1. Find the values of $x$ and $y$ that satisfy the following equations:
(i) $-4 x + 8 y i + 7 = 2 x - 3 y i + 7 i$.
(ii) $x + y i = \sqrt{a + b i}$.
1. (i) Solution:
$$-4 x+8 y i+7=2 x-3 y i+7 i$$

Rearranging terms
$$-6 x+11 y i=-7+7 i$$
Thus
$$x=\frac{7}{6}, \quad y=\frac{7}{11}$$
(ii) Solution:
$$x+y i=\sqrt{a+b i}$$

Squaring both sides gives
$$x^{2}-y^{2}+2 x y i=a+b i,$$

Therefore
$$\left\{\begin{array}{l}
x^{2}-y^{2}=a \\
2 x y=b
\end{array}\right.$$

Squaring and adding the two equations gives

That is $\square$
$$\begin{array}{l}
\left(x^{2}-y^{2}\right)^{2}+4 x^{2} y^{2}=a^{2}+b^{2} \\
\left(x^{2}+y^{2}\right)^{2}=a^{2}+b^{2} \\
x^{2}+y^{2}=\sqrt{a^{2}+b^{2}}
\end{array}$$

Solving (1) and (3) gives
$$x^{2}=\frac{\sqrt{a^{2}+b^{2}}+a}{2}, y^{2}=\frac{\sqrt{a^{2}+b^{2}}-a}{2},$$

Thus
$$x= \pm \sqrt{\frac{\sqrt{a^{2}+b^{2}}+a}{2}}, y= \pm \sqrt{\frac{\sqrt{a^{2}+b^{2}}-a}{2}}$$

From (2), when $b>0$, $x$ and $y$ have the same sign, and when $b<0$, $x$ and $y$ have opposite signs.