Chapter 2 Section 17 Problem 13 - Mary L Boas. 
Find real $x$ and $y$ for which $|z+3| = 1 - iz$, where $z=x+iy$.

My attempt:
$\require{cancel}\begin{align} |z+3|\;&=\;1-iz \\  \sqrt{(x+3)^2+y^2}\;&=\;1-iz \\ (x+3)^2+y^2\;&=\;(1-iz)^2 \\ x^2 +6x+9+y^2 &= 1-2iz-z^2 \\ x^2 +6x+9+y^2 &= 1-2i(x+iy)-(x+iy)^2 \\ x^2 +6x+9+y^2 &= 1-2ix+2y-x^2-2xiy+y^2 \\ x^2 +6x+9+\cancel{y^2} &= 1-2ix+2y-x^2-2xiy+\cancel{y^2} \\ (x^2 +6x+9) &= (1+2y-x^2)+i(-2x-2xy)
\end{align}$
Is this the correct way to keep going further or initially errors were made?

$$iz=1-|z+3|$$ which is real, so $z$ must be purely imaginary
Set $z=iy$ where $y$ is real
$$1+y=|3+iy|\ge0\implies y\ge-1$$
$$(1+y)^2=3^2+y^2$$
