Does $|\mathfrak{Re}((8+i)^n+(4+7i)^n)| \to \infty$ when $n\to\infty$?

Question

Let $u_n=\mathfrak{Re}((8+i)^n+(4+7i)^n)$ ; thus $u_1=12,u_2=30,u_3=-36$ etc. Is it known whether $\lim_{n\to\infty}|u_n|=\infty$ ?

My thoughts : clearly $\lim_{n\to\infty}\frac{|u_n|}{r^n}=0$ when $r \gt \sqrt{65}$. Perhaps one can show that $\lim_{n\to\infty}\frac{|u_n|}{r^n}=\infty$ when $r\lt \sqrt{65}$ ?

Answer 1

The factorizations $$8+i=(3+2i)(2-i) \qquad\text{ and }\qquad 4+7i=(3+2i)(2+i),$$ show that $$\operatorname{Re}((8+1)^n+(4+7i)^n) =\operatorname{Re}\Big((3+2i)^n\big((2-i)^n+(2+i)^n\big)\Big),$$ where of course $$(2-i)^n+(2+i)^n=2\cdot\operatorname{Re}\Big((2+i)^n\Big),$$ so the expression in your question simplifies to $$\operatorname{Re}((8+i)^n+(4+7i)^n) =2\cdot\operatorname{Re}\Big((2+i)^n\Big) \cdot\operatorname{Re}\Big((3+2i)^n\Big).$$ Because $N(2+i)=5$ and $N(3+2i)=13$ this diverges.