To view mathematical objects as alive not only helps our intuition but also gives us a better way to remember the main idea of the proof. In the following example we want to prove that every non-constant polynomial $f(z)$ with complex coefficients has a (complex) root. But before this, let us look at an old lady who takes her dog for a daily morning walk.
She always makes $n$ rounds around the flagpole (point) $O$. Because the lead is short the dog is forced to do $n$ rounds as well. One morning, the lady was very tired and decided to make the circles smaller.
It is time to name our players. The lady’s name is $z^n$. If $z = Re^{i\varphi}$, where $R$ is a constant and $\varphi$ changes from $0$ to $2\pi$ then the lady $z^n$ indeed makes $n$ rounds of radius R n around the origin $O$. The dog's name is $f (z) = z^n + a_1 z^{n-1} + \dots + a_n$. The lead is therefore $a_1 z^{n-1} + \dots + a_n$ and for sufficiently large $R$ the length of the lead is much smaller than $R^n$, and the dog is indeed forced to make $n$ rounds around $O$ as well. What happened when the lady was tired and changed $R$ to almost zero?
If we suppose that $a_n \ne 0$ then for a very small $R$, the dog will mostly be concentrated near $a_n$ and therefore makes no rounds around $O$ at all. Now let us change $R$ from large to small $R$ continuously. How could it happen that the dog decreased the number of rounds from $n$ to zero? There is only one way to do this: for some $R$ and some $\varphi$ the dog has to go through $O$, in other words $f(Re^{i\varphi}) = 0$ and this is exactly what we wanted to prove. Amazingly easy! The reader should not have any difficulties in completing this to a rigorous proof (for example estimating more exactly what large and small mean and what to do if $a_0 = 0$) but will hopefully never forget the idea of the proof.
If the coefficiens had been real numbers, I would see no problem. But as it is right now, how can one be sure that that a complex polynomial always rotates? When the length of the lead is the same as the length of $z^n$, why is it obvious that there is an angle such that $z^n=-$lead?
Perhaps I should clearify The question is about a proof, which starts at row 3 in the picture. I have already got several explanations, so it's clearer now. At the moment I'm downloading Hardy's book, to read the proof given there. Thanks everbody!
If we define $$ f(z)=z^n+\sum_{k=1}^na_kz^{n-k}\tag{1} $$ then $$ \arg(f(z))-\arg\left(z^n\right)=\arg\left(1+\sum_{k=1}^na_kz^{-k}\right)\tag{2} $$ For any $\{a_k\}\subset\mathbb{C}$, we can find an $R$ big enough so that when $|z|\ge R$, $$ \begin{align} \left|\sum_{k=1}^na_kz^{-k}\right| &\le\sum_{k=1}^n|a_k||z|^{-k}\\ &\le\sum_{k=1}^n|a_k|R^{-k}\\ &\le\frac12\tag{3} \end{align} $$ $(3)$ implies that if $|z|\ge R$, $$ -\frac\pi6\le\arg\left(1+\sum_{k=1}^na_kz^{-k}\right)\le\frac\pi6\tag{4} $$ as demonstrated in the image below
Then by $(2)$ and $(4)$, we have that $\arg(f(z))$ and $\arg\left(z^n\right)$ are within $\frac\pi6$ of each other at all times.