What is the shape of the graph $|z-1|+|z+i|=2$ in the complex plane? What is the shape of the graph $|z-1|+|z+i|=2$ in the complex plane?
$(A)\text{two points}\hspace{1cm}(B)\text{a line}\hspace{1cm}(C)\text{a parabola}\hspace{1cm}(D)\text{an ellipse}$
Let us take $z=x+iy$
$|(x-1)+iy|+|x+i(y+1)|=2$
$\sqrt{(x-1)^2+y^2}+\sqrt{x^2+(y+1)^2}=2$
Upon simplifying,$3x^2+3y^2-4x+4y-2xy=0$
But with this equation, i cannot tell the shape of the graph.Please help me.
Hint: $|a-b|$ represents the distance between the two points a and b in the complex plane. What geometric shape is defined by the sum of two distances being constant ? :-$)$
$|z+i|+|z-i|=2 \iff \max\{|(z+i)-(z-1)|;|(z+i)+(z-i)|\}=2$,
$\max\{|2z|;|2i|\}=2$,
because for sure $|2i|=2$, so $|2z|\leq2$,
$|z|\leq 1$  and there is just a circle.

When i'm using $\max\{\cdots\}$ i mean the highest number contained in $\{\cdots\}$

I used here theorem that: $|a|+|b|=c \iff  c=\max\{|a+b|;|a-b|\}$.

Proof: Considering $a&lt;0,b&lt;0; a&gt;0,b&gt;0$ we get$-a-b=c$ or $a+b=c$, but $c=|a|+|b| \geq 0$, so we can easy generalize it to $|a+b|=c$, when $a&lt;0,b&gt;0; a&gt;0;b&lt;0$, similar generalize $|a-b|=c$, but when $c=|a+b|, |a-b|$ is obviously smaller, and if $c=|a-b|, |a+b|$ is obviously smaller, so generalizing all combinations we get $c=\max\{|a+b|;|a-b|\}$.