ok i have to divide complex numbers:
question is: 1-2i+i^3/1+i
this is what i did:
1-2i+i(-i)/1+i
1-2i-i^2/1+i
1-2i-i(-1)/1+i
1-2i+i/1+i
1-3i/1+i*1-i/1-i
1-i-3i+3i^2/1-i+i-i^2
1-i-3i+3(-1)/1-i^2
1-i-3i-3/1-(-1)
1-4i-3/2
-2-4i/2
my answer is -2i-1
is that right?
2 answers
Be more specific with parentheses. Does the 1+i divide into only the i^3 term? Or does it divide into 1 - 2i + i^3?
from what you showed in your work I will asssume you meant
(1-2i+i^3)/(1+i)
you messed up in your first line.
remember i^3 = i(i^2) = i(-1) = -i
so
(1-2i+i^3)/(1+i)
= (1-2i - i)/1+i)
= (1-3i)/(1+i)
now multiply top and bottom by 1-i to get
(1-4i - 3i^2)/(1 - i^2)
= (4-4i)/2
= 2 - 2i
(1-2i+i^3)/(1+i)
you messed up in your first line.
remember i^3 = i(i^2) = i(-1) = -i
so
(1-2i+i^3)/(1+i)
= (1-2i - i)/1+i)
= (1-3i)/(1+i)
now multiply top and bottom by 1-i to get
(1-4i - 3i^2)/(1 - i^2)
= (4-4i)/2
= 2 - 2i