Thread:

1. Loci In complex plane

Show that in an Argand diagram the equation

$\arg (z-2) - \arg (z-2i) = \frac{3 \pi}{4}$

represents an arc of a circle and that $\frac{|z-4|}{|z-1|}$ is constant on this circle.

Find values of z corresponding to the points in which the circle is cut by the curve given by

$|z-1| + |z-4| = 5$
I am okay with getting the that the first arg equation describes the arc of a circle, but i am a bit stuck on how to interpret the next part.

anyone willing to give me a hand?

thanks, Bobak

2. Originally Posted by bobak
I am okay with getting the that the first arg equation describes the arc of a circle, but i am a bit stuck on how to interpret the next part.

anyone willing to give me a hand?

thanks, Bobak
The locus of $|z-1| + |z-4| = 5$ is an ellipse with centre at (5/2, 0), major axis of length 5 and vertices at (0, 0) and (5, 0).

There are several ways to see this - the simplest is if you understand the locus definition of an ellipse ..... Alternatively, you can sub z = x + iy, but there are a couple of twists and turns that will catch the unwary.

This gives a geometric picture of the lay of the land - there's only one intersection point.

3. Originally Posted by mr fantastic
The locus of $|z-1| + |z-4| = 5$ is a line segment joining z = 1 and z = 4. There are several ways to see this - believe it or not, the simplest is if you understand the locus definition of an ellipse ..... Alternatively, you can sub z = x + iy, but there are a couple of twists and turns that will catch the unwary.

Does this give you a nudge?
My apologies - this is wrong. It's an ellipse, not a line segment. I misread. It's an ellipse, with centre at (5/2, 0) and major axis of length 5. Vertices at (0, 0) and (5, 0).

I'm a bit short on time now - I'll drop by later unless someone beats me to the punch.

4. Originally Posted by bobak
I am okay with getting the that the first arg equation describes the arc of a circle, but i am a bit stuck on how to interpret the next part.

anyone willing to give me a hand?

thanks, Bobak
OK, from what I posted earlier you can hopefully see that there's only one intersection point of the arc with the ellipse.

How to find it:

You know from the earlier part that $|z - 1| = \lambda |z - 4|$, where you'll have found the value of $\lambda$.

Sub $|z - 1| = \lambda |z - 4|$ into $|z - 1| + |z - 4| = 5$:

$\lambda |z - 4| + |z - 4| = 5 \Rightarrow |z - 4| (\lambda + 1) = 5 \Rightarrow |z - 4| = \frac{5}{\lambda + 1}$.

This is a circle of known radius $\left( \frac{5}{\lambda + 1} \right)$ and centre at z = 4.

Now sub $|z - 4| = \frac{1}{\lambda} |z - 1|$ into $|z - 1| + |z - 4| = 5$:

$|z - 1| + \frac{1}{\lambda} |z - 1| = 5 \Rightarrow |z - 1| \left( 1 + \frac{1}{\lambda} \right) = 5 \Rightarrow |z - 1| = \frac{5}{1 + \frac{1}{\lambda}} = \frac{5 \lambda}{\lambda + 1}$.

This is a circle of known radius $\left( \frac{5 \lambda}{\lambda + 1} \right)\,$ and centre at z = 1.

Now find the intersection of these two circles (but only keep one of the solutions, the one that is the intersection of the arc with the ellipse ......):

$(x - 4)^2 + y^2 = r_1^2\,$, where $r_1 = \frac{5}{\lambda + 1}$ .... (1)

$(x - 1)^2 + y^2 = r_2^2\,$, where $r_2 = \frac{5 \lambda}{\lambda + 1}$ .... (2)

Equation (1) - Equation (2): $\, (x - 4)^2 - (x - 1)^2 = r_1^2 - r_2^2$

$\Rightarrow (x - 4 + x - 1)(x - 4 -[x - 1]) = (r_1 + r_2)(r_1 - r_2)$

$\Rightarrow 3(2x - 5) = \frac{25(\lambda - 1)}{\lambda + 1}$

$\Rightarrow x = ......$. Remember that you know the value of $\lambda$ from the first part.

Substitute x into either equation (1) or (2) and solve for y: y = ......

Therefore the intersection point is z = x + iy = ........

Let me know if you have further trouble with this.

5. Thank you a lot for your response Mr Fantastic.

I think it is only appropriate that I post all my working.

For the first part.

$Part (1)$
$\arg (z-2) - \arg (z-2i) = \frac{3 \pi}{4}<br />$

rewriting arg(Z) in arctan from where $z = x +iy$

$\arctan \left ( \frac{y}{x-2} \right ) - \arctan \left ( \frac{y-2}{x} \right ) = \frac{3 \pi}{4}$

.... $\tan \left (\arctan \left ( \frac{y}{x-2} \right ) - \arctan \left ( \frac{y-2}{x} \right ) \right )= \tan \left ( \frac{3 \pi}{4} \right )$

then using compound angle formula.

