how do you properly distribute 1/3(1-3i)?

Question

it is apart of 3i+[1/2]-[1/3](1-3i)+2

Answer 1

As can be seen by the grouping in the

original problem, and taking into account

the minus sign before this, this is supposed to be

 

-(1/3)(1-3i)

= -[(1/3)(1) + (1/3)(-3i)]

= -(1/3 - i)

= -1/3 + i

 

3i + [1/2] - [1/3](1-3i) + 2

= 3i + 1/2 - 1/3 + i + 2

= 2 + 1/2 - 1/3 + 3i + i

= 5/2 - 1/3 + 4i

= 13/6 + 4i

 

or:  2 1/6 + 4i

 

 

Answer 2

1/3(1-3i) = 1/(3-9i)