<s>[INST] Given that $z=2-i$ and $z^2=3-4i$ find the roots of the equation $(z+i)^2=3-4i$ Given that $z=2-i$ and $z^2=3-4i$ find the roots of the equation $(z+i)^2=3-4i$
How do you use the given properties to find the roots? I can only obtain them the long way by working through 
$$\begin{align}
&(z+i)(z+i)&=3-4i\\
&z^2+2zi+i^2&=3-4i\\
&z^2+2zi-1-3+4i&=0\\
&z^2+2zi+4i-4&=0\\
&z^2+2z-2z+2iz+4i-4&=0\\
&(z-(2-2i))(z+2)&=0\\ 
\end{align}$$
$$z=2-2i, z=-2$$
 [/INST] \begin{align*}
(z+i)^2 &= 3 -4i\\
(z + i)^2 &= (2 -i)^2\\
(z+i)^2 - (2-i)^2 &= 0\\ 
[(z+i) - (2-i)][(z+i) + (2-i)] &= 0\\
[z - (2 - 2i)][z - 2] &= 0
\end{align*}
Therefore $z = 2 -2i$ or $z = 2$.
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