + (z2 + z3) (associative law for addition) and (z1 z2) z3 = z1 (z2 z3) (associative law for multiplication).

• z1(z2 + z3) = z1 z2 + z1 z3 (distributive law).

• The sum and product of two conjugate complex quantities are both real.

• Proof:

Let z = x + iy be a complex number where x, y are real.

Then, conjugate of z is = x − iy.

Now, z + = x + iy + x − iy = 2x, which is real.

And z. = (x + iy)(x − iy) = x2 − i2y2 = x2 + y2 which is also real.

• If the sum and product of two complex quantities are both real then the complex quantities are conjugate to each other.

• Proof:

Let, z1 = a + ib and z2 = c + id be two complex quantities where a, b, c, d are real and b ≠ 0,

d ≠0.

By hypothesis, z1 + z2 = a + ib + c + id = (a + c) + i(b + d) is real.

Hence b + d = 0 or d = -b

And z1. z2 = (a + ib)( c + id) = (ac − bd) + i(ad + bc) is real.

Hence ad + bc = 0 or −ab + bc = 0 (Since d = -b)

Or b(c − a) = 0 or c = a (Since b ≠ 0)

Hence z2 = c + id = a + i(-b) = a − ib = , which proves that z1 and z2 are conjugate to each other.

• For two complex quantities z1 and z2, show that

• |z1+ z2 | ≤ |z1 | + |z2 |

Proof:

Let z1 = r1(cosθ1 + isinθ1 ) and z2 = r2(cosθ2 + isinθ2 ).

Hence |z1 | = r1 and |z2 | = r2

Now

z1 + z2 = r1(cosθ1isinθ1) + r2(cosθ2 + isinθ2)

= (r1cosθ1+ r2cosθ2 )+ i(r1sinθ1+ r2sinθ2)

Hence |z1+ z2 | = √(r1cosθ1+ r2cosθ2)2 + (r1sinθ1+ r2sinθ2)2

= √r12(cos2θ1+ sin21) + r22(cos2θ2+ sin2θ2) + 2r1r2 (cosθ1 cosθ2+ sinθ1 sinθ2)

= √r12 + r22 + 2r1r2cos (θ1- θ2)

Now, |cos(θ1- θ2)| ≤ 1

Hence |z1+ z2| ≤ √r12 + r22 + 2r1r2 or |z1+ z2 | ≤ |z1| + |z2 |

Example 1:

If x, y are real and (x + 3i) and (-2 + iy) are conjugate of each other, find x and y.

Solution:

Since (x + 3i) is conjugate of complex quantity (-2 + iy)

Hence (x + 3i) = = -2 − iy

Hence x = -2 and y = -3. (Answer)

Example 2: If z = x + iy and |2z- 1| = |z- 2| then prove that x2 + y2 = 1.

Solution:

|2z- 1| = |z- 2| ⇒ |2(x+iy)- 1| = |x+iy- 2|

⇒ |(2x-1)+2iy| = |(x-2)+iy|

⇒ √(2x − 1) 2 + 4y2 = √(x − 2)2+ y2

⇒ 4x2 - 4x + 1 + 4y2 = x2 − 4x + 4 + y2 (squaring on both sides)

⇒ 3x2 + 3y2 - 3 = 0

⇒ 3x2 + 3y2 = 3

⇒ x2 + y2 = 1. (Proved)

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