Properties of Complex Numbers
- If x, y are real and x + iy = 0 then x = 0, y = 0.
- If x, y, p, q are real and x + iy = p + iq then x = p and y = q.
- Like real numbers, the set of complex numbers also satisfies the commutative, associative and distributive laws i.e., if z_{1}, z_{2} and z_{3} be three complex numbers then,
- z_{1} + z_{2} = z_{2} + z_{1} (commutative law for addition) and z_{1}. z_{2} = z_{2}. z_{1} (commutative law for multiplication).
- (z_{1} + z_{2}+ z_{3} = z_{1} + (z_{2} + z_{3}) (associative law for addition) and (z_{1} z_{2}) z_{3} = z_{1} (z_{2} z_{3}) (associative law for multiplication).
- z_{1}(z_{2} + z_{3}) = z_{1} z_{2} + z_{1} z_{3} (distributive law).
- The sum and product of two conjugate complex quantities are both real.
- If the sum and product of two complex quantities are both real then the complex quantities are conjugate to each other.
- For two complex quantities z_{1} and z_{2}, show that
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.
Proof:
Since x + iy = p + iq
Hence x − p = -i(y − q) ⇒ (x − p)^{2} = i^{2} (y − q)^{2}
⇒ (x − p)^{2} + (y − q)^{2} = 0 (Since i^{2} = -1) (1)
Since x, y, p, q are real, (x − p)^{2} and (y − q)^{2}are 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.
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) = x^{2} − i^{2}y^{2} = x^{2} + y^{2} which is also real.
Proof:
Let, z_{1} = a + ib and z_{2} = c + id be two complex quantities where a, b, c, d are real and b ≠ 0,
d ≠0.
By hypothesis, z_{1} + z_{2} = a + ib + c + id = (a + c) + i(b + d) is real.
Hence b + d = 0 or d = -b
And z_{1}. z_{2} = (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 z_{2} = c + id = a + i(-b) = a − ib = , which proves that z_{1} and z_{2} are conjugate to each other.
|z_{1}+ z_{2} | ≤ |z_{1} | + |z_{2} |
Proof:
Let z_{1} = r_{1}(cosθ_{1} + isinθ_{1} ) and z_{2} = r_{2}(cosθ_{2} + isinθ_{2} ).
Hence |z_{1} | = r_{1} and |z_{2} | = r_{2}
Now
z_{1} + z_{2} = r_{1}(cosθ_{1}isinθ_{1}) + r_{2}(cosθ_{2} + isinθ_{2})
= (r_{1}cosθ_{1}+ r_{2}cosθ_{2} )+ i(r_{1}sinθ_{1}+ r_{2}sinθ_{2})
Hence |z_{1}+ z_{2} | = √(r_{1}cosθ_{1}+ r_{2}cosθ_{2})_{2} + (r_{1}sinθ_{1}+ r_{2}sinθ_{2})_{2}
= √r_{1}2(cos_{2}θ_{1}+ sin2_{1}) + r_{2}2(cos2θ_{2}+ sin2θ_{2}) + 2r_{1}r_{2} (cosθ_{1} cosθ_{2}+ sinθ_{1} sinθ_{2})
= √r1_{2} + r2_{2} + 2r_{1}r_{2}cos (θ_{1}- θ_{2})
Now, |cos(θ_{1}- θ_{2})| ≤ 1
Hence |z_{1}+ z_{2}| ≤ √r1_{2} + r2_{2} + 2r_{1}r_{2} or |z_{1}+ z_{2} | ≤ |z_{1}| + |z_{2} |
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 x^{2} + y^{2} = 1.
Solution:
|2z- 1| = |z- 2| ⇒ |2(x+iy)- 1| = |x+iy- 2|
⇒ |(2x-1)+2iy| = |(x-2)+iy|
⇒ √(2x − 1) ^{2} + 4y^{2} = √(x − 2)^{2}+ y^{2}
⇒ 4x^{2} - 4x + 1 + 4y^{2} = x^{2} − 4x + 4 + y^{2} (squaring on both sides)
⇒ 3x^{2} + 3y^{2} - 3 = 0
⇒ 3x^{2} + 3y^{2} = 3
⇒ x^{2} + y^{2} = 1. (Proved)
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 x^{2} + y^{2} = 1.
Solution:
|2z- 1| = |z- 2| ⇒ |2(x+iy)- 1| = |x+iy- 2|
⇒ |(2x-1)+2iy| = |(x-2)+iy|
⇒ √(2x − 1) ^{2} + 4y^{2} = √(x − 2)^{2}+ y^{2}
⇒ 4x^{2} - 4x + 1 + 4y^{2} = x^{2} − 4x + 4 + y^{2} (squaring on both sides)
⇒ 3x^{2} + 3y^{2} - 3 = 0
⇒ 3x^{2} + 3y^{2} = 3
⇒ x^{2} + y^{2} = 1. (Proved)
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