A problem on Schwarz Lemma I am currently solving problems on Schwarz's Lemma from Conway's book and stumbled in one problem (prob-3,Page-133). Let $D=\{z\in \mathbb{C} \mid |z|&lt;1\}$ be the open unit disk and $f:D\to \mathbb{C}$ be an analytic function such that $Ref(z)\geq 0$ for all $z\in D$.Then show that  Re$f(z)&gt;0$ for all $z\in D$.  If $f(0)=1$,use appropriate Möbius Transformation and Schwarz's lemma to show that $|f(z)|\leq \frac{1+|z|}{1-|z|}$ for all $z\in D$.  Also show that $\big|f(z)\big|\geq \frac{1-|z|}{1+|z|}$ for all $z\in D$.   I could solve the first two problems but can't solve the third. The first one I solved by Using Open Mapping theorem and the second one by using the Möbius Transform $T$ which maps the imaginary axis to $\partial D$ and right half plane to $D$ given by $T(z)=\frac{z-1}{z+1}$ and applying Schwarz's lemma to $T\circ f$. But can't think of the third one. Any suggestion will be appreciated. Thanks.
Schwarz's Lemma and functions with positive real part Problem:     Let $D = \{ z \in \mathbb C : |z| &lt; 1 \}$. Suppose $f : D \to \mathbb C$ is a non-constant analytic function such that $\mathrm{Re} f(z) \geq 0$ for all $z \in D$.      (a) Show that $\mathrm{Re} f(z) \neq 0$ for all $z \in D$.      (b) Using a Möbius transformation and Schwarz's lemma, prove that if   $f(0) = 1$, then $$ |f(z)| \leq {1+|z| \over 1-|z|}. $$   What can be said if $f(0) \neq 1$?      (c) Show that if $f(0) = 1$, then $f$ also satisfies $$ |f(z)| \geq {1-|z| \over 1+|z|}. $$   Progress:  (a) We can use the open mapping theorem.  (b) We can compose a Möbius transformation $m$ that sends the right-half plane to $D$ with $f$. Then $h = m \circ f$ satisfies the hypotheses of Schwarz's lemma by construction. Simple manipulations of the inequality from Schwarz's lemma give the result.  (c) Looking at part (a), we know that $\mathrm{Re} f(z) \neq 0$, and so $f(z) \neq 0$. Thus, we can use the function $g = 1/f$ in the statement of part (b).  Question:  What about when $f(0) \neq 1$? I don't think you can get anything from $f(z)-f(0)+1$ unless $\mathrm{Re} f(0) &lt; 1$. And you can't in general rearrange the inputs of $f$ because $1$ isn't necessarily in the image of $f$. Also, I don't think you can get anything from $f(z)/f(0)$ unless $f(0)$ is real, because otherwise the range of $f(z)/f(0)$ could extend into the left-half plane.
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