Quantum Computing 1.1 -- Qubit measurement

The absolute squares $| \alpha | ^ 2$ and $| \beta | ^ 2$ of the vector components of the state $(\alpha, \beta)$ must be proportional to the probabilities $80 \%$ and $20 \%$ . Their ratio then results $\frac{|\alpha|^2}{|\beta|^2} = \frac{80\%}{20\%} = 4$ We calculate this ratio for all vectors: $\begin{aligned} & \left(\begin{array}{c} 2 - 2i \\ 1 +i \end{array} \right) & \frac{|2-2i|^2}{|1 + i|^2} &= \frac{(2 - 2i)(2 + 2i)}{(1 + i)(1 - i)} = \frac{2^2 + 2^2}{1^2 + 1^2} = 4 \\ & \left(\begin{array}{c} 4 \\ -1 \end{array} \right) & \frac{|4|^2}{|-1|^2} &= 16 \\ & \left(\begin{array}{c} 2 \\ 1 + i \end{array} \right) & \frac{|2|^2}{|1 + i|^2} &= \frac{4}{(1 + i)(1 - i)} = \frac{4}{1^2 + 1^2} = 2\\ & \left(\begin{array}{c} -i \\ 2 \end{array} \right) & \frac{|-i|^2}{2^2} &= \frac{(-i)\cdot i}{4} = \frac{1}{4} \end{aligned}$ Thus, only the vector $(2-2i,1+i)$ corresponds to the measured probabilities.