$\frac{ \frac{y}{x-2} - \frac{y-2}{x} }{1 + \frac{y}{x-2} \cdot \frac{y-2}{x}} = -1$

$\frac{y-2}{x} - \frac{y}{x-2} = 1 + \frac{y}{x-2} \cdot \frac{y-2}{x}$
.... skipping some step
giving the equation of he circular arc as $x^2 + y^2 = 2^2$

Now I am not too sure about how I should restrict the domain of the circle to give the equation of the arc.

$Part (2)$
$\frac{|z-4|}{|z-1|} \Rightarrow \frac{|x-4 + iy|}{|x-1 + iy|}$
$\Rightarrow \frac{(x-4)^2 + y^2}{(x-1)^2 + y^2}$
$\Rightarrow \frac{x^2 + y^2 + 16 - 8x}{x^2 + y^2 +1 -2x}$
using $x^2 + y^2 = 2^2$
$\Rightarrow \frac{20 - 8x}{5 -2x}$
$\Rightarrow \frac{20 - 8x}{5 -2x}$
$\Rightarrow \frac{4(5 -2x)}{5 -2x}$
therefore $\frac{|z-4|}{|z-1|} = 4$ on the circle.

$Part (3)$

$|z - 1| = \frac{1}{4} |z - 4|$
Subbing into $|z - 1| + |z - 4| = 5<br />$
$\frac{5}{4} |z - 4| = 5$
$|z - 4| = 4$

so we have a circle $(x-4)^2 +y^2 = 2^2$

we also get $|z-1| = 1$
giving the circle $(x-1)^2 +y^2 = 1$

So the system is
$(x-4)^2 +y^2 = 2^2 (1)$
$(x-1)^2 +y^2 = 1 (2)$
$(1) - (2)$
$(x-4)^2 - (x-1)^2 = 3$
$-3(2x-5) = 3$
$x = 2$

using equation 2
$(2-1)^2 +y^2 = 1$
$y = 0$

so $z = 2$

which is wrong!

6. Originally Posted by bobak
Thank you a lot for your response Mr Fantastic.

I think it is only appropriate that I post all my working.

For the first part.

$Part (1)$
$\arg (z-2) - \arg (z-2i) = \frac{3 \pi}{4}<br />$

rewriting arg(Z) in arctan from where $z = x +iy$

$\arctan \left ( \frac{y}{x-2} \right ) - \arctan \left ( \frac{y-2}{x} \right ) = \frac{3 \pi}{4}$

.... $\tan \left (\arctan \left ( \frac{y}{x-2} \right ) - \arctan \left ( \frac{y-2}{x} \right ) \right )= \tan \left ( \frac{3 \pi}{4} \right )$

then using compound angle formula.

$\frac{ \frac{y}{x-2} - \frac{y-2}{x} }{1 + \frac{y}{x-2} \cdot \frac{y-2}{x}} = -1$

$\frac{y-2}{x} - \frac{y}{x-2} = 1 + \frac{y}{x-2} \cdot \frac{y-2}{x}$
.... skipping some step
giving the equation of he circular arc as $x^2 + y^2 = 2^2$

Now I am not too sure about how I should restrict the domain of the circle to give the equation of the arc.

[snip]
To get the restriction, note that:

$\arg (z-2) - \arg (z-2i) = \frac{3 \pi}{4} \Rightarrow \arg \left( \frac{z - 2}{z - 2i} \right) = \frac{3 \pi}{4}$.

But $\frac{z - 2}{z - 2i}$ cannot have an argument of $\frac{3 \pi}{4}$ unless its real part is less than zero AND its imaginary part is greater than zero ..... The latter restriction is the one that is easiest to apply. The former restriction doesn't affect things (as you can check).

Substitute z = x + iy and simplify to get cartesian form:

$\frac{[(x - 1)^2 + (y - 1)^2 - 2] + (2y + 2x - 4)i}{x^2 + (y - 2)^2} = \frac{3 \pi}{4}$.

So the restriction $Im \left( \frac{z - 2}{z - 2i} \right) > 0$ is equivalent to $2y + 2x - 4 > 0 \Rightarrow y + x - 2 > 0 \Rightarrow y > -x + 2$.

So you want the part of the circle $x^2 + y^2 = 2^2$ that lies above the line $y = - x + 2$. The endpoints do NOT get included since arg(0) is not defined.

7. Originally Posted by bobak
[snip]
$Part (2)$
$\frac{|z-4|}{|z-1|} \Rightarrow \frac{|x-4 + iy|}{|x-1 + iy|}$
$\Rightarrow \frac{(x-4)^2 + y^2}{(x-1)^2 + y^2}$
$\Rightarrow \frac{x^2 + y^2 + 16 - 8x}{x^2 + y^2 +1 -2x}$
using $x^2 + y^2 = 2^2$
$\Rightarrow \frac{20 - 8x}{5 -2x}$
$\Rightarrow \frac{20 - 8x}{5 -2x}$
$\Rightarrow \frac{4(5 -2x)}{5 -2x}$
therefore $\frac{|z-4|}{|z-1|} = 4$ on the circle. *
[snip]
Great work on this part (and part 1), bobak. You really nailed the algebra.

