I agree with FlippingBits. Just complementing the anwser

A single qubit state $\ket \psi$ is a superposition in respect to a given basis $\{\ket K , \ket{K_\perp}\}$ if it can be written as a **non trivial** linear combination of the basis (as stated in the second paragraph of FlippingBit)

To find a basis for which $\ket \psi$ is not a superposition, all one needs to do is find a orthogonal state $\ket {\psi_\perp}$

$\braket \psi {\psi_\perp} = 0$, and the new basis is of course $\{\ket \psi , \ket{\psi_\perp}\}$

**a)** As explained by FlippingBit

**b)**

$\frac{1}{\sqrt{2}}(\ket + + \ket -) \Rightarrow \frac{1}{\sqrt{2}}( \frac{1}{\sqrt{2}}(\ket 0 + \ket 1) + \frac{1}{\sqrt{2}}(\ket 0 - \ket 1)) \Rightarrow \frac{1}{2}(\ket 0 + \ket 1) + \frac{1}{2}(\ket 0 - \ket 1)$

$\frac{1}{\sqrt{2}}(\ket + + \ket -) = \ket 0$

"b)" is also not a superposition (non trivial linear combination)

**c)**

$\frac{1}{\sqrt{2}}(\ket + - \ket -) \Rightarrow \frac{1}{\sqrt{2}}( \frac{1}{\sqrt{2}}(\ket 0 + \ket 1) - \frac{1}{\sqrt{2}}(\ket 0 - \ket 1)) \Rightarrow \frac{1}{2}(\ket 0 + \ket 1) - \frac{1}{2}(\ket 0 - \ket 1)$

$\frac{1}{\sqrt{2}}(\ket + - \ket -) = \ket 1$

"c)" is not a superposition (non trivial linear combination)

**d)**

$\frac{\sqrt3}{2}\ket + - \frac{1}{2}\ket - \Rightarrow \frac{\sqrt3}{2}( \frac{1}{\sqrt{2}}(\ket 0 + \ket 1) ) - \frac{1}{2} (\frac{1}{\sqrt{2}}(\ket 0 - \ket 1))$

$\frac{1}{\sqrt{2}}(\ket + - \ket -) = \frac{\sqrt3 - 1}{2\sqrt{2}}\ket 0 + \frac{\sqrt{3} + 1}{2\sqrt{2}}\ket 1$ which is a trivial case,

therefore "d) is a superposition"

In the basis $\{\frac{\sqrt3 - 1}{2\sqrt{2}}\ket 0 + \frac{\sqrt{3} + 1}{2\sqrt{2}}\ket 1 ,\frac{\sqrt3 - 1}{2\sqrt{2}}\ket 0 - \frac{\sqrt{3} + 1}{2\sqrt{2}}\ket 1\}$ this state is not in superposition. as explained by FlippingBit

One may check by doing $\braket {(\frac{\sqrt3 - 1}{2\sqrt{2}}\ket 0 + \frac{\sqrt{3} + 1}{2\sqrt{2}}\ket 1)} {(\frac{\sqrt3 - 1}{2\sqrt{2}}\ket 0 - \frac{\sqrt{3} + 1}{2\sqrt{2}}\ket 1\})} = 0$

**e)**

$\frac{1}{\sqrt{2}}(\ket i - \ket{-i}) \Rightarrow \frac{1}{\sqrt{2}}( \frac{1}{\sqrt{2}}(\ket 0 + i\ket 1) - \frac{1}{\sqrt{2}}(\ket 0 - i\ket 1)) \Rightarrow \frac{1}{2}(\ket 0 + i\ket 1) - \frac{1}{2}(\ket 0 - i\ket 1)$

$\frac{1}{\sqrt{2}}(\ket i - \ket{-i}) = i\ket 1$

"e)" is not a superposition (non trivial linear combination)

**f)**

as explained by FlippingBit