<s>[INST] 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.
 [/INST] 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
$$z_1=\sqrt{65}e^{\arctan(7/4)i/2},\quad    z_2=\sqrt{65}e^{(\arctan(7/4)/2+\pi)i} .$$
</s>