I have to graph the inequality: $$|z-1|+|z-i| \leqslant 4$$
My calculation: $$\sqrt{(x-1)^2+y^2}+\sqrt{x^2+(y-1)^2}\leqslant4$$ $$\vdots$$ $$-19-2x+2y\leqslant-8\sqrt{x^2+(y-1)^2}$$
This is where the problem comes in. I tried:
$$\frac{-19-2x+2y}{-8\sqrt{x^2+(y-1)^2}}\geqslant1$$
Here I got stuck...
$|z-1|$ is the distance between the point, $z$, and $1$. $|z-\mathrm{i}|$ is the distance between the point, $z$, and $\mathrm{i}$. You want the closed region whose boundary is the locus of points whose sum of distances to $1$ and $\mathrm{i}$ is $4$. That locus is an ellipse with foci at $1$ and $\mathrm{i}$. Now that we know what we're going to draw, how do we get its coordinates?
(There are standard methods from the theory of conic sections to go from the description above to various measurements of the ellipse and locations of points on the ellipse, useful for drawing it. Let's assume we aren't using those methods.)
We square both sides and collect non-radical terms to the right, cancelling as possible. Twice. $$ \sqrt{(x-1)^2 + y^2} + \sqrt{x^2 + (y-1)^2} \leq 4 $$ $$ \sqrt{(x-1)^2 + y^2}\sqrt{x^2 + (y-1)^2} \leq 7 - x^2 - y^2 +x +y $$ $$ 0 \leq -15 x^2 - 2 xy - 15 y^2 + 16 x + 16 y + 48 $$ Notice that $x = y = 0$ satisfies this inequality, so our graph will indicate that whichever region contains $(0,0)$ satisfies the inequality.
To find the boundary of the region, we replace the inequality with an equality and solve. Here, we view the expression on the right as a quadratic in $y$ and apply the quadratic formula. $$ y = \frac{8 - x \pm 4 \sqrt{7}\sqrt{-2x^2 + 2x + 7}}{15} $$ Let $y_\text{upper}$ be the expression obtained when we choose "$+$" for $\pm$ and $y_\text{lower}$ be obtained when we choose "$-$". The graph of the expression $-2x^2 + 2x + 7$ is a parabola that opens down. The square root of that expression is not real if that expression is negative, so we find the roots of that expression. (If there are none, the graph is empty. If there is only one, the graph is a single point. If there are two the graph is two arcs on the interval bounded by the two roots.) There are two roots $x = \frac{1 \pm \sqrt{15}}{2}$.
Let's find the left-most and right-most endpoints of our graph. $$ \left. \phantom{\frac{1}{1}}(x,y) \right|_{x = \frac{1 \pm \sqrt{15}}{2}} = \left\{ \left( \frac{1 - \sqrt{15}}{2} , \frac{15 + \sqrt{15}}{30}\right), \left( \frac{1 + \sqrt{15}}{2} , \frac{15 - \sqrt{15}}{30}\right) \right\} $$
The maximum of the upper curve is found from $y_\text{upper}'(x) = \frac{-1}{15}\left( \frac{2 \sqrt{7} (4x-2)}{\sqrt{-2x^2 +2x +7}} + 1 \right) = 0$, giving the point $\left( \frac{15-\sqrt{15}}{30}, \frac{1+\sqrt{15}}{2} \right)$. Likewise the minimum on the lower curve is $\left( \frac{15+\sqrt{15}}{30}, \frac{1-\sqrt{15}}{2} \right)$, obtained from $y_\text{lower}'(x) = \frac{-1}{15}\left( \frac{2 \sqrt{7} (2-4x)}{\sqrt{-2x^2 +2x +7}} + 1 \right) = 0$.
Sketching arcs between these points, we obtain
Then, recalling that the region containing the origin is what we are to graph,