I need to find where $f(x+iy)=-6(\cos x+i\sin x)+(2-2i)y^3+15(y^2+2y)$ is complex differentiable.
I first rearranged the function into its real and imaginary parts: $f(x+iy)=(-6\cos x+2y^3+15y^2+30y)+i(-6\sin x-6y^2)$
That means $u(x,y)=-6\cos x+2y^3+15y^2+30y$ and $v(x,y)=-6\sin x-6y^3$.
Then, if we take the partial derivative of u and v in terms of x and y:
Then, by the Cauchy-Riemann equations, $u_x=v_y$ and $u_y=-v_x$.
This means that: $6\sin x=-18y^2$ and $6y^2+30y+30=6\cos x$.
This is where I am stuck. How do I solve for x and y? I was thinking that I could proceed in this way:
$\sin^2 x + \cos^2 x=1 \Rightarrow (-3y^2)^2+(y^2+5y+5)^2=1 \Rightarrow 10y^4+10y^3+35y^2+50y+24=0$
However, from here, how do I solve for y and then solve for x? I'd appreciate any tips. Thanks for your help in advance!