They want to find the values of a where g(t) can be divided by f(t).
$f(t) = t^2 + it − ai$
$g(t) = t^4 + (1 − i)t^3 + (1 − 2i)t^2 − 3at − (4 + 2i)a$
Euclidean algorithm:
$g(t) = f(t)q(t) + r(t)$
The mark scheme say that they have found the values of q and r. How did they manage to get the values below?? And why does r have to be 0??
$q(t) = t^2 + (1 − 2i)t + (−1 − 3i + ai)$
$r(t) = (−3 + i + ai)t + a(−1 − a − 3i)$
g(t) can be divided by f(t) if and only if r(t) = 0. Meaning (−3 + i + ai) = 0 and a(−1 − a − 3i) = 0 which gives a = −1 − 3i. So g(t) can be divided by f(t) if and only if a = −1 − 3i
PLEASE HELP!!
The division algorithm is a bit wasteful since here we need only the remainder, not the quotient. Compute $\,g\,$ mod $\,f = t^2\! + it -ai\,$ using $\,\color{#c00}{t^2 \equiv -it+ai}\,$ to eliminate all powers of $\,t\,$ higher than $2$. In particular we can write $\, t^4\! + b t^3\! + c t^2 = \color{#c00}{t^2}(\color{#c00}{t^2} + b t + c)\,$ so replacing $\color{#c00}{t^2}$ by $\,\color{#c00}{-it+ai}\,$in both factors yields a product of linear factors. Expand that then, again, replace $\,\color{#c00}{t^2}$ to get a linear result. Adding that to the linear part of $\,g\,$ yields the claimed remainder $\,r(t).$
Why must $\,r = 0?\,$ Notice that $\, f\mid q f + r\iff f\mid r\iff r = 0,\,$ since $\,\deg r < \deg f.\,$ Said in words, $\,g\,$ is divisible by $\,f\,$ iff $\,g\,$ leaves remainder $0$ when divided by $\,f.$