Suppose z_1,...,z_n all lie in an open half-plane bounded by…

Suppose $z_1,\ldots,z_n$ all lie in an open half-plane bounded by a line through the origin. Show that $\dfrac1{z_1},\ldots,\dfrac1{z_n}$ also lie in such a half-plane, and that both $\ z_1+\cdots+z_n\neq0\ $ and $\ \dfrac1{z_1}+\cdots+\dfrac1{z_n}\neq0$.

Part 1 · Inversion maps half-planes to half-planes

Step 1 · Describe the half-plane by argument

An open half-plane through the origin, bounded by the line at angle $\alpha$, consists of all $z=re^{i\theta}$ ($r>0$) with $\theta\in(\alpha,\alpha+\pi)$.

Step 2 · Track the argument under inversion

For $z=re^{i\theta}$: $\ \dfrac1z=\dfrac1re^{-i\theta}$, so $\arg\!\left(\dfrac1z\right)=-\theta$. If $\theta\in(\alpha,\alpha+\pi)$, then $-\theta\in(-\alpha-\pi,-\alpha)$ — again an open interval of length $\pi$.

Step 3 · Conclusion

Since every $z_k$ has argument in $(\alpha,\alpha+\pi)$, every $\dfrac1{z_k}$ has argument in $(-\alpha-\pi,-\alpha)$ — an interval of length $\pi$, hence itself an open half-plane through the origin (bounded by the line at angle $-\alpha$). So the reciprocals all lie in a common half-plane too.

Part 2 · Neither sum can vanish

Step 4 · An open half-plane is convex and avoids the origin

An open half-plane $H$ through the origin is a convex set, and by definition the boundary line — hence the origin itself — is excluded from $H$.

Step 5 · The centroid of points in $H$ lies in $H$

If $z_1,\ldots,z_n\in H$, then the average $\dfrac{z_1+\cdots+z_n}n$ is a convex combination of points of $H$, so (by convexity) it also lies in $H$. Since $0\notin H$:

$$ \frac{z_1+\cdots+z_n}n\neq0 \quad\Longrightarrow\quad z_1+\cdots+z_n\neq0 $$

Step 6 · Apply the same argument to the reciprocals

By Part 1, $\dfrac1{z_1},\ldots,\dfrac1{z_n}$ all lie in some open half-plane through the origin, so Step 5 applies to them directly:

$$ \frac1{z_1}+\cdots+\frac1{z_n}\neq0 $$

Verification

Take $z_1=1,\ z_2=i,\ z_3=1+i$, all with argument in $\left(-\frac\pi4,\frac{3\pi}4\right)$ (an open half-plane). Sum $=2+i\neq0$ ✓. Reciprocals: $1,\ -i,\ \frac{1-i}2$, all with argument in $\left(-\frac{3\pi}4,\frac\pi4\right)$ (a half-plane) — sum $=1-i+\frac{1-i}2=\frac{3-3i}2\neq0$ ✓.