The problem asks to graph $|z-1| + |z-i| < 2 \sqrt{2}$.
So I can tell this is an ellipse and that the distance from $(1,0)$ and $(0,1)$ sum to less than $2\sqrt{2}$. However, I can't see to get this into $\left (\frac{x}{a} \right )^2 + \left (\frac{y}{b} \right )^2 < 1$ form.
$|z-1| + |z-i| < 2 \sqrt{2}$
$\sqrt{(a-1)^2 + b^2 }+ \sqrt{(a^2 + (b-1)^2)} < 2 \sqrt{2}$
$2 - 2a + 2a^2 - 2b + 2b^2 + 2 \sqrt{a^2 + (b-1)^2} \sqrt{(a-1)^2 + b^2} < 2 \sqrt{2}$
But this doesn't help me collect similar terms.
Further algebraic manipulation seems to go nowhere.
Is there a different approach I should be trying?
The form $(x/a)^2+(y/b)^2=1$ will only produce ellipses whose major and minor axes are parallel to the coordinate axes.
Your ellipse has foci at $1$ and $i$, so its major axis is not parallel to either of the coordinate axes. Therefore it cannot end up in the form you're seeking.
Its major axis is $2\sqrt2$, twice the focal distance, so the major axis goes between $\frac32 - \frac12i$ and $-\frac12+ \frac32i$.
Since the major axis length is twice the focal distances, the ends of the minor axis will be at the tips of the two equilateral triangles that have the foci as two of their corners, which are $$ \frac12+\frac12i \pm \frac{\sqrt3}2(1+i) $$