Intuitive proof for the Schwarz inequality (Complex Analysis) I have been following Serge Leng's Introduction to Complex Analysis at a graduate level and I feel a bit stuck on one of the exercises. The exercise in question is

I can't seem to prove N3 without assuming the Schwarz inequality. On the other hand, I am able to prove the Schwarz inequality but I would have to assume N3.
There have been published solutions for these questions:

however, this is not the slitest bit intuitive (atleast to me). Can anybody help me/explain to me an intuitive way of looking at the problem?
 A: The intuition behind the Schwarz inequality is perhaps easily seen for an inner product over a real vector space;  for any vectors $u, v$,
\begin{align}
\langle u \pm v, u \pm v\rangle \geqslant 0.
\end{align}
Now take $u, v$ to have unit length and expand the product to obtain,
\begin{align}
2 \geqslant \pm 2 \langle u, v \rangle
\end{align}
so that $\lvert \langle u,v \rangle \rvert \leqslant 1$.  This is extended to general vectors $u$ and $v$ by dividing each by its norm and using the bi-linearity of the inner product.
The argument is also extended to a complex linear space by taking again unit vectors $u$ and $v$ but now multiply $u$ by $e^{i\theta}$ where $\theta$ is chosen to make  $\langle u,v \rangle$  real and positive.  This is always possible.
Then
\begin{align}
\langle e^{i\theta}u - v, e^{i\theta} u - v \rangle &= \langle e^{i\theta} u, e^{i \theta}u\rangle - e^{i\theta} \langle u, v\rangle
 - e^{-i\theta}\langle v,u\rangle  +\langle v,v \rangle \\
&=\langle u,u \rangle - 2 \Re \big( e^{i\theta}\langle u,v \rangle\big) + \langle v,v \rangle
\end{align}
Now the choice of $\theta$ comes into play. We have made $e^{i\theta}\langle u,v\rangle$  real and positive, so as before we now obtain
\begin{align}
\Re \big( e^{i\theta} \langle u,v \rangle \big) &= e^{i\theta} \langle u,v \rangle \\
&= \lvert e^{i\theta} \langle u,v \rangle \rvert \\ 
&= \lvert \langle u,v \rangle \rvert
\end{align}
leading neatly to $\lvert \langle u,v \rangle \rvert \leqslant 1$.  This is extended to general $u$ and $v$ as before.
