I Help evaluating complex function in form m+ni?

[NotASmurf]

Hey all, I need the complex version of the sigmoid function in standard form, that is to say $$f(\alpha) =\frac{1}{1+e^{-\alpha}} , \hspace{2mm}\alpha = a+bi , \hspace{2mm} \mathbb{C} \to \mathbb{C}$$ in the simplified form: $$f = m+ni$$ but found this challenging, for some reason i assumed there was an identity for $$e^{e^{x} }, \hspace{2mm} x \in \mathbb{C}$$, so wasted my time with
$$e^{-a-bi}= e^{e^{tan^{-1}\frac{b}{a}i + ln[\sqrt{ a^{2} + b^{2} }]}}$$ and tried from there, (just showing I did make an attempt, no matter how abysmal). Any help appreciated as I am not too familiar with complex numbers outside of the basics needed for transformation matrices.

 

[blue_leaf77]

so wasted my time with

e−a−bi=eetan−1bai+ln[√a2+b2]e−a−bi=eetan−1bai+ln[a2+b2]​

e^{-a-bi}= e^{e^{tan^{-1}\frac{b}{a}i + ln[\sqrt{ a^{2} + b^{2} }]}} and tried from there

No need to go that far.
Try multiplying $f(\alpha) =\frac{1}{1+e^{-\alpha}}$ with $\frac{1+e^{-\alpha^*}}{1+e^{-\alpha^*}}$ and then use Euler formula.

 

[NotASmurf]

Thanks :D ,Comes to $$ \frac{1}{2cos(b) e^{-a}}[1+e^{-\overline{\alpha}]}$$ right (before further simplification)?
Then $$ =[ \frac{1}{2cos(b) e^{-a}} +\frac{1}{2}] + [\frac{1}{2} tan(b)]i $$ ? Or have I screwed up? (can't exactly substitute in numbers as easily in this case to test)

 

[mathman]

\frac{1}{1+e^{-\alpha}}=\frac{1}{1+e^{-a}(cosb-isinb)}=\frac{1+e^{-a}(cosb+isinb)}{(1+e^{-a}cosb)^2+(e^{-a}sinb)^2}=\frac{1+e^{-a}(cosb+isinb)}{1+2e^{-a}cosb+e^{-2a}}

\alpha and a look alike in itex.

 

[NotASmurf]

Thanks mathman but since my previous answer is more computationally efficient (has to run vast iterations for the program) my previous answer correct?

 

[NotASmurf]

Turns out this was a fruitless exercise since $$\frac{\partial u }{\partial x} \neq \frac{\partial v }{\partial y} $$ , so it doesn't conform with the Cauchy -Riemann equations D: (It needs to be differentiable)

 

[FactChecker]

Turns out this was a fruitless exercise since $$\frac{\partial u }{\partial x} \neq \frac{\partial v }{\partial y} $$ , so it doesn't conform with the Cauchy -Riemann equations D: (It needs to be differentiable)

I think you should check that again. 1/(1+e^-α) is analytic in the complex plane except where (1+e^-α) = 0.

 

[NotASmurf]

well when I reworked it I got $$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} $$ BUT $$\frac{\partial u}{\partial y} = \frac{\partial v}{\partial x} $$ when it should be equal to the negative does that mean its diffentiable, but only for certain regions? (havn't had to do complex differentiation before), also quick question , the Riemann - Cauchy equations essentially say $$\frac{\partial f}{\partial z} = 0 $$ must be true for it do be differentiable, this confuses me, can someone elucidate on the intuition here?

 

[FactChecker]

$$\frac{\partial u}{\partial y} = \frac{\partial v}{\partial x} $$ when it should be equal to the negative

I think there must be a sign problem somewhere.
The series of operations α => -α => e^-α => 1 + e^-α gives an entire function (analytic in the entire complex plane)
Then 1/(1 + e^-α) is analytic for α in ℂ except where it is a division by 0.

the Riemann - Cauchy equations essentially say $$\frac{\partial f}{\partial z} = 0 $$ must be true for it do be differentiable,

This is not right. I'm not familiar with the Wirtinger derivatives (https://en.wikipedia.org/wiki/Wirtinger_derivatives), but apparently this should be the partial wrt z conjugate (see equation 3 of https://en.wikipedia.org/wiki/Cauchy–Riemann_equations )

 

[mathman]

Thanks mathman but since my previous answer is more computationally efficient (has to run vast iterations for the program) my previous answer correct?

I doubt it. It is different from mine - unlikely to be correct. Check out the denominator.

 

[Svein]

the Riemann - Cauchy equations essentially say

∂f∂z=0​

\frac{\partial f}{\partial z} = 0 must be true for it do be differentiable, this confuses me, can someone elucidate on the intuition here?

No, you are wrong. An analytic function is characterized by \frac{\partial f}{\partial \bar{z}}=0 (you need to define \frac{\partial f}{\partial \bar{z}} in a sensible manner, but that should not be too hard).