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Starting at: $ a+\left(\left(\frac{i-a}{\gamma+1}\right)(\gamma+1)(i-a)^{\gamma}\right)^{\frac{1}{\gamma+1}} $

gives:

$ a+\left(\frac{(i-a)(\gamma+1)(i-a)^{\gamma}}{(\gamma+1)}\right)^{\frac{1}{\gamma+1}} $

which reduces to:

$ a+(i-a)^{\frac{\gamma+1}{\gamma+1}} $

and finally becomes:

$ i $

However, inserting values like: $a=1$, $i=-3$, and $\gamma=-3$ gives $5$ rather than $-3$ (for $i$) as the result. Interestingly with the "+'' between the terms changed to "-'', it works:

$a-\left(\frac{(i-a)(\gamma+1)(i-a)^{\gamma}}{(\gamma+1)}\right)^{\frac{1}{\gamma+1}}=-3$

I guess I did something in the above symbolic simplification that is not allowed (such that I miss a change in the "+" sign). However, I have no idea what! The parameters are in general constrained like this: $a=\{0,1\},$ $i<0$ (if $a=0$) and $i<1$ (if $a=1$), $\gamma<-1$.

Fabian
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  • This looks like a power law issue with complex numbers similar to this. – Klaus Jan 22 '20 at 16:22
  • I suspect your problems are essentially about how you evaluate $a+\Big((i-a)^2\Big)^{1/2}$. If you take the positive square root you get $a+|i-a|$; if you take the negative square root you get $a-|i-a|$. When using numerical values one usually take the positive square roots, so in your examples you would get $2a-i$. The issue is blurred when one uses symbolic values. – almagest Jan 22 '20 at 16:31
  • @almagest I'm a bit confused. I always thought that this is only the case if solving an equation for $(i-a)$, but here I insert values. Is there also two "solutions" for a square root if inserting values? – Fabian Jan 22 '20 at 16:41

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