Suppose $f(-1 + i) = 2 + 5i$ and $f(-2 - i) = -3$ determine the remainder of $f(x)$ divided by $(x + 1 - i)(x + 2 + i)$.
I don't really know where to start any help would be great. Thanks :)
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If $f(a)=s$ and $f(b)=t$, then by the polynomial division algorithm there are polynomials $q$ and $r$ such that $f(x)=(x-a)(x-b)q(x) + r(x)$ where $\deg r<2$.
This means $s=f(a)=r(a)$ and $t=f(b)=r(b)$, and since $r(x)=Ax+B$ for some constants $A$ and $B$, one can write by inspection that
Plug in your values $a=-1+i,b=-2-i,s=2+5i,t=-3$ and simplify.
Write $f(z) = h(z)*(z - z_1) + a$, and $f(z) = k(z)*(z - z_2) + b$, multiply the 1st equation by $z - z_2$, and the 2nd equation by $z - z_1$, and subtract the two new equations to get: $(z_1 - z_2)*f(z) = (h(z) - k(z))*(z - z_1)*(z - z_2) + a(z - z_1) - b(z - z_2)$, from this we can find the remainder.