$\psi$ Is Attractive

From the post "Not Exponential, But Hyperbolic And Positive Gravity!" dated 21 Nov 2014,

$F=e^{ i3\pi /4 }D\sqrt { 2{ mc^{ 2 } } } .tanh\left( \cfrac { { D } }{ \sqrt { 2{ mc^{ 2 } } }  }( x-x_o).e^{ i\pi /4 } \right)$

If $G=D.e^{ i\pi /4 }$ is real then

$F=e^{i\pi/2}\sqrt { 2{ mc^{ 2 } } }\,G.tanh\left( \cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (x-x_z) \right)$

$F=i\sqrt { 2{ mc^{ 2 } } }\,G.tanh\left( \cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (x-x_z) \right)$

If however we allow the augment to the function $tanh(x)$ to be complex, and interpret $e^{ i\pi /4 }$ as rotation about the origin anti-clockwise by $\pi/4$ from $e^{ i\pi /4 }$ , and then a further rotation by $3\pi/4$ from $e^{ i3\pi /4 }$, $F$ is rotated by $\pi$ about the origin in total.

$F=-\sqrt { 2{ mc^{ 2 } } }\,G.tanh\left( \cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (x-x_z) \right)$

$F$ is then along the negative $x$ direction, towards the center of the particle which is consistent with the expression

$4\pi { \cfrac { \dot { x } ^3 }{ a_{ \psi  } } m }=-q$

that a negative charge is normally associated with particle.

But, this interpretation of $e^{ i\pi /4 }$ in the augment of $tanh(x)$ is odd because $tanh(x)$ does not accept complex augments.  In this way, the modulus of a complex augment serves as the real input to the function and the argument of the complex augment rotates the x axis of the graph of the function about the origin anti-clockwise.

If this is true,

$F=-\sqrt { 2{ mc^{ 2 } } }\,G.tanh\left( \cfrac { G }{ \sqrt { 2{ mc^{ 2 } } }  } (x-x_z) \right)$

without $i$.  Force density $F$, points towards the center of the particle, along the negative $x$ direction.

And we have a new class of real functions with complex augments.

Good night.

Note: This suggests $e^{ia}$ has immunity, that

$F=e^{ i3\pi /4 }D\sqrt { 2{ mc^{ 2 } } } .tanh\left( \cfrac { { D } }{ \sqrt { 2{ mc^{ 2 } } }  }( x-x_o).e^{ i\pi /4 } \right)$

$F=e^{ i3\pi /4 }.e^{ i\pi /4 } D\sqrt { 2{ mc^{ 2 } } } .tanh\left( \cfrac { { D } }{ \sqrt { 2{ mc^{ 2 } } }  }( x-x_o)\right)$

$F=e^{ i\pi } D\sqrt { 2{ mc^{ 2 } } } .tanh\left( \cfrac { { D } }{ \sqrt { 2{ mc^{ 2 } } }  }( x-x_o)\right)$

$F=- D\sqrt { 2{ mc^{ 2 } } } .tanh\left( \cfrac { { D } }{ \sqrt { 2{ mc^{ 2 } } }  }( x-x_o)\right)$

 $e^{ia}$ passes through a function as a constant coefficient passes through an integral.

Warning, over generalization!!