Question

i'm asked to draw graph for following equality:

|z-1|+|z+1|=3

note that: z=x+yi

Answer 1

you can think of z - 1 as z - (1+0i)
and | z - (1+0i) | is the distance between point z and 1+0i on the complex plane
similarly | z +1| is the distance between z and -1

so this question is asking for the locus of points whose distance to 1 and -1 sum to a constant 3

maybe an ellipse?

Answer 2

yes it is an ellipse, but i don't know how to mathematically prove it.
i have instructions in this way:

\[|z+1|=|(x+1)+yi|=\sqrt{(x+1)^{2}+y^2}\]
\[|z-1|=|(x-1)+yi|=\sqrt{(x-1)^{2}+y^2}\]

Now I somehow have to play around with this:
\[\sqrt{(x+1)^{2}+y^2}+\sqrt{(x-1)^{2}+y^2}=3\]
to get ellipse equation which is :
\[\frac{ 4x^2 }{9 }+\frac{ 4y^2 }{ 5 } = 1\]

Answer 3

by definition it is an ellipse
http://www.mathsisfun.com/definitions/ellipse.html

Answer 4

I assume you made progress ?
\[ \sqrt{(x+1)^{2}+y^2}+\sqrt{(x-1)^{2}+y^2}=3 \\
\sqrt{(x+1)^{2}+y^2}= 3 - \sqrt{(x-1)^{2}+y^2}
\]
square both sides
\[ (x+1)^{2}+y^2 = 9 + (x-1)^{2}+y^2 -6 \sqrt{(x-1)^{2}+y^2}\\
6 \sqrt{(x-1)^{2}+y^2}= 9 + (x-1)^{2}+\cancel{y^2} - (x+1)^{2}-\cancel{y^2} \\
6 \sqrt{(x-1)^{2}+y^2}= 9 + x^2 -2x + 1 - ( x^2 +2x +1)\\
6 \sqrt{(x-1)^{2}+y^2}= 9 -4x
\]
square both sides
\[ 36\left((x-1)^2 + y^2\right) = 81 +16x^2 -72x \\

36\left(x^2 -2x +1 + y^2\right) = 81 +16x^2 -72x\\
36 x^2 -72x + 36 +36y^2 =81 +16x^2 -72x\\
20 x^2 + 36y^2= 45
\]
which simplifies to
\[ \frac{4}{9}x^2 + \frac{4}{5} y^2 =1 \]