Question: Map $\Re(z) > 1$ under $f(z) = z^2 + 2z + 1$
What I've Tried: Using algebraic manipulation, I've gotten the equations:
$(w)^2 - 1 = z$ and by extension, $(u + iv)^{1/2} -1 = x + iy$ (Assuming $w = u+iv$ and $z=x+iy$)
I tried expanding this and found that:
$$u = x^2+y^2+2x+1 \;\text{ and }\; v = 2xy + 2y$$
Taking $x$ as $1$ (since $\Re(z)$ represents the half plane beginning at $x = 1$), I found: $4y = v$ $y= (u-3)^{1/2}$
I'm not really certain where to go from here. Should I have used polar coordinates? Thanks in advance to anyone willing to help