Claim: For every $f \in L^1[0,1]$, the set $A_f$ is convex.

Proof: Let $\mu_1=\mu$ be Lebesgue measure on $[0,1]$ and consider the signed measures $\mu_2,\mu_3$ on $[0,1]$ defined using the real and imaginary parts of $f$ by $\mu_2(E)= \mu(E)^{-1}\int_E {\mathrm Re} (f) \,d\mu$ and $\mu_3(E)= \mu(E)^{-1}\int_E {\mathrm Im} (f) \,d\mu$.

We are given Lebesgue measurable sets $D,E$ in $[0,1]$ and $a,b>0$ with $a+b=1$. We must show there is a measurable set $G$ with $$(*) \quad z_G=a \cdot z_D+b \cdot z_E \, . $$ We may assume that $r:=\mu(D)/\mu(E) \le 1$. By the Lyapunov theorem on convexity of the range of nonatomic vector measures (see references below; it applies to signed measures, see e.g. [4]) there is a measurable set $E_1$ in $[0,1]$ with $$(\mu_1,\mu_2,\mu_3)(E_1)=r \cdot (\mu_1,\mu_2,\mu_3)(E) +(1-r) \cdot(\mu_1,\mu_2,\mu_3)(\emptyset) \, , $$ so $\mu(E_1)=\mu(D)$ and $z_{E_1}=z_E$. By another application of Lyapunov's theorem, there is a measurable set $G$ in $[0,1]$ with $$ (\mu_1,\mu_2,\mu_3)(G)=a \cdot(\mu_1,\mu_2,\mu_3)(D)+b \cdot (\mu_1,\mu_2,\mu_3)(E_1) \,. $$ Clearly $\mu(G)=\mu(D)=\mu(E_1)$ and $G$ satisfies (*).

References:

[1] A. Liapounoff, Sur les fonctions-vecteurs compl6tement additives, Bull. Acad. Sci. URSS S6r. Math. [Izvestia Akad. Nauk SSSR] 4 (1940) 465-478.

[2] J. Lindenstrauss, A short proof of Liapounoff's convexity theorem, J. Math. Mech. 15 (1966) 971-972

[3] https://en.wikipedia.org/wiki/Vector_measure#Lyapunov

[4] Artstein, Zvi. "Yet another proof of the Lyapunov convexity theorem." Proceedings of the American Mathematical Society 108, no. 1 (1990): 89-91. https://www.ams.org/journals/proc/1990-108-01/S0002-9939-1990-0993737-0/S0002-9939-1990-0993737-0.pdf

[5] Ross, David A. "An elementary proof of Lyapunov's theorem." The American Mathematical Monthly 112, no. 7 (2005): 651-653.

Claim: For every $f \in L^1[0,1]$, the set $A_f$ is convex.

Proof: Let $\mu_1=\mu$ be Lebesgue measure on $[0,1]$ and consider the signed measures $\mu_2,\mu_3$ on $[0,1]$ defined using the real and imaginary parts of $f$ by $\mu_2(E)= \int_E {\mathrm Re} (f) \,d\mu$ and $\mu_3(E)= \int_E {\mathrm Im} (f) \,d\mu$.

We are given Lebesgue measurable sets $D,E$ in $[0,1]$ and $a,b>0$ with $a+b=1$. We must show there is a measurable set $G$ with $$(*) \quad z_G=a \cdot z_D+b \cdot z_E \, . $$ We may assume that $r:=\mu(D)/\mu(E) \le 1$. By the Lyapunov theorem on convexity of the range of nonatomic vector measures (see references below; it applies to signed measures, see e.g. [4]) there is a measurable set $E_1$ in $[0,1]$ with $$(\mu_1,\mu_2,\mu_3)(E_1)=r \cdot (\mu_1,\mu_2,\mu_3)(E) +(1-r) \cdot(\mu_1,\mu_2,\mu_3)(\emptyset) \, , $$ so $\mu(E_1)=\mu(D)$ and $z_{E_1}=z_E$. By another application of Lyapunov's theorem, there is a measurable set $G$ in $[0,1]$ with $$ (\mu_1,\mu_2,\mu_3)(G)=a \cdot(\mu_1,\mu_2,\mu_3)(D)+b \cdot (\mu_1,\mu_2,\mu_3)(E_1) \,. $$ Clearly $\mu(G)=\mu(D)=\mu(E_1)$ and $G$ satisfies (*).

References:

[1] A. Liapounoff, Sur les fonctions-vecteurs compl6tement additives, Bull. Acad. Sci. URSS S6r. Math. [Izvestia Akad. Nauk SSSR] 4 (1940) 465-478.

[2] J. Lindenstrauss, A short proof of Liapounoff's convexity theorem, J. Math. Mech. 15 (1966) 971-972

[3] https://en.wikipedia.org/wiki/Vector_measure#Lyapunov

[4] Artstein, Zvi. "Yet another proof of the Lyapunov convexity theorem." Proceedings of the American Mathematical Society 108, no. 1 (1990): 89-91. https://www.ams.org/journals/proc/1990-108-01/S0002-9939-1990-0993737-0/S0002-9939-1990-0993737-0.pdf

[5] Ross, David A. "An elementary proof of Lyapunov's theorem." The American Mathematical Monthly 112, no. 7 (2005): 651-653.