posted by Tony on .
If f(x) is a polynomial with real coefficients such that f(4+3i)=31+16i, then f(4−3i)=a+bi. What is the value of a+b
f(x) = ax^n + bx^n-1 + ...
f(h+ki) = a(h+ki)^n + b(h+ki)^n-1 + ..
f(h-ki) = a(h-ki)^n + b(h-ki)^n-1 + ...
All of the even powers of i remain unchanged, and all of the odd powers of i change sign. So, the real part is unchanged, and the imaginary part changes sign.
f(4-3i) = 31-16i