Find all $∈ℂ$ such that $|-1|+|+i|=4$

Question

I'd like to find all points of the complex plane which satisfy

$|z − 1| + |z + i| = 4 $

I know that $z = x+yi$ and I know this is an ellipsis and that the answer is

$15x^2+15y^2-2xy-16x+16y-48=0$

But I don't know how to get right answer and how to plot a paragraph from the answer.

Answer 1

Part $1$: Getting to the Equation

Assume that $z=x+iy$. We can now substitute this into the equation:

\begin{align} |z-1|+|z+i|&=4\\ |x+iy-1|+|x+iy+i|&=4\\ |(x-1)+iy|+|x+(y+1)i|&=4\\ \sqrt{(x-1)^2+y^2}+\sqrt{x^2+(y+1)^2}&=4\\ \sqrt{x^2-2x+1+y^2}+\sqrt{x^2+y^2+2y+1}&=4\\ \end{align}

We can use a substituation $k=x^2+y^2+1$ (for convenience, use User Deepak's method if you want), and so

\begin{align} \sqrt{k-2x}+\sqrt{k+2y}&=4\\ \sqrt{k-2x}&=4-\sqrt{k+2y}\\ \left(\sqrt{k-2x}\right)^2&=\left(4-\sqrt{k+2y}\right)^2\\ k-2x&=16+(k+2y)-8\sqrt{k+2y}\\ x^2-2x+1+y^2&=16+(x^2+y^2+2y+1)-8\sqrt{x^2+y^2+2y+1}\\ x^2-2x+1+y^2-16-x^2-y^2-2y-1&=-8\sqrt{x^2+y^2+2y+1}\\ -2x-2y-16&=-8\sqrt{x^2+y^2+2y+1}\\ x+y+8&=4\sqrt{x^2+y^2+2y+1}\\ (x+y+8)^2&=\left(4\sqrt{x^2+y^2+2y+1}\right)^2\\ x^2+xy+8x+xy+y^2+8y+8x+8y+64&=16(x^2+y^2+2y+1)\\ x^2+y^2+2xy+16x+16y+64-16x^2-16y^2-32y-16&=0\\ -15x^2-15y^2+2xy+16x-16y+48&=0\\ 15x^2+15y^2-2xy-16x+16y-48&=0 \end{align}

This gives the equation mentioned in your question.

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Part $2$: Finding the $x$ and $y$ intercepts of the ellipse

Now, we can use Desmos for plotting, but by solving the equation for $x=0$ and $y=0$, you can also find the $x$ and $y$ intercepts easily (the method for figuring out the foci has been explained by the User Deepak).

If $y=0$,

\begin{align} 15x^2+15y^2-2xy-16x+16y-48&=0\\ 15x^2-16x-48&=0\\ x&=\frac{-(-16)\pm \sqrt{16^2-4\times15\times(-48)}}{2\times15}\\ x&=\frac{16\pm \sqrt{256+2880}}{30}\\ x&=\frac{16\pm \sqrt{3136}}{30}\\ x&=\frac{16\pm 56}{30}\\ x=\frac{-40}{30}=\frac{-4}{3}\ &\text{or } x=\frac{72}{30}=\frac{12}{5} \end{align}

The $y$-intercepts may be solved in a similar way, by substituting $x=0$ and using the quadratic formula.

I hope this helps!

Answer 2

We know that the ellipse is the locus of points whose sum of distances from two fixed points is constant.
If $z$ is any complex number on argand plane then $|z-1|$ denotes the distance of of $z$ from 1 and $|z+i|$ denotes the distance of $z$ from $-i$.
Hence, $|z+i|+|z-1|=4$ denotes the equation of the ellipse where the sum of distances from $-i$ and $1$ is constant.
We also know that the fixed points are the foci of the ellipse. Hence the foci of the given ellipse are $(1,0)$ and $(0,-1)$.
Therefore, by property of ellipse;
Solve this equation to get the answer
Click on above link for the equation.

Answer 3

I will work through the algebra carefully for your benefit. Some steps seemed natural for me to skip as they were obvious to me. If you can't follow along, I suggest you work through what you find "missing" with pen and paper.

$|z-1| + |z-i| = 4$

$|z-1| = 4 - |z-i|$

Convert to real Cartesian coordinates:

$\sqrt{(x-1)^2 + y^2} = 4 - \sqrt{x^2 + (y+1)^2}$

Square both sides,

$(x-1)^2 + y^2 = 16 + x^2 + (y+1)^2 - 8\sqrt{x^2 + (y+1)^2}$

Don't expand immediately. Exploit the difference of squares identity ($a^2 - b^2 = (a-b)(a+b)$)when rearranging,

$(2x-1) + (2y+1) + 16 = 8\sqrt{x^2 + (y+1)^2}$

$x + y + 8 = 4\sqrt{x^2 + (y+1)^2}$

Square again,

$(x+y)^2 + 16(x+y) + 64 = 16(x^2 + (y+1)^2)$

$x^2 + 2xy + y^2 + 16x + 16y + 64 = 16x^2 + 16y^2 + 32y + 16$

$15x^2 + 15y^2 - 2xy - 16x + 16y - 48 = 0$

exactly as required.

As far as plotting it on axes goes, you can use Desmos or some other computerised solution. Or do it tediously by hand.

Here is the plot from Desmos. It is a rotated ellipse, not centred around the origin, about as general as you can get.

But if you were just sketching the curve and wanted the key points, I would focus (no pun intended) on using the original complex locus equation to figure out the foci, which are quite obviously $(1,0)$ and $(0,-1)$. Then note that the axis of symmetry will be the line passing through these points, so $y = x-1$. The centre will be the midpoint of the foci (so that's $(\frac 12, -\frac 12)$). Finally, work out the $x$ and $y$ intercepts and the points of intersection with the axis of symmetry by solving the relevant quadratics, so that's $6$ points in total that define the curve, quite enough to get a good idea of the shape, especially with the symmetry about the axis.