# Subtraction of Complex Numbers

**The Subtraction of complex numbers:**We know that for any complex numbers z

_{1}and z

_{2}, there exists a complex number 'z' such that z

_{1}+ z = z

_{2}. This number 'z' is denoted by z

_{2}- z

_{1}.

Let z

_{1}= a + ib and z

_{2}= c + id and z = x + iy . Then

z

_{1}+ z = z

_{2}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 z

_{1}= a + ib and z

_{2}= c + id z

_{1}- z

_{2}= (a + ib) - (c +id) = (a - c) + i ( b - d) z

_{2}- z

_{1}= (c + id) - (a + ib)= (c - a) + i(d - b) But we know that, a - c ≠ c - a and b - d ≠ d - b ∴ z

_{1}- z

_{2}≠ z

_{2}- z

_{1}

**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 z

_{1}= 8 + i2 and z

_{2}14 - i

**Solution :**z

_{2}- z

_{1}= (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

From subtraction of complex numbers to Home

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