Problem: An $e^{-}$ exists in such a state that the probability of its spin aligning across the $x_{(+)}$ axis is $P_{x+}=1/2$ and across the $y_{(+)}$ axis is $P_{y+}=1/2$ as well. What is the spin wave function of the electron ?
Solution: Let $\psi_s$ be the spin wave function of the ${e^-}$: $\psi_s=\left(\begin{array}{c}a\\b\\\end{array}\right), a,b\in C$. Then: $P_{x+}=\left|\langle\psi_s, X_{x+}\rangle\right|^2$ and $P_{y+}=\left|\langle\psi_s, X_{y+}\rangle\right|^2$. As for the $X_{X_+}$ and $X_{y_+}$ they are calculated via the formula $X_{n_+}=\left(\begin{array}{c}\cos(θ/2)\\\sin(θ/2)e^{i\varphi}\\\end{array}\right)$, where $φ,θ$ are the angles of the unitary vector $\vec{n}$ with axes $x,z$ respectively:
$X_{X_+}=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\1\\\end{array}\right), X_{y_+}=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\i\\\end{array}\right)$
Therefore, we end up with 2 equations and 2 unknowns:
$\left|\langle\left(\begin{array}{c}a\\b\\\end{array}\right),\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\1\\\end{array}\right)\rangle\right|^2=1/2$ and
$\left|\langle\left(\begin{array}{c}a\\b\\\end{array}\right),\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\i\\\end{array}\right)\rangle\right|^2=1/2$
But I'm having some difficulties to solve the system:
$|a^*+b^*|^2=1, |a^*+i b^*|^2=1$
Any hints ? Also the fact that both $a=1,b=0$ and $a=0,b=1$ both satisfy the system worries me.
EDIT: The missing clue was $|a|^2 + |b|^2=1$.