Generalised Square of Sum of Modulus of Product of Complex Numbers A cursory search of the site showed no answer I was looking for. Essentially I have a few questions. 
I know already that for $z_1, z_2 \in \mathbb{C}$, with arguments $\theta_1$ and $\theta_2$ respectively, we have
$$
|z_1 + z_2|^2 = |z_1|^2 + |z_2|^2 + 2|z_1||z_2|\cos(\theta_1 - \theta_2)
$$
(for those who are interested, proof wiki has a simple proof here: https://proofwiki.org/wiki/Complex_Modulus_of_Sum_of_Complex_Numbers )
I have 2 questions. I think the first one is solved by me, but would like confirmation. The second question is a bit beyond me at the moment xD 


*

*How would one generalize this for $z_1,...,z_k \in \mathbb{C},$ where we know the arguments of $z_i$:
$$
\left| \sum_{i=1}^k z_i \right|^2?
$$

*How would one generalize this for $\alpha_1,...,\alpha_k,z_1,...,z_k \in \mathbb{C},$ where we know the arguments of $z_i$ but not of $\alpha_i$:
$$
\left| \sum_{i=1}^k \alpha_iz_i \right|^2?
$$


My attempt so far:
1.
Suppose for $z_1, ..., z_k \in \mathbb{C}$ with arguments $\theta_1, ...,\theta_k$ respectively, we have
$$
\left| \sum_{i=1}^k z_i \right|^2 = |z_1 + ... + z_k|^2 
\\
=(|z_1|]\cos(\theta_1) + ... + |z_k|\cos(\theta_k))^2 + (|z_1|\sin(\theta_1) + ... + |z_k|\sin(\theta_k))^2
\\
= |z_1|^2 \cos^2\theta_1 + |z_1||z_2|\cos\theta_1\cos\theta_2 + ... |z_1|z_k|\cos\theta_1\cos\theta_k + 
\\
|z_2|^2 \cos^2\theta_2 + |z_1||z_2|\cos\theta_1\cos\theta_2 + ... |z_2|z_k|\cos\theta_2\cos\theta_k + 
\\
... +
\\
|z_k|^2\cos^2\theta_k + |z_1||z_k|\cos\theta_1\cos\theta_k + ... + |z_{k-1}||z_k|\cos\theta_{k-1}\cos\theta_k +
\\
|z_1|^2 \sin^2\theta_1 + |z_1||z_2|\sin\theta_1\sin\theta_2 + ... |z_1|z_k|\sin\theta_1\sin\theta_k + 
\\
|z_2|^2 \sin^2\theta_2 + |z_1||z_2|\sin\theta_1\sin\theta_2 + ... |z_2|z_k|\sin\theta_2\sin\theta_k + 
\\
... +
\\
|z_k|^2\sin^2\theta_k + |z_1||z_k|\sin\theta_1\sin\theta_k + ... + |z_{k-1}||z_k|\sin\theta_{k-1}\sin\theta_k
\\
= |z_1|^2(\cos^2\theta_1 + \sin^2\theta_1) + |z_2|^2(\cos^2\theta_2 + \sin^2\theta_2) + ... + |z_k|^2(\cos^2\theta_k + \sin^2\theta_k) +
\\
2|z_1||z_2|(\cos\theta_1\cos\theta_2 + \sin\theta_1\sin\theta_2) +2|z_1||z_3|(\cos\theta_1\cos\theta_3 + \sin\theta_1\sin\theta_3) +
\\
... +
\\
2|z_{k-1}||z_k|(\cos\theta_{k-1}\cos\theta_k + \sin\theta_{k-1}\sin\theta_k)
$$
Now use the fact that $\cos^2\theta + sin^2\theta = 1$ and the fact that $\cos\theta_1\cos\theta_2 + \sin\theta_1\sin\theta_2 = \cos(\theta_1 - \theta_2)$:
$$
\left| \sum_{i=1}^k z_i \right|^2 = \sum_{i=1}^k |z_i|^2 + \sum_{i \neq j}|z_i||z_j|\cos(\theta_i - \theta_j)
$$
For 2. I have no idea where to start. I am asking because I am working through this paper on Quantum Clustering and Gaussian Mixtures https://arxiv.org/abs/1612.09199 and they give a simple answer:
$$
\left| \sum_{i=1}^k \alpha_iz_i \right|^2 = \sum_{i=1}^k \alpha_i|z_i| \sum_{j=1}^k \bar{\alpha_j}|z_j|\cos(\theta_i - \theta_j)
$$
is this true? How would I show it? 
For those who are interested, in the paper $|z_k|$ is the unnormalized Gaussian Distribution with parameters $(\mu_k, \mathbf{C}_k)$ given for some $d$-dimensional point $p$ by
$$
|z_k| = \frac{1}{\sqrt{Z_k}}e^{-\frac{1}{4}(p - \mu_k)^T \mathbf{C}^{-1}_k(p-\mu_k)}
\\
Z_k = (2\pi)^{\frac{d}{2}}|\mathbf{C}_k|^{\frac{1}{2}}
$$
whereas $z_k$, the complex number, is given by
$$
z_k = \frac{1}{\sqrt{Z_i}}e^{-\frac{1}{4}(p - \mu_k)^T \mathbf{C}^{-1}_k(p-\mu_k)}e^{-i\phi_k}
$$
When I programmed this up I encountered some errors (i am getting negative probabilities), and now I'm trying to determine if its because of my code or if there's an error in the paper. I think it might because the exponent with the phase is $-i\phi_k$ instead of $i\phi_k$. I'm checking out this hunch but it would be super helpful to know that the above summation of square modulus is correct. 
Edit history: 


*

*Removed coefficient of 2 from last term.

