Given statement is true.
Since, "f'(z)\\neq 0 \\ \\forall z\\in D"
Let "z_1, z_2\\in D" and "z_1\\neq z_2"
Now, since "f'(z)\\neq 0 \\implies \\frac{f(z_1)-f(z_2)}{z_1-z_2} \\neq 0" .
As "z_1\\neq z_2 \\implies f(z_1)\\neq f(z_2)"
Hence, f is univalent in D
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