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We can use a straight line segment because the function f is holomorphic on a simply connected domain D, which guarantees that the contour integral of f is path-independent. This allows us to strategically choose the simplest path—the straight line segment from a to z—to evaluate the difference $F(z) - F(a)$.

Conditions: The function f is holomorphic on the domain D.; The domain D is simply connected.; The points a and z are in D, and the straight line segment connecting them lies entirely within D.

We can use a straight line segment because the function f is holomorphic on a simply connected domain D, which guarantees that the contour integral of f is path-independent. This allows us to strategically choose the simplest path—the straight line segment from a to z—to evaluate the difference $F(z) - F(a)$.

Conditions: The function f is holomorphic on the domain D.; The domain D is simply connected.; The points a and z are in D, and the straight line segment connecting them lies entirely within D.

The factors cancel because the ML inequality introduces the length of the integration path, which is $|z-a|$ for the straight line segment from a to z. This length factor multiplies the supremum term, and it exactly cancels with the $1/|z-a|$ factor that comes from the denominator of the difference quotient.

Conditions: The integration path is the straight line segment from a to z.; The difference quotient has a denominator of $z-a$, whose absolute value is $|z-a|$.; The ML inequality is applied to the integral over this segment.

The factors cancel because the ML inequality introduces the length of the integration path, which is $|z-a|$ for the straight line segment from a to z. This length factor multiplies the supremum term, and it exactly cancels with the $1/|z-a|$ factor that comes from the denominator of the difference quotient.

Conditions: The integration path is the straight line segment from a to z.; The difference quotient has a denominator of $z-a$, whose absolute value is $|z-a|$.; The ML inequality is applied to the integral over this segment.