Let $X$ be a topological vector space and $E$ a subset of $X$. A point $x\in E$ is called a extreme point of $E$ if there is no proper line segment contained in $E$ which contains $x$.
Let $K$ be a compact Hausdorff space and $\mathrm{ball}M(K)$ the closed ball of (complex) regular measures on $K$. How to prove that $\alpha \delta_x$ is an extreme point of $\mathrm{ball}M(K)$, where $\delta_x$ is Dirac measure?
I understand how to prove that $\delta_x$ is an extreme point of $P(K)$ the set of probabilities on $K$. But in statement above difficulties appear due to $\alpha \in \mathbb{C}$
I assume you mean the unit ball, and that $|\alpha|=1$.
If $$\tag1 \alpha\delta_x=t\,\mu_1+(1-t)\,\mu_2, $$ That is, for any $f\in C(K)$ you have $$\tag2 f(x)=t\,\int_Kf\,d(\alpha^{-1}\mu_1)+(1-t)\int_Kf\,d(\alpha^{-1}\mu_2). $$ Using the real and imaginary parts and the Hahn decomposition, you can write $$ \alpha^{-1}\mu_j=\mu_j^1-\mu_j^2+i(\mu_j^{3}-\mu_j^{4}),\qquad\qquad j=1,2. $$ For $f\geq0$, we get $$\tag3 f(x)=t\,\int_Kf\,d\mu_1^1+(1-t)\int_Kf\,d\mu_2^1, $$ $$\tag4 0=t\,\int_Kf\,d\mu_1^k+(1-t)\int_Kf\,d\mu_2^k,\qquad\qquad k=2,3,4. $$ As this can be done for any $f\geq0$, from $(4)$ we get $\mu_j^k=0$ for $k=2,3,4$, and from $(3)$ we get (as in the probability case) $\mu_j^1=\delta_x$. Thus $$ \mu_j=\alpha\,\mu_j=\alpha\,\delta_x. $$