Drawing sets of Complex Variables

[MaestroBach]

Homework Statement

Draw the following set in the complex plane: |z - i| + |z + i| = 3

Relevant Equations

N/A

I tried saying z = x + iy, then squared both sides so that I would get something that looked like:
|z - i|^2 + |z + i|^2 + |z - i||z + i| = 3, where the first two terms are simple but the third term is what I don't know what to do with. I'm wondering if I'm using the wrong approach.

For that matter, on a simpler part, where I was told to draw |z - 1 + i| = 2, I said z = x + iy and then got |x - 1 + i + iy|^2 = 2^2 (I squared both sides and rewrote z), then said (x-1)^2 + (1 + y)^2 = 4 to get the equation of a circle (which I figured is what I would get from the equation). Did I even do this right?

Appreciate any help, my textbook is trash unfortunately.

 

[fresh_42]

Whenever you have an expression $a+b$ and want to know something about $a^2$ and $b^2$, which formula can you use?

 

[MaestroBach]

Whenever you have an expression $a+b$ and want to know something about $a^2$ and $b^2$, which formula can you use?

Umm, maybe I'm misunderstanding the intent behind your question but all I can think of is squaring a+b..

 

[fresh_42]

If you square $a+b$ then you get $2ab$ and nothing is won. I haven't done the exercise, but whenever I see an expression $a+b$ which in our case is $a+b=|z-i|+|z+i|=\sqrt{x^2+(y-1)^2}+\sqrt{x^2+(y+1)^2}$ then squaring doesn't remove the nasty roots. A better binomial formula is $(a+b)\cdot (a-b)=a^2-b^2$ where no mixed terms appear. So I would set $a:=|z-i|\, , \,b:=|z+i|\, , \,z=x+iy$ and work with that formula.

 

[MaestroBach]

If you square $a+b$ then you get $2ab$ and nothing is won. I haven't done the exercise, but whenever I see an expression $a+b$ which in our case is $a+b=|z-i|+|z+i|=\sqrt{x^2+(y-1)^2}+\sqrt{x^2+(y+1)^2}$ then squaring doesn't remove the nasty roots. A better binomial formula is $(a+b)\cdot (a-b)=a^2-b^2$ where no mixed terms appear. So I would set $a:=|z-i|\, , \,b:=|z+i|\, , \,z=x+iy$ and work with that formula.

But then, in the process of doing so, I would have to multiply the 3 on the other side by (a-b) too, and wouldn't that be just as problematic?

 

[fresh_42]

You can also draw a picture: Which points are $|z-i|=r$ and which $|z-(-i)|=s$? Those sets must intersect for $r+s=3$.

 

[fresh_42]

But then, in the process of doing so, I would have to multiply the 3 on the other side by (a-b) too, and wouldn't that be just as problematic?

The approach is a bit work to do, yes, so maybe a drawing is faster. But the multiplication yields two equations with two unknowns (o.k. $y$ will remain unknown, too, but resubstitution will get the answer):
$a+b=3$ and $(a^2-b^2)=3a-3b$.

 

[FactChecker]

Are you familiar with the relationship between an ellipse and its focal points?

 

[MaestroBach]

Are you familiar with the relationship between an ellipse and its focal points?

Yup, I am! Didn't recognize this as an ellipse though haha. I'm still having trouble getting it into the standard form for the equation of an ellipse as well

 

[MaestroBach]

The approach is a bit work to do, yes, so maybe a drawing is faster. But the multiplication yields two equations with two unknowns (o.k. $y$ will remain unknown, too, but resubstitution will get the answer):
$a+b=3$ and $(a^2-b^2)=3a-3b$.

The two equations only give me 1 = 1, because essentially, $(a^2-b^2)=3a-3b$ and $a+b=3$ are the same equation, with one just multiplied by a factor. Kinda like how if I had $x + y = 2$, having another equation $2x + 2y = 4$ wouldn't help.

 

[FactChecker]

Yup, I am! Didn't recognize this as an ellipse though haha. I'm still having trouble getting it into the standard form for the equation of an ellipse as well

For the standard Cartesian coordinate form, the length, $a$, of the semi-major axis is trivial and the length, $b$, of the semi-minor axis just requires the Pythagorean Theorem. (see https://en.wikipedia.org/wiki/Ellipse#In_Cartesian_coordinates )
Or once you have $a$ and $b$, you might prefer to use the standard parametric equation:
$(x,y) = (a*cos(t), b*sin(t)), 0 \le t \le 2 \pi$

 

[fresh_42]

The two equations only give me 1 = 1, because essentially, $(a^2-b^2)=3a-3b$ and $a+b=3$ are the same equation, with one just multiplied by a factor. Kinda like how if I had $x + y = 2$, having another equation $2x + 2y = 4$ wouldn't help.

No, it does not. $a^2-b^2=4y$ or $-4y$ depending on which is which. So we have $-4y=3(a-b)$ and the original $a+b=3$, which makes two linear independent equations for two variables. Then you solve $a=p_a(y),b=p_b(y)$, and calculate $|x+(y-1) i |+|x - (y+1) i |=3$ and get the equation for the ellipse.

Or you follow post #6: stick a needle in $(0,i)$ and a needle in $(0,- i )$, cut a string of length $3$, fix it between the needles and try to draw two circles of radius $r$ and $s$ with $r+s=3$ centered at the needles simultaneously.

 

[SammyS]

Homework Statement:: Draw the following set in the complex plane: |z - i| + |z + i| = 3
Relevant Equations:: N/A

I tried saying z = x + iy, then squared both sides so that I would get something that looked like:
|z - i|^2 + |z + i|^2 + |z - i||z + i| = 3, where the first two terms are simple but the third term is what I don't know what to do with. I'm wondering if I'm using the wrong approach.

For that matter, on a simpler part, {Maybe later.}​

There are at least a couple of problems with your result for squaring both sides.

1. Of course, $3^2=9$. not 3.
2. Also, $(a+b)^2=a^2+b^2+2ab$. You left out the coefficient, 2.

The problem can be worked out in this fashion, but it can be very messy. Keeping various quantities grouped together at critical steps can help immensely.

As @fresh_42 points out, squaring both sides in this fashion, still leaves you with one radical. But, that's better than the two you start with. Isolate that radical and square both sides again.

$\displaystyle \left( \sqrt{x^2+(y+1)^2}+\sqrt{x^2+(y-1)^2} \right)^2 = 3^2$

$\displaystyle x^2+(y+1)^2 + x^2+(y-1)^2 +2\sqrt{x^2+(y+1)^2}\sqrt{x^2+(y-1)^2} = 9$

Looks messy, but lots to simplify as well.

 

[ehild]

As @fresh_42 points out, squaring both sides in this fashion, still leaves you with one radical. But, that's better than the two you start with. Isolate that radical and square both sides again.

It will be simpler this way: (You do not get product of square roots)
$\left( \sqrt{x^2+(y+1)^2}\right)^2 =\left( 3-\sqrt{x^2+(y-1)^2}\right)^2$

 

[SammyS]

It will be simpler this way: (You do not get product of square roots)
$\left( \sqrt{x^2+(y+1)^2}\right)^2 =\left( 3-\sqrt{x^2+(y-1)^2}\right)^2$

Right you are. This works out so much better.

I had first tried this, but must have made a significant error.