*Using saulspatz idea:
$$
\left|\sum_{i=1}^k \alpha_iz_i \right|^2 = \sum_{i=1}^k \alpha_i z_i \sum_{j=1}^k \overline{\alpha_j z_j}
\\
= \sum_{i=1}^k \alpha_i |z_i|(\cos\theta_i + i\sin\theta_i) \sum_{j=1}^k \overline{\alpha_j} |z_j|(\cos\theta_j - i\sin\theta_j)
\\
= \sum_{i=1}^k \alpha_i |z_i| \sum_{j=1}^k \overline{\alpha_j} |z_j|(\cos\theta_i + i\sin\theta_i)(\cos\theta_j - i\sin\theta_j)
\\
= \sum_{i=1}^k \alpha_i |z_i| \sum_{j=1}^k \overline{\alpha_j} |z_j| (\cos\theta_i\cos\theta_j -i\cos\theta_i\sin\theta_j + i\cos\theta_j\sin\theta_i +\sin\theta_i\sin\theta_j)
\\
= \sum_{i=1}^k \alpha_i |z_i| \sum_{j=1}^k \overline{\alpha_j} |z_j| (\cos\theta_i\cos\theta_j + \sin\theta_i\sin\theta_j + i\cos\theta_j\sin\theta_i - i\cos\theta_i\sin\theta_j)
\\
= \sum_{i=1}^k \alpha_i |z_i| \sum_{j=1}^k \overline{\alpha_j} |z_j| 
(\cos(\theta_i - \theta_j) + i\sin(\theta_i - \theta_j))
\\
= \sum_{i=1}^k \alpha_i |z_i| \sum_{j=1}^k \overline{\alpha_j} |z_j|\cos(\theta_i - \theta_j) + \overline{\alpha_j} |z_j|i\sin(\theta_i - \theta_j)
\\
= \sum_{i=1}^k \alpha_i |z_i| \sum_{j=1}^k \overline{\alpha_j} |z_j|\cos(\theta_i - \theta_j) + i\sum_{i=1}^k \alpha_i |z_i| \sum_{j=1}^k \overline{\alpha_j} |z_j|\sin(\theta_i - \theta_j)
$$
Presumably this imaginary part equals 0, but I'm really at a loss as to how to show that >.< 


In any case, now I see that the real part must be as was shown in the paper, meaning I have a bug in my code... whoops. 
 A: Start with
$$\left|\sum_{i=1}^k{a_iz_i}\right|^2=\sum_{i=1}^k{a_i}{z_i}\sum_{j=1}^k\overline{{a_j}{z_j}}
$$
Note that the coefficient of $2$ is incorrect in the formula you give.  When you sum over $i\ne j$ you sum over $(2,3)$ and $(3,2)$ separately.
EDIT You're making things too hard. I'm going to change the index of summation from $i$ to $m$ to avoid confusion with the imaginary unit.
$$\left|\sum_{m=1}^k{a_mz_m}\right|^2=
\sum_{m=1}^k{a_mz_m}\sum_{j=1}^k\overline{{a_jz_j}}=
\sum_{m=1}^k{a_mz_m}\sum_{j=1}^k\overline{{a_jz_j}}=\\
\sum_{m=1}^k{a_m|z_m|e^{i\theta_m}}\sum_{j=1}^k{\overline{a_j}|z_j|e^{-i\theta_j}}=
\sum_{m=1}^k{a_m|z_m|}\sum_{j=1}^k{\overline{a_j}|z_j|e^{i(\theta_m-\theta_j)}}=\\
\sum_{m=1}^k{a_m|z_m|}\sum_{j=1}^k{\overline{a_j}|z_j|\cos(\theta_m-\theta_j)}+i\sum_{m=1}^k{a_m|z_m|}\sum_{j=1}^k{\overline{a_j}|z_j|\sin(\theta_m-\theta_j)}=\\
\sum_{m=1}^k{a_m|z_m|}\sum_{j=1}^k{\overline{a_j}|z_j|\cos(\theta_m-\theta_j)}
$$
because $$\left|\sum_{m=1}^k{a_mz_m}\right|^2\text{ is real.}$$  
So yes, the imaginary part is $0$, but it's usually easier to use $e^{i\theta}$ than $\cos\theta+i\sin\theta.$
