Proof. Both parts follow from the mean value property, which says that the value of an analytic function at the centre of a circle is the average of its values around the circle: \[f(a)=\frac {1}{2\pi }\int _0^{2\pi }f\!\left (a+re^{i\theta }\right )d\theta .\]

(i) An interior maximum forces \(f\) to be constant

Suppose \(\left |f\right |\) attains its maximum \(M\) at an interior point \(a\). Taking moduli in the mean value property and using \(\left |\int \right |\leq \int \left |\cdot \right |\), \[M=\left |f(a)\right |\leq \frac {1}{2\pi }\int _0^{2\pi } \left |f\!\left (a+re^{i\theta }\right )\right |d\theta \leq \frac {1}{2\pi }\int _0^{2\pi }M\,d\theta =M .\] The two ends are equal, so both inequalities are equalities. The second can be an equality only if \(\left |f\!\left (a+re^{i\theta }\right )\right |=M\) for every \(\theta \) — a continuous function whose average equals its maximum is constantly that maximum. Since \(r\) was any sufficiently small radius, \(\left |f\right |\equiv M\) on a disc about \(a\).

Now \(\left |f\right |\) constant forces \(f\) constant. If \(M=0\) then \(f\equiv 0\). If \(M>0\) then \(f\overline {f}=M^2\), so \(\overline {f}=M^2/f\) is analytic; a function analytic together with its conjugate has all four partial derivatives zero by the Cauchy–Riemann equations, hence is constant. So \(f\) is constant on that disc, and by the identity theorem constant throughout \(D\).

(ii) Otherwise the maximum sits on the boundary

Suppose \(f\) is not constant. Then by (i) \(\left |f\right |\) has no maximum at an interior point. But \(\left |f\right |\) is continuous on \(D\cup \Gamma \), which is closed and bounded, so it does attain a maximum somewhere. That point cannot be interior, so it lies on \(\Gamma \). □

Proof. Apply the maximum modulus principle to \(\, F(z) = \frac {1}{f(z)}\). □

Proof. The function \(f(z) = e^{u + iv}\) satisfies conditions of the maximum modulus principle where \(v\) is the harmonic conjugate of \(u\). Thus either \(\, \left |f(z)\right | = e^u\,\) constant throughout \(D\) or attains maximum of \(\Gamma \).
Now, \(\, \left |f(z)\right | = e^u\,\) constant in \(D\) implies \(u\equiv \) constant in \(D\).
If \(\, \left |f(z)\right | = e^u\,\) is non constant then it attains its maximum on \(\Gamma \). But since \(e^u\) is monotonic increasing. \(u\) also attains its maximum on \(\Gamma \).
Finally, since \(f(z) = e^{u + iv}\) is non vanishing the minimum principle holds. □

Solution. \(f(z) = \cos z\,\) is bounded, continuous, non vanishing and analytic in \(\, \left |z\right | \leq 1\).
Hence, by the maximum, minimum principle it attains its max/min on the boundary.
We show this by elementary means. Let \(z = x + iy\) and use the fact that \[\left |f(z)\right | = \left |\cos \big (x + iy\big )\right | = \Big [\cos ^2x + \sinh ^2y\Big ]^{\frac {1}{2}}\] and for simplicity, consider the real function \(\, w = \cos ^2x + \sinh ^2y\).
Now any stationary point of \(w\) inside \(\, \left |z\right | \leq 1\) occurs when \[\frac {\partial w}{\partial x} = \frac {\partial w}{\partial y} = 0\] AS \(\, \frac {\partial w}{\partial x} = -2\cos x\sin x = -\sin 2x = 0\)

\(\frac {\partial w}{\partial y } = 2\sinh y \cosh y = \sinh 2y = 0\)

\(\implies \quad \sin 2x = 0 \implies x = 0\)

\(\implies \quad \sinh 2y = 0 \implies y = 0\)

Thus the stationary point is on (0,0) inside \(\, \left |z\right | \leq 1\).
Now \begin {align*} \triangle & = \frac {\partial ^2w}{\partial x^2} \cdot \frac {\partial ^2w}{\partial y^2} - \frac {\partial ^2w}{\partial x \partial y}\\ & = -4\cos 2x \cosh 2y\Big |_{(0,0)}\\ & = -4 < 0 \end {align*}

Elementary real variable theory says.