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Addition and scalar multiplication of complex numbers obey the same familiar algebraic rules as real numbers. Under those rules, expressions can be rearranged and simplified without changing their value.
Let z 1 z_1 z 1 , , and be any complex numbers, and let and be any scalars (real numbers).
Changing the order of two complex addends does not change their sum.
Example: ( 2 + i ) + ( 1 − 3 i ) = ( 1 − 3 i ) + ( 2 + i ) = 3 − 2 i (2+i) + (1-3i) = (1-3i) + (2+i) = 3-2i ( 2 + i ) + ( 1 − 3 i ) = ( 1 − 3 i ) + ( 2 + i ) = 3 − 2 i
When adding three complex numbers, the grouping of the addition does not affect the result.
There exists a complex number 0 = 0 + 0 i 0 = 0 + 0i 0 = 0 + 0 i (zero) such that when added to any complex number z 1 z_1 z 1 , the result is z 1 z_1 z 1 itself.
Every complex number z 1 = x + i y z_1 = x + iy z 1 = x + i y has an additive inverse (opposite), denoted by − z 1 = − x − i y -z_1 = -x - iy − z 1 = − x − i y , such that their sum is the zero element (0 0 0 ).
If z 1 = 5 − 2 i z_1 = 5-2i z 1 = 5 − 2 i , then − z 1 = − 5 + 2 i -z_1 = -5+2i − z 1 = − 5 + 2 i .
Then ( 5 − 2 i ) + ( − 5 + 2 i ) = ( 5 − 5 ) + i ( − 2 + 2 ) = 0 + 0 i = 0 (5-2i) + (-5+2i) = (5-5) + i(-2+2) = 0 + 0i = 0 ( 5 − 2 i ) + ( − 5 + 2 i ) = ( 5 − 5 ) + i ( − 2 + 2 ) = 0 + 0 i = 0 .
The grouping of scalar multiplication does not affect the result.
Multiplying a complex number by a sum of scalars is equivalent to multiplying it by each scalar and then adding the results.
A scalar can be distributed over the addition of complex numbers.
Multiplying a complex number by the scalar 1 1 1 does not change the complex number.
Multiplying a complex number by the scalar 0 0 0 results in the complex number zero.
The next example uses distributivity and multiplication by the zero scalar to verify an identity.
Show that for any complex number z z z , 4 z + ( − 4 ) z = 0 4z + (-4)z = 0 4 z + ( − 4 ) z = 0 holds.
Use distributivity to combine the scalar coefficients, then use the fact that multiplying by the zero scalar gives zero.
It is proven that 4 z + ( − 4 ) z = 0 4z + (-4)z = 0 4 z + ( − 4 ) z = 0 .
Using the properties above, prove that 3 z − 1 2 ( 2 z ) = 2 z 3z - \frac{1}{2}(2z) = 2z 3 z − 2 1 ( 2 z ) = 2 z for any complex number z z z .
It is proven that 3 z − 1 2 ( 2 z ) = 2 z 3z - \frac{1}{2}(2z) = 2z 3 z − 2 1 ( 2 z ) = 2 z .
Published: May 1, 2025 . Updated: Sep 1, 2026 .