Except for the line marked with an * (see post #9)

8. Originally Posted by bobak
[snip]
$Part (3)$

$|z - 1| = \frac{1}{4} |z - 4|$
Subbing into $|z - 1| + |z - 4| = 5<br />$
$\frac{5}{4} |z - 4| = 5$
$|z - 4| = 4$

so we have a circle $(x-4)^2 +y^2 = 2^2$

we also get $|z-1| = 1$
giving the circle $(x-1)^2 +y^2 = 1$

So the system is
$(x-4)^2 +y^2 = 2^2 (1)$
$(x-1)^2 +y^2 = 1 (2)$
$(1) - (2)$
$(x-4)^2 - (x-1)^2 = 3$
$-3(2x-5) = 3$
$x = 2$

using equation 2
$(2-1)^2 +y^2 = 1$
$y = 0$

so $z = 2$

which is wrong!
Hmmmmm ... That's a shame. It's a nice approach and should work ...... there's a mistake somewhere and I will find it.

In the meantime, you can get brutal and find the intersection of the arc of $x^2 + y^2 = 2^2$ with the ellipse. Can you get the cartesian equation of the ellipse .....

The equation is of the form $\frac{(x - h)^2}{a^2} + \frac{y^2}{b^2} = 1$ where h = 5/2, giving centre at (5/2, 0), a = 5/2, giving major axis of length 5, and b = 2 (which can be got from geometric considerations). I've used the locus definition of the ellipse to get this ......

9. I've found the problem!
Originally Posted by bobak
[snip]
$Part (2)$
$\frac{|z-4|}{|z-1|} \Rightarrow \frac{|x-4 + iy|}{|x-1 + iy|}$
$\Rightarrow \frac{(x-4)^2 + y^2}{(x-1)^2 + y^2}$
$\Rightarrow \frac{x^2 + y^2 + 16 - 8x}{x^2 + y^2 +1 -2x}$
using $x^2 + y^2 = 2^2$
$\Rightarrow \frac{20 - 8x}{5 -2x}$
$\Rightarrow \frac{20 - 8x}{5 -2x}$
$\Rightarrow \frac{4(5 -2x)}{5 -2x}$
therefore $\frac{|z-4|}{|z-1|} = 4$ on the circle. Mr F says: This is the mistake that causes the trouble with part 3 ...... What you've actually found is that $\frac{|z-4|^2}{|z-1|^2} = 4\,$ ! Therefore $\frac{|z-4|}{|z-1|} = 2$ ...... So $\frac{1}{\lambda} = 2$, NOT 4 ....
[snip]
Part 3 looks OK now using the original method. I get $x = \frac{10}{9}$ and $y^2 = +\frac{\sqrt{224}}{9}$. I haven't checked whether this point lies on the ellipse (I leave that little chore for you), but it does lie on the arc - good news.

And you could use the brutal method as a check - good practice. I've used the wonders of hindsight to edit post #8.

10. By the way ..... did you notice that the locus defined by $|z - 4| = 2 |z - 1|$ is the circle $x^2 + y^2 = 2^2$ ....?

A circle defined by $|z - z_1| = \lambda |z - z_2|$, $\lambda > 0 \,$ and $\, \lambda \neq 1 \,$, is called an Apollonius Circle.

11. Okay picking up where i left off.
using
$\frac{|z-4|}{|z-1|} = 2$
and the same method I applied earlier for part 3 I get
$x= \frac{10}{9}$
and $y = \pm \frac{ 4 \sqrt{14}}{9}$

But we must neglect the negative solution of y as it lies underneath, the line $y = - x + 2$

so answering the initial question $z = \frac{10}{9} + \frac{ 4 \sqrt{14}}{9}i$

The book however wrote $z = \frac{10}{9} \pm \frac{ 4 \sqrt{14}}{9}i$
But I guess they are paying less attention than we are.

I'll use the brute force method to check as you suggested.

I need to transform this $|z-1| + |z-4| = 5<br />$ into an ellipse equation first, From the definition of the elpise i know the the sum distance of a point on the loci to each focus is 2a so $2a = 5$ so $a= \frac{5}{2}$ and the mid point of the two foci is clearly $\frac{5}{2}$

as we know the position of the foci we can use either foci to find e
I did $ae+h = 4$
giving e = $\frac{3}{5}$
then using $b^2 =a^2(1-e^2)$
I got $b=2$

so the equation of our elipse is $\frac{(x - \frac{5}{2})^2}{\frac{5}{2}^2}+ \frac{y^2}{2^2} = 1<br />$

messing around a bit i get
$x^2 - 5x + \frac{25}{16} y^2 = 0 <br />$

then subbing $y^2 = 4 - x^2$

i get a quadratic in x $9x^2 +80x -100 = 0$
$(9x - 10)(x+10) = 0$
x = -10 can clearly be rejected by geometric considerations, we get $x = \frac{10}{9}$ which is very reassuring.

Mr F, i didn't notice that last part about the Apollonius Circle. never come across that before, shall do some reading. Thanks for help, i am very surprised that you thanked me for asking a question.

12. Originally Posted by bobak
[snip]
i am very surprised that you thanked me for asking a question.
Ha ha. Don't be. Sometimes questions can be useful too .....
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 141, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9363727569580078, "perplexity_flag": "middle"}