homework and exercises - Multipole expansion relative to point $z=-d$

I'm attempting to solve a problem where the solution involves the multipole expansion of a the relative vector between a point at $z=-d, x=y=0$, i.e., $r = -d, \theta = \pi$ and a point at r.

We know that in this system, the relative vector |r - r'| where r' is a point at $z=d$ is equal to:

$$ \frac{1}{\sqrt{x^2 + y^2 + (z-d)^2}} = \frac{1}{\sqrt{r^2 + d^2 - 2rd\cos(\theta)}}$$

Employing standard multipole expansion, we find that this is equal to:

$$\sum_{l =0}^{\infty}\frac{r_<^l}{r_>^{l+1}}P_l(\cos\theta)$$

Where $r_<$ is less than $r_>$ are either $r$ or $d$.

My question is on how to do the same expansion where $r'$ is at $z=-d$

$$ \frac{1}{\sqrt{x^2 + y^2 + (z+d)^2}}] = - \frac{1}{\sqrt{r^2 + d^2 + 2rd\cos(\theta)}} = \frac{1}{\sqrt{r^2 + d^2 - 2rd\cos(\theta')}} $$

In terms of $\theta'$ we can expand this as $$\sum_{l =0}^{\infty}\frac{r_<^l}{r_>^{l+1}}P_l(\cos\theta')$$

Since $\theta$ is simply $-\theta'$, does this imply that this expansion in the original coordinate system is actually equal to $$\sum_{l =0}^{\infty}\frac{r_<^l}{r_>^{l+1}}P_l(\cos(-\theta))$$

Answer

No, the mistake is that $\theta'\neq-\theta$. The correct transformation is $\theta' = \pi - \theta$ leading to: $$\sum_{\ell=0}^\infty \frac{r_<^\ell}{r_>^\ell} P_\ell(-\cos \theta) = \sum_{\ell=0}^\infty (-1)^\ell\frac{r_<^\ell}{r_>^\ell} P_\ell(\cos \theta).$$