Why is $f=0$ in this case? Conway- functions of one complex variable p.137
Let $a,b\in\mathbb{R}$ such that $a<b$.
Let $G$ be the vertical strip $\{x+iy:a<x<b\}$.
Suppose $f:\overline{G}\rightarrow \mathbb{C}$ is continuius and $f$ is holomorphic on $G$ and $f$ is bounded on $\overline{G}$.
Define $M(x)=sup\{|f(x+iy)| : y\in\mathbb{R}\}$.
In my text it is written that "if either $M(a)$ or $M(b)$ is zero then $f\equiv 0$", but how?
If $M(x)=0$ for some $a<x<b$, then $f$ is zero on the straight line $\{x+iy : y\in \mathbb{R}\}$. Since this line is contained in $G$ and $f$ is holomorphic on $G$, $f\equiv 0$ on $G$.
However, neither $\{a+iy : y\in\mathbb{R}\}$ nor $\{b+iy : y\in\mathbb{R}\}$ are contained in $G$, we cannot apply the above argument to derive that $f\equiv 0$ on $G$.
Why is $f\equiv 0$ if $M(a)$ or $M(b)$ is zero?
 A: If $M(a) = 0$ then, in particular, $f$ takes real values on the line
$L := \{a+iy : y\in\mathbb{R}\}$. It follows from the
Schwarz reflection principle
that $f$ can be continued analytically across that line to the larger strip
$\tilde G \cup L \cup G$ where $\tilde G$ is the reflection of $G$
at the line $L$. This continuation is holomorphic and identically 
to zero on $L$, and it follows that $f \equiv 0$.

More elementary proof using  Morera's theorem: Without loss
of generality we can assume that $a = 0$. Define $F$ on the strip
$H := \{x+iy:-b<x<b\}$ by
$$
F(z) = \begin{cases}
f(z) & \text{ if } 0 < \text{Re}(z) < b \\
0 & \text{ if }  \text{Re}(z) = 0 \\
\overline{f(-\overline z)} & \text{ if } -b < \text{Re}(z) < 0
\end{cases}
$$
$F$ is continous in $H$. Now show that $\int_\gamma f(z)\, dz = 0$
for each "triangle path" in $H$. This is clear if $\gamma $ 
does not intersect the imaginary axis, and otherwise follows from
the symmetry of $F$. Morera's theorem implies that $F$ is analytic
in $H$. (Remark: That's how the Schwartz reflection principle
is proved.) Since $F(z) = 0$ on the imaginary axis, $F \equiv 0$
and therefore $f \equiv 0$. 
A: This is related to Hadamard's three-lines lemma.

Lemma. Let $f$ be a bounded holomorphic function on $G$ that is continuous on $\overline{G}$. Then for $0\le t\le 1$,
  $$ M(a(1-t)+bt) \le M(a)^{1-t} M(b)^t.$$

As a consequence, if $M(a)=0$ or $M(b)=0$, then $M(\theta)=0$ for all $a<\theta<b$, so $f=0$ by the identity theorem.
However, one should keep the following in mind. The way this lemma is usually proven requires $M(a), M(b)$ to be positive. But you can always make the temporary assumptions $M(a)>0, M(b)>0$.
Then go through the proof of the lemma (see below).
Finally if you proved the statements for all $M(a),M(b)$ positive you can see that it must also hold for $M(a)=0$ or $M(b)=0$ (simply take the limit on the right hand side of the inequality).
Addendum: Here is the proof of Hadamard's lemma.
First of all, whenever we talk about $z$ we only mean $z$ in the strip.
Set $$F_n(z)=f(z)M(a)^{z-1}M(b)^{-z}e^{z^2/n}.$$ Observe that 
$|F_n(z)|\to 0$ as $|z|\to \infty$. Here is why: Compute $$|e^{z^2/n}|=e^{\mathrm{Re}(z^2)/n}=e^{(\mathrm{Re}(z)^2-\mathrm{Im}(z)^2)/n}.$$ The real part $\mathrm{Re}(z)$ is bounded in the strip $G$. As the imaginary part goes to $\infty$ in absolute value, this converges to $0$. 
Now that means if we make $|z|$ large enough, then $|F_n(z)|\le 1$ on $|z|\ge R$. In particular, if we take the axis-aligned rectangle bounded by the strip boundaries and the lines $\mathrm{Im}(z)=\pm R$, then we can apply the maximum modulus principle to $F_n$ on that domain to see that $|F_n(z)|\le 1$ on all of the rectangle. Then let $R\to \infty$ to see that $|F_n(z)|\le 1$ must hold throughout the strip.
Finally, $$F_n(z)\to f(z)M(a)^{z-1}M(b)^{-z}$$ for every $z$ as $n\to\infty$, so we must also have $|f(z)M(a)^{z-1}M(b)^{-z}|\le 1$ on the whole strip. But this means for $z=t+iu$, where $a\le t\le b$ that
$$|f(z)|\le |M(a)^{1-z}M(b)^{z}| = M(a)^{1-t} M(b)^{t}$$
Taking the supremum over the imaginary part $u$ we get the claim.
A: WLOG $G$ is the strip $0<x<1.$ Suppose $Mf(0) = 0.$ Then for $g(z) = f(z)f(1-z),$ $Mg(0)= Mg(1) = 0.$
For $R>0,$ think of the boundary of the rectangle $[0,1]\times [-R,R]$ as a contour $C_R,$ oriented positively. Fix $z\in G.$ Using the continuity of $g$ on $\overline G,$ Cauchy's formula gives
$$g(z) = \frac{1}{2\pi i}\int_{C_R}\frac{g(w)}{w-z}\,dw.$$
Because $g$ vanishes on the vertical sides of $C_R,$ the above equals the sum of the integrals along the top and bottom of $C_R.$ But $g$ is bounded, so a simple estimate shows each of those integrals $\to 0$ as $R\to \infty.$ Therefore $g(z) = 0.$ Since $z $ was arbitrary, $g\equiv 0,$ hence $f\equiv 0.$
