FP2: Very confused by these answers

Question 1 (part d)

Question 2 (part bii)

You can spot that $z=-1$ is a root, otherwise, the fact that $\beta$ is real is important but not so much here. Alternative approach is the following:

$\beta^2 -2i \beta - (1+2i) = 0$ which is the same as $(\beta^2-1) -2(\beta + 1)i = 0$ which is only true if both real/im parts are 0, and obviously from the im part, $\beta = -1$ which can be checked with the real part as well.



Use bi) and reciprocate both sides. Multiply top and bottom by $i$

Yeah I just realised shortly after posting that I could use the fact that $\beta + \alpha = 2i$, with $\beta$ represented by $a+bi$ and $\alpha$ by $c$. Doing the same with the product of the roots just leaves you to easily find a, b and c by comparing real and Im parts.

For whatever reason I didn't think of that, despite messing around with the products and roots of the cubic to try and find them...

Yeah, though I think you meant to say $\beta + \gamma = 2i$ then let $\gamma = a+bi$ and $\beta = c$ so $(a+c)+bi = 2i$ and from product $ac+bci = -(1+2i)$

There are a number of ways to do this question, I'd stick with my method as it's the nicest - no need for introducing new constants and solving for them tbh.

As for the denominator with i, it is very common to get rid off it from there in many questions to a point where you should be able to recognise it and do it as naturally as rationalising the denominator due to some surd.

They simply rationalized the denominator