Which Area Is $|z-i|+|z+i|<4$?

Question

We can set $z=x+yi$

and get the expression $2x^2+2y^2+2<(-x^2-y^2+7)^2$

But how to continue from here? or is there a simpler way?

Answer 1

Hint:

in the Argand Plane it is the ellipse ( with foci at $\pm i$) of equation $$\frac{x^2}{3}+\frac{y^2}{4}=1$$

can you prove this? So the area is $2\pi \sqrt{3}$

Answer 2

Hint:

This is the interior of an ellipse, centred at the origin, with foci the images in the Argand-Cauchy plane of $i$ and $-i$. So we only have to find its semi-axes $a$ and $b$: if its reduced equation is $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$$ the area will be $\pi ab$.

Can you points on the $x$ axis and on the $y$ axis, to determine the semi-axes?

Answer 3

$|z-i|+|z+i|=4 $ is an ellipse with foci at $i$ and $-i$ and the sum of the distances from the foci is $4$.

If the maximum $y$ value ($x$ real, $y$ imaginary) is $b$, then $b-1+b+1 = 4$ so $b = 2$.

If the maximum $x$ value is $a$, then $2\sqrt{a^2+1} = 4$ or $a^2+1 = 4$ or $a = \sqrt{3}$.

The area is $\pi a b = 2\pi \sqrt{3}$.

Answer 4

This is an ellipse centered at origin and foci located on the y-axis at $(0,\pm 1)$ and the major axis length of 4

The equation of such ellipse is $$ \frac{x^2}{3}+\frac{y^2}{4}=1$$

Answer 5

Perhaps you know that a defining feature of an ellipse is that from any point on its boundary, the sum of the distance from that point to one focus with the distance from that point to the other focus is constant. The equation $$|z-i| + |z+i| = 4$$ tells you to find all points so that summing the distance to $i$ with the distance to $-i$ is a constant ($4$). It has $i$ and $-i$ as the foci.

With this mapped out you can see the major and minor axes coincide with the imaginary and real axes. And the ellipse is centered at the origin. And its northernmost point is $2i$ and southernmost point is $-2i$, so its major radius is $2$. It takes a little Pythagorean calculation to find its easternmost point is $\sqrt{3}$ and westernmost point is $-\sqrt{3}$. So its minor radius is $\sqrt{3}$.

The area formula for an ellipse is $\pi ab$, where $a$ and $b$ are the minor and major radii.

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Side note: the inequality you converted to is quartic in $x$ and $y$, and therefore does not form an ellipse. So I suspect there is some mistake in obtaining that inequality.