The locus of Z satisfying $\left|z\right| + \left|z-1\right| =3$ is

$$\begin{gathered} \left|z\right| + \left|z-1\right| =3,\text{ Let, }\mathrm{z}=\mathrm{x}+\text{ iy and }\left|z\right|=\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2} \\ \sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}+\sqrt{(\mathrm{x}-1{)}^2+{\mathrm{y}}^2}=3 \end{gathered}$$

$$\begin{gathered} \mathrm{Squaring} \mathrm{on} \mathrm{both} \mathrm{sides} \\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2+9-6\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}={\mathrm{x}}^2+1-2\mathrm{x}+{\mathrm{y}}^2 \\ \Rightarrow \mathrm{x}+4=3\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2} \end{gathered}$$

$\Rightarrow \sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}-3=-\sqrt{(\mathrm{x}-1{)}^2+{\mathrm{y}}^2}$

$$\begin{gathered} \mathrm{Squaring} \mathrm{on} \mathrm{both} \mathrm{sides} \\ \Rightarrow {\mathrm{x}}^2+16+8\mathrm{x}=9\left({\mathrm{x}}^2+{\mathrm{y}}^2\right) \\ \Rightarrow 8{\mathrm{x}}^2+9{\mathrm{y}}^2-8\mathrm{x}-16=0 \\ \Rightarrow \begin{array}{r}\Rightarrow 8\left[{\left(\mathrm{x}-\frac{1}{2}\right)}^2-\frac{1}{4}\right]+9{\mathrm{y}}^2=16\\ \Rightarrow 8{\left(\mathrm{x}-\frac{1}{2}\right)}^2-2+9{\mathrm{y}}^2=16\\ \Rightarrow 8{\left(\mathrm{x}-\frac{1}{2}\right)}^2+9{\mathrm{y}}^2=18\\ \therefore \frac{{\left(\mathrm{x}-\frac{1}{2}\right)}^2}{\frac{9}{4}}+\frac{{\mathrm{y}}^2}{2}=1\end{array} \\ \text{ it represents an equation of ellipse. } \end{gathered}$$