If $f(z)$ is a function with a Taylor series expansion $f(z)=\sum _{ n=0 }^{ \infty }{c_n z^n }$, then we define $f(M)=\sum _{ n=0 }^{ \infty }{c_n M^n }$ First consider $M=\sigma_2=\sigma_y$. Using the definition for the function of an operator, show that $e^{i\sigma_y\theta}$ (here $\theta$ is any number) takes the form $$e^{i\sigma_y\theta}=A(\theta)1+B(\theta)\sigma_y$$ where $1$ is the identity operator, and $A(\theta)$ and $B(\theta)$ are functions that you are to determine. Repeat the same exercise, but now with $\sigma_x$ and $\sigma_z$.
Can anyone explain this question? Just a hint on how to solve it would be sufficient thanks!
I will just outline the approach, I would take, and then the details should be providable. I take it that $\sigma_x$, $\sigma_y$, and $\sigma_z$ are the Pauli matrices, which I state here for completeness:
$$\sigma_x = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix},$$ $$\sigma_y = \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix},$$ $$\sigma_z = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}.$$
The idea is first to consider the Taylor expansion of the map $z\mapsto e^{iz\theta}$: It is an expression of the form $\sum_{n=0}^{\infty} c_nz^n$. Here the coefficients $c_n$ of course depend on $\theta$.
Then you plug in one of your operators in this expansion, say the first one, so you plug $M=\sigma_x$ into the expansion for $f$ to get the expression $$e^{iM\theta}=\sum_{n=0}^{\infty} c_nM^n.$$ Now, what is the matrix $M$ to different powers? For $M=\sigma_x$, you have in fact that $M$ to all even powers is the identity matrix! So each even power in the Taylor expansion gives you something that involves the identity operator, and each odd power gives you something that involves $\sigma_x$ itself.
This means that you can separate the two parts, the part involving only the identity operator, and the part involving $M=\sigma_x$ itself, so that you can express the exponential as
$$e^{iM\theta} = A(\theta)1+B(\theta)M,$$
where the functions $A(\theta)$ and $B(\theta)$ are found by inspecting the coefficients of the power series expansion above.
I hope that this helps.