Equivalence between complex and real operator norms

Since you seem really interested in the inequality $\rho(A)\le\|A\|$ ($\rho$ the spectral radius), here is a simple and elegant proof. In the 2nd edition of my book Matrices (Springer Verlag GTM216), it is Proposition 7.6.

Take any operator norm $N$ on ${\bf M}_n({\mathbb C})$ ; you may take $N=\|\cdot\|_c$, but this is not necessary. Because of finite dimension, the restriction of $N$ to ${\bf M}_n({\mathbb R})$ is equivalent to $\|\cdot\|$ : there exists a finite constant $C$ such that $N(M)\le C\|M\|$ for every $M\in {\bf M}_n({\mathbb R})$. On the other hand, $N$ satisfies $\rho(M)\le N(M)$ for every $M\in {\bf M}_n({\mathbb R})$. Apply these to a power of $A$ : $$\rho(A)^m=\rho(A^m)\le N(A^m)\le C\|A^m\|\le C\|A\|^m.$$ Take the $m$-root in the inequality above $$\rho(A)\le C^{1/m}\|A\|,$$ and let $m\rightarrow+\infty$.

Remark that this proof is valid for every norm over ${\bf M}_n({\mathbb R})$ satisfying the inequality $\|A^m\|\le\|A\|^m$. Such a norm is called super-stable. Super-stable norms include all the operator norms. The numerical radius $$r(A)=\sup_{x\in{\mathbb C}^n\,;\,\|x\|_2=1}|x^*Ax|$$ is a super-stable norm, though not an operator norm. Actually, there exist two matrices such $r(A)r(B)<r(AB)$.