# Subtraction of Complex Numbers

The Subtraction of complex numbers: We know that for any complex numbers z1 and z2, there exists a complex number 'z' such that z1 + z = z2. This number 'z' is denoted by z2 - z1.
Let z1 = a + ib and z2 = c + id and z = x + iy . Then
z1 + z = z2 or (a + ib) +(x + iy) = (c + id)
⇒ (a +x) +i(b + y) = c + id
⇒ a + x = c and b + y = d
This system of equation has a unique solution
x = c - a and y = d - b
Thus, x = (c - a) + i(d - b)

## Properties of subtraction of complex numbers

Closure : The subtraction of two complex-numbers is , by definition , a complex number. Hence, the set of complex numbers is closed under subtraction.

Example : (6 + i5) - 8i = 6 + i(5 - 8) = 6 - i3 < From the above we can see that 6 + i5 is a complex number, -i8 is a complex number and the subtraction of these two numbers is 6 - i3 is again a complex number.

Commutative property: For two complex numbers z1 = a + ib and z2 = c + id z1 - z2 = (a + ib) - (c +id) = (a - c) + i ( b - d) z2 - z1 = (c + id) - (a + ib)= (c - a) + i(d - b) But we know that, a - c ≠ c - a and b - d ≠ d - b ∴ z1 - z2 ≠ z2 - z1

Associative property does not hold true for subtraction of complex-numbers.

Examples on subtraction of complex-numbers

1) Find the difference of the complex numbers z1= 8 + i2 and z2 14 - i
Solution : z2- z1 = (14 - i ) - ( 8 + i2)
= (14 -8) + (-i - i2)
= 6 + (-3i)
= 6 - i3

2) Subtract the following complex numbers:
a) (5+4i)−(2+9i)
= (5 + 4i) + (-2 - 9i) ------(Distribute the negative sign)
= ( 5 + - 2) +( 4i + -9i)
= 3 + (-5i)
= 3 - 5i

b) (18+14i)−(3−3i)
= (18 +14i) + (-3+ 3i) -----(distribute the negative sign)
= (18 + -3) + (14i + 3i)
= 15 + 17i

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