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.

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