There is a solution which starts by taking $\langle 2+i\rangle$ as a subset of $\langle a+ib\rangle $ and then proceeds as, Hence $(2+i)$ can be represented as multiple of $(a+ib)$. I understand till this part.
Next it proceeds, $(2+i)=(a+ib)(c+id)$. I don't understand this line.
Shouldn't it proceed by multiplying $(a+ib)$ with an integer. Why is $(a+ib)$ multiplied with another complex number?