Example 3. Let $w, z$ be complex numbers, and the equation of $w$ is $w^{2}+z \cdot w$ $+z i=0$ which always has a real root $\alpha$ ( $\alpha$ is a real parameter). Try to find the locus of point $Z$.

Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
Given that the equation of $w$ is $w^{2}+z \cdot w+z \boldsymbol{i}=0$ always has a real root $\ell$, hence
$$
a^{2}+r_{i} z+i=0 \text {. }
$$

Let $z=x+y i(x, y$ be real numbers $)$,
we get: $\left(a^{2}+a x-y\right)+i(a y+x)=0$.
From the equality of complex numbers, we have: $\left\{\begin{array}{l}x^{2}+a x-y=0, \\ a y+x=0 .\end{array}\right.$
At the origin;
When $y \neq 0$, $a=-\frac{x}{y}$, substituting into (2) we get
$$
\left(-\frac{x}{y}\right)^{2}+\left(-\frac{x}{y}\right)-y=r_{0}
$$

which is $x^{2}=\frac{y^{3}}{1-y}$. This is the equation of the trajectory of point $Z$.
Example 3 Explanation:
$1^{\circ}$ To find the trajectory of point 乙 in the complex plane, in essence,
$2^{\circ}$ When finding the trajectory of a point, fully utilize the conditions given in the problem to find the relationship between the real and imaginary parts.

Example 4. What does the equation $|z+i|+|z-i|=3(z=x+y i)$ represent in the $x y$ plane?
Solution 1: From $|z+i|=|x+y i+i|$
$$
\begin{array}{l}
\quad=\sqrt{x^{2}+(y+1)^{2}}, \\
|z-i|=\sqrt{x^{2}+(y-1)^{2}},
\end{array}
$$

According to the problem, we have:
$$
\sqrt{x^{2}+(\overline{y+1})^{2}}+\sqrt{x^{2}+(y-1)^{2}}=3 .
$$

Simplifying, we get: $\frac{x^{2}}{5}+\frac{y^{2}}{9}=1$. This represents an ellipse.
Solution 2. According to the problem, on the $x y$ plane, the sum of the distances from a moving point to the two fixed points $(0,1),(0,-1)$ is 3. Therefore, the trajectory of the moving point $Z$ is an ellipse.
$$
\begin{array}{l}
\because 2 c=2, \quad 2 a=3, \\
\therefore b^{2}=a^{2}-c^{2}=\frac{9}{4}-1=\frac{5}{4} .
\end{array}
$$

Since the foci are on the $y$-axis, the equation of the trajectory in the $x y$ plane is: $\frac{x^{2}}{\frac{5}{4}}+\frac{y^{2}}{\frac{9}{4}}=1$.