In the complex plane, what does $abs(z+1 ) +abs(z-1)= 8$ look like?

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Answer 1 · 2016-06-14T11:45:04

The ellipse $15 x^2 + 16 y^2 = 240$

Making $z = x + i y$ we have

$abs(z+1)=abs(x+1+i y)=sqrt((x+1)^2+y^2)$

also

$abs(z-1)=abs(x-1+i y)=sqrt((x-1)^2+y^2)$

then

$abs(z+1)+abs(z-1) = 8$ is equivalent to

$sqrt((x+1)^2+y^2) +sqrt((x-1)^2+y^2)=8$

squaring

$(sqrt((x+1)^2+y^2) +sqrt((x-1)^2+y^2))^2=8^2$

and again

$(2 sqrt[(-1 + x)^2 + y^2] sqrt[(1 + x)^2 + y^2])^2 = (8^2-(2 + 2 x^2 + 2 y^2))^2$

Finally

$15 x^2 + 16 y^2 = 240$

an ellipse.

enter image source here

In the complex plane, what does abs(z+1 ) +abs(z-1)= 8 abs(z+1 ) +abs(z-1)= 8 look like?