Complex number cannot arive at $\frac{9}{2}-\frac{9}{2}i$ with problem $\frac{4+i}{i}+\frac{3-4i}{1-i}$ I am asked to evaluate: $\frac{4+i}{i}+\frac{3-4i}{1-i}$
The provided solution is: $\frac{9}{2}-\frac{9}{2}i$
I arrived at a divide by zero error which must be incorrect. My working:
$\frac{4+i}{i}$, complex conjugate is $-i$ so:
$\frac{-i(4+i)}{-i*i}$ 
= $\frac{-4i+i^2}{i^2}$ 
= $\frac{-4i--1}{-1}$ 
= $-4i+1$
Then the next part:
$\frac{3-4i}{1-i}$ complex conjugate is $1+i$ so:
$\frac{(1+i)(3-4i)}{(1+i)(1-i)}$
= $\frac{3-4i+3i-4i^2}{1-i^2}$
= $\frac{7-i}{0}$ # 1 + -1 = 0
How can I arrive at $\frac{9}{2}-\frac{9}{2}i$?

Here is your error $1-i^2=1-(-1)=2\ne 0$
