Subtraction of Complex Numbers
The Subtraction of complex numbers:
We know that for any complex numbers z1
, there exists a complex number 'z' such that z1
+ z = z2
. This number 'z' is denoted by z2
= a + ib and z2
= c + id and z = x + iy . Then
+ 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
The subtraction of two complex-numbers is , by definition , a complex number. Hence, the set of complex numbers is closed under subtraction.
(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.
For two complex numbers z1
= a + ib and z2
= c + id
= (a + ib) - (c +id) = (a - c) + i ( b - d)
= (c + id) - (a + ib)= (c - a) + i(d - b)
But we know that, a - c ≠ c - a and b - d ≠ d - b
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
= (14 - i ) - ( 8 + i2)
= (14 -8) + (-i - i2)
= 6 + (-3i)
= 6 - i3
2) Subtract the following complex numbers:
= (5 + 4i) + (-2 - 9i) ------(Distribute the negative sign)
= ( 5 + - 2) +( 4i + -9i)
= 3 + (-5i)
= 3 - 5i
= (18 +14i) + (-3+ 3i) -----(distribute the negative sign)
= (18 + -3) + (14i + 3i)
= 15 + 17i
11th grade math
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