For limlimits_(n→∞)z_n=a≠0 to exist, it is necessary and…

For $\lim\limits_{n\to\infty}z_n=a\neq0$ to exist, it is necessary and sufficient that $\lim\limits_{n\to\infty}|z_n|$ (nonzero) and $\lim\limits_{n\to\infty}\arg z_n$ both exist.

Necessity ($\Rightarrow$): convergence of modulus

From $|z_n-a|\to0$ and the reverse triangle inequality $\big||z_n|-|a|\big|\le|z_n-a|$:

$$ |z_n|\to|a| $$

Necessity: convergence of argument

Expand using the law of cosines for $|z_n-a|^2=|r_ne^{i\theta_n}-re^{i\theta}|^2$:

$$ |z_n-a|^2=r_n^2+r^2-2r_nr\cos(\theta_n-\theta) $$

Since $r_n\to r$ and $|z_n-a|\to0$, taking the limit gives $2r^2-2r^2\cos(\theta_n-\theta)\to0$, i.e. $\cos(\theta_n-\theta)\to1$ (dividing by $2r^2\neq0$, using $r=|a|\neq0$). This forces $\theta_n-\theta\to0$ along a continuous branch, i.e. $\arg z_n\to\arg a$.

Sufficiency ($\Leftarrow$)

Given $r_n\to r\neq0$ and $\theta_n\to\theta$, set $a=re^{i\theta}$. Writing the real and imaginary parts separately:

$$ r_n\cos\theta_n-r\cos\theta=(r_n-r)\cos\theta_n+r(\cos\theta_n-\cos\theta) $$

$$ r_n\sin\theta_n-r\sin\theta=(r_n-r)\sin\theta_n+r(\sin\theta_n-\sin\theta) $$

Each term on the right vanishes as $n\to\infty$ (since $r_n\to r$, $\cos\theta_n\to\cos\theta$, $\sin\theta_n\to\sin\theta$ by continuity of $\cos,\sin$), so both the real and imaginary parts of $z_n-a$ vanish, hence $|z_n-a|\to0$, i.e. $z_n\to a$.

A necessary caveat

Since $\arg$ is multi-valued, "$\arg z_n$ converges" must mean along a fixed continuous branch (otherwise jumps of $2\pi$ could be mistaken for divergence). The theorem also excludes $a=0$: $\arg(0)$ is undefined, and indeed $z_n\to0$ only requires $|z_n|\to0$, with no constraint on argument at all.