$$y^{-i2ln(a)/\pi}=a^{\frac{1}{n}}$$
$$e^{i\theta-i2ln(a)/\pi}=a^{\frac{1}{n}}$$
This is wrong! So have been,
$$\LARGE{e^{(i\theta)^{-i2ln(a)/\pi}=\,\LARGE{{a}^{\small\frac{1}{n}}}}}$$
Which is bad...
$$\LARGE{e^{{(i\theta)^{-i2ln(a)/\pi}}=\,\LARGE{{a}^{\small\frac{1}{n}}}}}$$
$$\LARGE{e^{{(ie)^{-i2ln(\theta)ln(a)/\pi}}=\,\LARGE{{a}^{\small\frac{1}{n}}}}}$$
$$\LARGE{e^{{(i^i)^{-2ln(\theta)ln(a)/\pi}.(e)^{-i2ln(\theta)ln(a)/\pi}}=\,\LARGE{{a}^{\small\frac{1}{n}}}}}$$
$$\LARGE{e^{{(i^i)^{-2ln(\theta)ln(a)/\pi}}.e^{-i2ln(\theta)ln(a)/\pi}=\,\LARGE{{a}^{\small\frac{1}{n}}}}}$$
\(\theta=\cfrac{\pi}{2n}\), and \(i^i=0.20788\lt 1\)
What about,
$$\LARGE{e^{(i^i)^{-\frac{2}{\pi}ln(a+\frac{\pi}{2n})}}}$$
$$\LARGE{e^{e^{-i\frac{2}{\pi}ln(a+\frac{\pi}{2n})}}}$$
And what the f**k for?
$$\LARGE{|a^{\frac{1}{n}}|=e^{(i^i)^{-\frac{2}{\pi}ln(a+\frac{\pi}{2n})}}}.{e^{{cos\left(\frac{2}{\pi}ln(a+\frac{\pi}{2n})\right)}}}$$
Since the imaginary part has is a sine function,
$$-1 \le \angle arg\le 1$$
and as only positive values are considered, and the stated condition,
$$\cfrac{1}{m}+\cfrac{1}{n}+\cfrac{1}{k}\lt 1$$
$$|\angle arg |\lt 1$$
Still not going to heaven.
