Real proofs with shorter equivalent proofs in Complex numbers? Are there any proofs about Real numbers that have shorter equivalent proofs going through Complex numbers?
Are there proofs about Integers going through Reals, with longer equivalent proofs using pure Integers?
 A: Many many interesting definite integrals of functions whose indefinite integrals have no closed form are derived using contour integration in the complex plane. Some of these can, however, be derived by far messier work in the real line. 
Almost everything involving computational complexity in computer science ends up using the logarithm, which can be defined for only integer arguments if you choose to do so, but whose properties, as a function on the reals, are of great interest and use in complexity analysis. 
A: Another nice one is in working out which integers may be expressed as the sum of two square integers. Dedekind's proof of the sum-of-two-squares theorem relies on factorising $x^2 + 1$ as $(x+i)(x-i)$.
A: Some time ago I had answered a nice simple question that turned out to be a good example for the Integers/Reals part (I couldn't find simpler discrete proof).
Problem:

Let $G = (U \uplus V, E)$ be a connected bipartite graph such that $n_U = |U| < |V| = n_V$. Then there exists an edge $\{u,v\} \in E$ such that $\deg(u) > \deg(v)$.

Discrete proof:
Sort the vertices such that 
\begin{align}
\deg(u_1) &\geq \deg(u_2) \geq &\ldots &&\geq \deg(u_{n_U}), \\
\deg(v_1) &\geq \deg(v_2) \geq &\ldots &&\geq \deg(v_{n_V}). 
\end{align}
Observe that because there are no isolated vertices in $G$ we have $\deg(v_{n_V}) \geq 1$, thus $$\sum_{i=1}^{n_U}\deg(u_i)-\deg(v_i) > \sum_{i=1}^{n_U}\deg(u_i)-\sum_{i=1}^{n_V}\deg(v_i) = 0.$$
Let $k$ be the smallest natural number with this property, that is,
$$k = \min\Bigg\{k \in \mathbb{N} \ \Bigg|\ \sum_{i=1}^{k}\deg(u_i)-\deg(v_i) > 0\Bigg\}.$$
It follows that 


*

*$\deg(u_k) > \deg(v_k)$, otherwise $k$ would not be the smallest number with the above property. 

*$N\big(\{u_1,\ldots,u_k\}\big) \not\subseteq\big\{v_1,\ldots,v_k\big\}$, because the sum of degrees in $\{v_1,\ldots,v_k\big\}$ is too small to exhaust  all the degrees of $\{u_1,\ldots,u_k\big\}$.

*There exists an edge $\{u_i,v_j\} \in E$ such that $i \leq k \leq j$.

*$\deg(u_i) \geq \deg(u_k) > \deg(v_k) \geq \deg(v_j)$.


Non-discrete proof:
Let $$f\big(\{u,v\}\big) = \frac{1}{\deg(u)} - \frac{1}{\deg(v)}.$$ 
This is a proper definition, because there are no isolated vertices in $G$. Now observe that $$\sum_{e \in E}f(e) = \sum_{u \in U}\deg(u)\cdot\frac{1}{\deg(u)} - \sum_{v \in V}\deg(v)\cdot\frac{1}{\deg(v)}
 = |U|-|V| < 0.$$
Therefore, there exists an edge $\{u,v\}$ such that $f\big(\{u,v\}\big) < 0$, but that means $\deg(u) > \deg(v)$.
I hope this helps $\ddot\smile$
A: A nice one is: determination of the radius of convergence of a real power series by analyzing the singularities of the complex function defined by the series.   
And more generally: analyzing the asymptotic properties of a sequence $a_n$ by examining the complex singularities of the series $\sum a_n z^n$, the so-called "generating function".
