Deriving Similarity Transformations for SU(2) Algebra

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I'm working on a project that requires similarity transformations for certain spin operators, but the book I'm working with Mathematical Methods of Quantum Optics doesn't include all the ones I need. From (2.41) the book it has this identity:

$$\exp(-i\theta \textbf{n}\cdot\hat{\textbf{S}})\textbf{a}\cdot\hat{\textbf{S}}\exp(i\theta \textbf{n}\cdot\hat{\textbf{S}}) $$

$$= \cos(\theta)\textbf{a}\cdot\hat{\textbf{S}}\ + (\textbf{n} \times \textbf{a})\cdot\hat{\textbf{S}}\sin(\theta)+[1-\cos(\theta)](\textbf{n}\cdot\hat{\textbf{S}})(\textbf{n} \cdot \textbf{a}) $$

It then goes on to enumerate a few special cases:

$$\hat{S_{\pm}}(\theta) = \exp(-i\theta\hat{S_z})\hat{S_{\pm}}\exp(i\theta\hat{S_z}) = \exp(\mp i\theta)\hat{S_{\pm}}$$

Which is 'derived by noting $d\hat{S_{\pm}}(\theta)/d\theta = \mp i \hat{S_{\pm}}(\theta).$

or: $$\hat{S_{z}}(\theta) = \exp(-i\theta\hat{S_+})\hat{S_z}\exp(i\theta\hat{S_+}) = \hat{S_{z}} + i\theta\hat{S_{+}}$$

which states 'This is derived by noting that $d\hat{S_{z}}(\theta)/d\theta = i \hat{S_{+}}$. Multiply [the equation] on the left by $\exp(i\theta \hat{S_+})$, differentiate it m times with respect to $\theta$ at $\theta = 0$ and show that $[\hat{S_z},\hat{S_+^m}] = m\hat{S_+^m}$.

I'd like to be able to derive the special cases: $$\exp(-i\theta\hat{S_-})\hat{S_z}\exp(i\theta\hat{S_-})$$ and $$\exp(-i\theta\hat{S_-})\hat{S_+}\exp(i\theta\hat{S_-}) $$ which aren't present in the book, but I am having difficulty. What would these special cases be? Any assistance would be appreciated.

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First rewrite the product using a commutator: $$\begin{align} \exp(-i\theta\hat S_+) \, \hat S_z \, \exp(i\theta \hat S_+) &= \exp(-i\theta\hat S_+) \, \left( [\hat S_z, \exp(i\theta \hat S_+)] + \exp(i\theta \hat S_+) \, \hat S_z \right) \\ &= \exp(-i\theta\hat S_+) \, [\hat S_z, \exp(i\theta \hat S_+)] + \hat S_z \end{align}$$

Then calculate the commutator: $$\begin{align} [\hat S_z, \exp(i\theta \hat S_+)] &= [\hat S_z, \sum_{n=0}^{\infty} \frac{(i\theta)^n}{n!} \hat S_+^n] = \sum_{n=0}^{\infty} \frac{(i\theta)^n}{n!} [\hat S_z, \hat S_+^n] = \sum_{n=0}^{\infty} \frac{(i\theta)^n}{n!} n \hat S_+^n \\ &= \sum_{n=1}^{\infty} \frac{(i\theta)^n}{(n-1)!} \hat S_+^n = \sum_{n=0}^{\infty} \frac{(i\theta)^{n+1}}{n!} \hat S_+^{n+1} \\ &= \left(\sum_{n=0}^{\infty} \frac{(i\theta)^n}{n!} \hat S_+^n\right) ( i\theta \hat S_+) = \exp(i\theta \hat S_+) ( i\theta \hat S_+ ) \end{align}$$

Inserting this into the first result gives $$ \exp(-i\theta\hat S_+) \, \hat S_z \, \exp(i\theta \hat S_+) = i\theta \hat S_+ + \hat S_z. $$