Show that $\mathrm{Re} (z)>0$

Question

Let $$z=(1^2+i)(2^2+i)\cdots(n^2+i).$$ I have to show that $\mathrm{Re} (z)>0$, where $i\in\mathbb {C} $, $i^2=-1$ and $\mathrm{Re} (z) $ is the real part of $z $.

I noticed that $$z=(1+a)(2+a)\cdots(n+a)(1-a)(2-a)\cdots(n-a),$$ where $a\in\mathbb {C} $, $a^2=i $.

Now I denote $$P (x)=(1+x)(2+x)\cdots(n+x).$$ Then $z=P (a)\cdot P (-a) $ and $$\mathrm{Re} (z)=\frac {1}{2}(P (a)\cdot P (-a)+P (ia)\cdot P (-ia) ).$$

I am stuck. I want just a hint.

Answer 1

If you look at the arguments of those complex numbers ($\arctan\frac{1}{k^2}$ for $k=1,2,\ldots,n$), it is enough to show that the sum of those arguments is $\le\frac{\pi}{2}$. Use $\arctan x\le x$ and $\sum_{n=1}^{\infty}=\frac{\pi^2}{6}$ to conclude:

$$\sum_{k=1}^{\infty}\arctan\frac{1}{k^2}=\arctan(1)+\sum_{k=2}^{\infty}\arctan\frac{1}{k^2}\le\frac{\pi}{4}+\sum_{k=2}^{\infty}\frac{1}{k^2}=\frac{\pi}{4}+\left(\frac{\pi^2}{6}-1\right)\lt\frac{\pi}{2}$$

so the whole infinite sum is smaller than $\frac{\pi}{2}$ - the finite sum must be even smaller.

Answer 2

Another approach: Note that Each factor has the argument $\arctan(1/k^2)$, so we want to show that $\sum_{k=1}^n \arctan(1/n^2) <\pi/2$.

It turns out the argument of the product is

$$0 < \sum_{k=1}^n\arctan\left(\frac{1}{k^2}\right) \leq \sum_{k=1}^\infty \arctan\left(\frac{1}{k^2}\right) = \arctan\left(\frac{1-\cot(\pi/\sqrt{2})\tanh(\pi/\sqrt{2})}{1+\cot(\pi/\sqrt{2})\tanh(\pi/\sqrt{2})} \right) = 1.42... < 1.57... = \pi/2$$

So the real part must be positive.