Separate imaginary and real parts from complex expression I learned about complex numbers after I was trying to create a fractal object.
The main problem is that I have an equation with complex numbers and I have to separate their parts (real & imaginary) to calculate the next iteration.
Some equations like the $f(p) = p^2 + c$ are obviously and easy to solve them. But some of them are using the exponential form of the complex numbers, that really bothers me.
Take for example this equation:
\begin{equation*}
F(P) = c*e^{-p} + p*p~\text{Where}~c = u + v*i~\text{and}~p = a + b*i.
\end{equation*}
I solved it partly by this way - express $e^{-p}$ like $1 / (e^a \ast (\cos(b) + \sin(b) * i))$ and so on... But in the end I couldn't see the relation to separate the groups.
Any help and advice how to proceed in this or similar case will be greatly appreciated.
 A: \begin{equation*}
F(P) = c*e^{-p} + p*p~\text{Where}~c = u + v*i~\text{and}~p = a + b*i.
\end{equation*}
\begin{equation*}
F(P) = e^{-p}(c+e^p*p^2)~\text{Where}~c = u + v*i~\text{and}~p = a + b*i.
\end{equation*}
\begin{equation*}
F(P) = \frac{1}{e^p}(c+e^p*p^2)~\text{Where}~c = u + v*i~\text{and}~p = a + b*i.
\end{equation*}
\begin{equation*}
F(P) = \frac{(c+e^p*p^2)}{e^p}~\text{Where}~c = u + v*i~\text{and}~p = a + b*i.
\end{equation*}
\begin{equation*}
F(P) = \frac{c}{e^p}+p^2~\text{Where}~c = u + v*i~\text{and}~p = a + b*i.
\end{equation*}
$------------$
$$F(P)=\frac{u+vi}{e^{a+bi}}+(a+bi)^2=$$
$$F(P)=e^{-a-bi}(u+vi)+(a+bi)^2$$
$$F(P)=e^{|-a-bi|e^{\arg(-a-bi)i}}(u+vi)+(a+bi)^2=$$
$$F(P)=e^{\sqrt{a^2+b^2}e^{\arg(-a-bi)i}}(u+vi)+(a+bi)^2=$$
$$F(P)=\left|e^{\sqrt{a^2+b^2}e^{\arg(-a-bi)i}}(u+vi)\right|e^{\arg\left(e^{-a-bi}(u+vi)\right)i}+\left(\left|a+bi\right|e^{\arg\left(a+bi\right)i}\right)^2=$$
$$F(P)=\left(\sqrt{u^2+v^2}e^{\Im(b)-\Re(a)}\right)e^{\arg\left(e^{-a-bi}(u+vi)\right)i}+\left(\sqrt{a^2+b^2}e^{\arg\left(a+bi\right)i}\right)^2=$$
$$F(P)=\left(\sqrt{u^2+v^2}e^{\Im(b)-\Re(a)}\right)e^{\arg\left(e^{-a-bi}(u+vi)\right)i}+(a^2+b^2)e^{2\arg\left(a+bi\right)i}=$$
SO:
$$\Re=(a^2+b^2)\cosh(2\arg(a+bi))+\sqrt{u^2+v^2}e^{\Im(b)-\Re(a)}\cosh(\arg(e^{-a-bi}(u+vi)))$$
$$\Im=\left((a^2+b^2)\sinh(2\arg(a+bi))+\sqrt{u^2+v^2}e^{\Im(b)-\Re(a)}\sinh(\arg(e^{-a-bi}(u+vi)))\right)i$$
