Perform the indicated operation and write the answer in standard form.
1)(-3+8i)+(3-6i)
Treat just like any binomials, but remember that i^2 = -1
(-3+8i)(3-6i)
-9 + 18i + 24i - 48i^2
-9+48 + 42i
39+42i
add the terms ... like to like
-3 + 3 = 0 ... 8i + -6i = 2i
2i
oops. I missed that + sign ... heh heh
To perform the indicated operation (-3+8i)+(3-6i), we simply add the real parts separately and the imaginary parts separately.
The real part of the sum (-3+8i) + (3-6i) is (-3 + 3) = 0.
The imaginary part of the sum (-3+8i) + (3-6i) is (8i - 6i) = 2i.
So, the sum of (-3+8i) and (3-6i) is 0 + 2i, which can be written in standard form as 2i.