If $A=4-3i$, $M=-4+i$, $S=i$, and $P=2$, find $A-M+S-P$.
Adding real parts and imaginary parts separately, we have $(4-(-4)+0-2)+(-3-1+1-0)i=\boxed{6-3i}$.
6-3i
Level 3
Algebra
Problem: If $A=4-3i$, $M=-4+i$, $S=i$, and $P=2$, find $A-M+S-P$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve for $A-M+S-P$, we first need to substitute the given values into the expression. 

1. Identify the values: 
   - $A = 4 - 3i$
   - $M = -4 + i$
   - $S = i$
   - $P = 2$

2. Substitute these values into the expression:
   - $A - M + S - P = (4 - 3i) - (-4 + i) + i - 2$

3. Simplify the expression step-by-step:
   - Start with $A - M$: 
     - $A - M = (4 - 3i) - (-4 + i) = 4 - 3i + 4 - i = 8 - 4i$
   - Now add $S$: 
     - $8 - 4i + i = 8 - 4i + 1i = 8 - 3i$
   - Finally, subtract $P$: 
     - $8 - 3i - 2 = 6 - 3i$

4. The final result after performing all operations is $6 - 3i$. 

Thus, the final answer is $6 - 3i$.<|END_OF_LATENT|> The final answer is $6-3i$.