So, Not Going To Heaven

  $$y^{-i2ln(a)/\pi}=a^{\frac{1}{n}}$$

$$e^{i\theta-i2ln(a)/\pi}=a^{\frac{1}{n}}$$

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.