The substitution x=cisα, y=cisβ automatically satisfies x+x1=2cosα and y+y1=2cosβ (since cisα+cis(−α)=2cosα), turning every part of this question into a direct application of the product/quotient/power rules for cis expressions.
Step 1. Justify the substitution. Take x=cisα. Then x1=cis(−α), so x+x1=cisα+cis(−α)=2cosα, matching the given 2cosα=x+x1. Similarly take y=cisβ, so y+y1=2cosβ.
Step 2. Part (i): compute yx+xy.
yx=cisβcisα=cis(α−β),xy=cis(β−α)=cis(−(α−β)).
yx+xy=cis(α−β)+cis(−(α−β))=2cos(α−β).
Step 3. Part (ii): compute xy−xy1.
xy=cisα⋅cisβ=cis(α+β),xy1=cis(−(α+β)).
xy−xy1=cis(α+β)−cis(−(α+β))=2isin(α+β),
using cisϕ−cis(−ϕ)=2isinϕ.
Step 4. Part (iii): compute ynxm−xmyn. By de Moivre's theorem, xm=cis(mα) and yn=cis(nβ), so …