# 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?

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.