Complex impedence between 2 terminals I have this problem to solve.

I have this so far. I am struggling with this so looking for some help please.
$$\frac{1}{5j}+\frac{1}{5+8.66j}+\frac{1}{15}+\frac{1}{-10j}=-\frac{j}{5}+\frac{5-8.66j}{5^2+8.66^2}+\frac{1}{15}+\frac{j}{10}=a-bj$$
$$a:=\frac{5}{5+8.66}+\frac{1}{15},\,b=\frac{1}{5}+\frac{8.66}{5^2+8.66^2}-\frac{1}{10}=\frac{1}{10}+\frac{8.66}{5^2+8.66^2},$$
$$\frac{1}{a-bj}=\frac{a+bj}{a^2+b^2}.$$
I know i need a imaginary number for the x and real number for the y axis.
I an struggling because thier is more than 2 branches. I know:
$$ A + B = (4 + j1) + (2 + j3)
A + B = ( 4+2 ) + j(1+3)$$
So i added like this:
$$ j + 5 - 8.66j + 1 + j = -2.66j $$
But i guess this isn't correct?
This is for cartesian form.

Impedances sum in series; admittances (reciprocals of impedances) sum in parallel. Thus the total admittance in $\Omega^{-1}$ is $$\frac{1}{5j}+\frac{1}{5+8.66j}+\frac{1}{15}+\frac{1}{-10j}=-\frac{j}{5}+\frac{5-8.66j}{5^2+8.66^2}+\frac{1}{15}+\frac{j}{10}=a-bj$$with $$a:=\frac{5}{5+8.66}+\frac{1}{15},\,b=\frac{1}{5}+\frac{8.66}{5^2+8.66^2}-\frac{1}{10}=\frac{1}{10}+\frac{8.66}{5^2+8.66^2},$$while the total impedance in $\Omega$ is $$\frac{1}{a-bj}=\frac{a+bj}{a^2+b^2}.$$I'll leave arithmetic to you.
