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

To find the value of a and b, we need to use the concept of complex conjugates.

Given that f(x) is a polynomial with real coefficients, this means that if a complex number is a root of the polynomial, its complex conjugate must also be a root. In other words, if f(4+3i)=31+16i, then f(4-3i) must be equal to the complex conjugate of 31+16i.

The complex conjugate of a complex number a+bi is obtained by changing the sign of the imaginary part, which means it is a-bi.

So, to find f(4-3i):

Step 1: Take the complex conjugate of 31+16i.
The complex conjugate of 31+16i is 31-16i.

Step 2: Use the fact that if a+bi is a root of the polynomial, then a-bi must also be a root.

Therefore, f(4-3i) must be equal to 31-16i.

Finally, the value of a+b is obtained by adding the real parts and the imaginary parts of the number we obtained:
a = 31
b = -16

So, a+b = 31 + (-16) = 15.