Prove that…

Prove that $\ \left|\displaystyle\sum_{k=1}^nz_k\right|=\displaystyle\sum_{k=1}^n|z_k|\ $ (for nonzero $z_k$) if and only if $\arg z_k=\arg z_1$ for $k=2,\ldots,n$.

Step 1 · Write each $z_k$ in polar form

$$ z_k=r_ke^{i\theta_k}, \qquad r_k=|z_k|>0,\ \ \theta_k=\arg z_k $$

Step 2 · Square the assumed equality

$$ \left|\sum_{k=1}^nr_ke^{i\theta_k}\right|^2=\left(\sum_{k=1}^nr_k\right)^2 $$

Step 3 · Expand both sides

Left side: $\ \displaystyle\sum_{k,j}r_kr_je^{i(\theta_k-\theta_j)}=\sum_kr_k^2+\sum_{k\neq j}r_kr_j\cos(\theta_k-\theta_j)$ (the imaginary parts cancel in pairs $k,j$ and $j,k$). Right side: $\ \displaystyle\sum_kr_k^2+\sum_{k\neq j}r_kr_j$.

Step 4 · Subtract

$$ \sum_{k\neq j}r_kr_j\big(1-\cos(\theta_k-\theta_j)\big)=0 $$

Every term $r_kr_j\big(1-\cos(\theta_k-\theta_j)\big)$ is $\ge0$ (since $r_k,r_j>0$ and $1-\cos(\cdot)\ge0$), so a sum of nonnegative terms equal to $0$ forces each term to vanish:

$$ \cos(\theta_k-\theta_j)=1 \ \Longrightarrow\ \theta_k=\theta_j \pmod{2\pi} \qquad \text{for every } k\neq j $$

In particular $\theta_k=\theta_1$ for all $k$, i.e. $\arg z_k=\arg z_1$.

Step 5 · Converse

If $\arg z_k=\theta$ for every $k$, write $z_k=r_ke^{i\theta}$, so $\ \sum z_k=e^{i\theta}\sum r_k$, and since $\sum r_k\ge0$:

$$ \left|\sum_{k=1}^nz_k\right|=|e^{i\theta}|\left|\sum_{k=1}^nr_k\right|=\sum_{k=1}^nr_k=\sum_{k=1}^n|z_k| $$

Verification, $n=3$

$z_1=1,\ z_2=2,\ z_3=3$ (same argument $0$): $|1+2+3|=6=1+2+3$ ✓. Changing $z_3=3i$ (different argument): $|1+2+3i|=\sqrt{13}\approx3.606\neq6=|z_1|+|z_2|+|z_3|$, confirming the equality fails once arguments differ.