Given statement is true.
Since, f′(z)≠0 ∀z∈Df'(z)\neq 0 \ \forall z\in Df′(z)=0 ∀z∈D
Let z1,z2∈Dz_1, z_2\in Dz1,z2∈D and z1≠z2z_1\neq z_2z1=z2
Now, since f′(z)≠0 ⟹ f(z1)−f(z2)z1−z2≠0f'(z)\neq 0 \implies \frac{f(z_1)-f(z_2)}{z_1-z_2} \neq 0f′(z)=0⟹z1−z2f(z1)−f(z2)=0 .
As z1≠z2 ⟹ f(z1)≠f(z2)z_1\neq z_2 \implies f(z_1)\neq f(z_2)z1=z2⟹f(z1)=f(z2)
Hence, f is univalent in D
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