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

Question

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

1 Answer

Impact of this question

10133 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-14T11:45:04

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

Explanation:

Making $z = x + i y$ $z = x + i y$ we have

$| z + 1 | = | x + 1 + i y | = \sqrt{{(x + 1)}^{2} + y^{2}}$ $abs(z+1)=abs(x+1+i y)=sqrt((x+1)^2+y^2)$

also

$| z - 1 | = | x - 1 + i y | = \sqrt{{(x - 1)}^{2} + y^{2}}$ $abs(z-1)=abs(x-1+i y)=sqrt((x-1)^2+y^2)$

then

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

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

squaring

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

and again

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

Finally

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

an ellipse.

enter image source here

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question

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