Idk if it's time to talk about continuous iterations for non-analytic functions that can still be defined on the whole complex plane, but these ideas did disastrously harm to my brain lol.

I worked out some cases last year, these are my results.

Consider any function g(x):R->R, then we consider a function f(z), which geometrically rotates the original plane in a way that cut the plane into infinitely many slices in the shape of ring, all locates its center at the origin, each ring-shaped slice has its own rotation angle, which can be written as:

$$f(z)=ze^{ig(\|z\|)}$$

Then, we can easily prove and get a continuous iteration(but only to real orders) of f(z) following these steps:

$$\text{For real t, }f^t(z)=ze^{itg(\|z\|)}$$

Remember abs(e^(it))=1 for real t,

$$\text{proof: For real s and t, }f^{s+t}(z)=ze^{i(s+t)g(\|z\|)}=ze^{itg(\|z\|)}e^{isg(\|z\|)}=\bigg(ze^{itg(\|z\|)}\bigg)e^{isg\big(\|ze^{itg(\|z\|)}\|\big)}=f^s(f^t(z))$$

it's easy to prove, and also satisfies the geometric intuition that the t-th iteration of a rotation is a rotation whose angular velocity is t times the original rotation's.

Another idea, came by trying to figure out whether there's continuous iteration of the conjugate function conj(z)

We start by conj(z)=abs(z)*e^(-i*arg(z)), consider a set of functions $$f_u(z)=\|z\|e^{iu\arg(z)}$$ where u is real,

Then $$\text{For real t and u^t, }f_u^t(z)=\|z\|e^{iu^t\arg(z)}$$

Because $$\text{For real s and t, }f_u^{s+t}(z)=\|z\|e^{iu^{s+t}\arg(z)}=\|ze^{iu^tz}\|e^{iu^s\arg(\|z\|e^{iu^tz})}=f_u^s(f_u^t(z))$$

if u^t and u^s are both real.

But since u=-1, t=1/2, u^t is not real, we cannot find a functional square root of conj(z) through this.

Do you have more examples, I basically based all this on the Polar decomposition of complex numbers and don't have so much time to take more investigation into it.

I worked out some cases last year, these are my results.

Consider any function g(x):R->R, then we consider a function f(z), which geometrically rotates the original plane in a way that cut the plane into infinitely many slices in the shape of ring, all locates its center at the origin, each ring-shaped slice has its own rotation angle, which can be written as:

$$f(z)=ze^{ig(\|z\|)}$$

Then, we can easily prove and get a continuous iteration(but only to real orders) of f(z) following these steps:

$$\text{For real t, }f^t(z)=ze^{itg(\|z\|)}$$

Remember abs(e^(it))=1 for real t,

$$\text{proof: For real s and t, }f^{s+t}(z)=ze^{i(s+t)g(\|z\|)}=ze^{itg(\|z\|)}e^{isg(\|z\|)}=\bigg(ze^{itg(\|z\|)}\bigg)e^{isg\big(\|ze^{itg(\|z\|)}\|\big)}=f^s(f^t(z))$$

it's easy to prove, and also satisfies the geometric intuition that the t-th iteration of a rotation is a rotation whose angular velocity is t times the original rotation's.

Another idea, came by trying to figure out whether there's continuous iteration of the conjugate function conj(z)

We start by conj(z)=abs(z)*e^(-i*arg(z)), consider a set of functions $$f_u(z)=\|z\|e^{iu\arg(z)}$$ where u is real,

Then $$\text{For real t and u^t, }f_u^t(z)=\|z\|e^{iu^t\arg(z)}$$

Because $$\text{For real s and t, }f_u^{s+t}(z)=\|z\|e^{iu^{s+t}\arg(z)}=\|ze^{iu^tz}\|e^{iu^s\arg(\|z\|e^{iu^tz})}=f_u^s(f_u^t(z))$$

if u^t and u^s are both real.

But since u=-1, t=1/2, u^t is not real, we cannot find a functional square root of conj(z) through this.

Do you have more examples, I basically based all this on the Polar decomposition of complex numbers and don't have so much time to take more investigation into it.

Regards, Leo