# 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.

• if you want to go by your method, Rotate the axes by $45^{\circ}$ Counter clockwise, then you will get to know the shape Nov 30, 2015 at 11:05
• The answer given in the book is ''two points''.@EkaveeraKumarSharma
– diya
Nov 30, 2015 at 11:59
• The given answer is incorrect (not unheard of in textbooks, for a variety of reasons). The separation of $\ z \ = \ 1 \$ and $\ z \ = \ -i \$ is plainly less than $\ 2 \ \ , \$ so there can easily be more than two points for which the sum of the distances from those points is equal to $\ 2 \ \ . \$ The points $\ z \ = \ 0 \$ and $\ z \ = \ 1 - i \ \$ satisfy the equation, but so do $\ z \ = \ \frac43 \ \ , \ \ z \ = \ -\frac43 i \ \ , \ \ z \ = \ -\frac13 - i \ \$ and a host of other points. The correct answer is $\ \mathbf{(D)} \ \ .$
– user882145
Nov 10, 2022 at 9:21

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 ? :-$)$

• The answer given in the book is ''two points'',and i suppose you are suggesting ellipse.@Lucian
– diya
Nov 30, 2015 at 12:00
• @diya: $3x^2+3y^2-4x+4y-2x y=0$ represents an ellipse. Even if that ellipse were degenerate $($i.e., one of its axes is $0,$ which here simply isn't the case, since the distance between $1=(1,0)$ and $-i=(0,-1)$ is $\sqrt2,$ not $2),$ we'd still be dealing with and entire line segment, definitely not with a finite number of points $($unless that finite number of points happens to be $0,$ which occurs when the sum is less then the actual distance$)$. So even if by mistake they forgot a radical sign above the $2,$ the answer would never be A. Nov 30, 2015 at 22:29

Let two fixed points $$F_1,F_2$$ which are called the foci and $$a \in \mathbb {R_+}$$, $$2a>F_1 F_2$$ be given. An ellipse $$E$$ is the set of the points $$P$$, such that $$E=\left\{P\in \mathbb {R} ^{2}: \,|PF_1|+|PF_2|=2a\right\}$$ . According to this definition, $$F_1=1, \, F_2=-i, \, a=1$$ and $$2a>F_1 F_2$$, where $$F_1 F_2=\sqrt 2$$. That is this shape is an ellipse.

$|z+i|+|z-i|=2 \iff \max\{|(z+i)-(z-1)|;|(z+i)+(z-i)|\}=2$, $\max\{|2z|;|2i|\}=2$, 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 $a<0,b<0; a>0,b>0$ 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 $a<0,b>0; a>0;b<0$, 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 $c=\max\{|a+b|;|a-b|\}$.

• Confusing. On the one hand we're talking about modulus in $\mathbb C$ and on the other .. $\mathbb R$? Is the claim also true in $\mathbb C$? Sep 30, 2020 at 8:08