For complex $z$, find the value of $z$; $z^2 -2z+( 3-4i) =( 6+3i)$

\begin{array}{l} \boldsymbol{z^{2} -2z+( 3-4i) =( 6+3i)}\\\\ z^{2} -2z+( 3-4i) =( 6+3i)\\ x^{2} -y^{2} +2xyi-2x-2yi-3-7i=0\\ \left( x^{2} -y^{2} -2x-3\right) +i( 2xy-2y-7) =0\\ x^{2} -y^{2} -2x-3=0\rightarrow ( 1)\\ 2xy-2y-7=0\rightarrow ( 2)\\ ( 2) \ rewrite\ as\ x=\frac{7+2y}{2y}\\ \\ ( 1) \Longrightarrow \frac{( 7+2y)^{2}}{4y^{2}} -y^{2} -2\frac{( 7+2y)}{2y} -3=0\\ 4y^{4} +16y^{2} -49=0\\ considering,\ y^{2} =p;\\ 4p^{2} +16p-49=0\\ p=-2\mp \frac{\sqrt{65}}{2}\\ \therefore y=\mp \sqrt{-2\mp \frac{\sqrt{65}}{2}} \end{array} I got stuck at this point, is this the correct way or have another way to solve this equation. Thank you.

Yo can do this: \begin{align*} z^2 -2z+( 3-4i) &= 6+3i \\ z^2 -2z+1+2-4i & = 6+3i \\ z^2 -2z+1 & = 4+7i \\ (z-1)^2 & = 4+7i \\ z-1 & =(4+7i)^{1/2}\end{align*} Remember that for complex numbers, the square root has two values. And so $$z_1=\sqrt{65}e^{\arctan(7/4)i/2},\quad z_2=\sqrt{65}e^{(\arctan(7/4)/2+\pi)i} .$$