1
$\begingroup$

Question:

If argument of $\frac{z - z_1}{z-z_2}$ is $\pi\over4$, find the locus of $z$. $$z_1 = 2 + 3i$$$$z_2 = 6 + 9i$$

Approach: I tried to solve the equation using diagram, basically plotting the points on the Argand plane. What I got is a circle with center $7 + 4i$ and a radius of $\sqrt{26}$ units. The two complex numbers given lie on this circle, and form a chord. Any point lying on the major arc of this chord satisfies the condition.

How exactly would I represent this as a locus of the point? And is there any other method that I can use that does not involve a diagram?

$\endgroup$
4
  • $\begingroup$ this is an arc of a circle. you can use rule of $\sin$ to find the center and the radius. $\endgroup$
    – abel
    Jan 13, 2015 at 16:38
  • $\begingroup$ I have the center and the radius. However, I need to write the locus as only that arc of a circle. How would I do that? @abel $\endgroup$
    – Gummy bears
    Jan 13, 2015 at 16:41
  • $\begingroup$ depends on whether you want to do it in polar coordinates(complex numbers) which is easier or in cartesian which is harder. did you get two centers? $\endgroup$
    – abel
    Jan 13, 2015 at 17:21
  • $\begingroup$ $z_1, z_2,$ and the two centers form a square od length $\sqrt{26}$ you can just draw the picture and it will make it easier to see. $\endgroup$
    – abel
    Jan 13, 2015 at 17:25

2 Answers 2

Reset to default
1
$\begingroup$

the angle subtended by the chord $z_1z_2$ at the center is $2 \pi/4 = \pi/2$ so the radius is $\frac{|z_1-z_2|}{\sqrt 2} = \sqrt{26}$ the center of the chord is $4 + 3i$ you add or subtract $\dfrac{-6+4i}{2}$ so that you will get two centers. the two centres, $z_1$ and $z_2$ form a square of side $\sqrt{ 26}.$

$\endgroup$
1
  • $\begingroup$ Well it's true that we get two centers, but we cannot take both the centers due to the condition given. The argument of the complex number given implies that we are measuring the angle from $z_2$ to $z_1$ and the positive sign on the angle indicates that we are doing so clockwise. On drawing the figure it becomes apparent that we can only take the lower circle. $\endgroup$
    – Gummy bears
    Jan 14, 2015 at 16:08
0
$\begingroup$

Put $\;z=x+iy\;,\;\;x,y\in\Bbb R\;$ , so

$$\frac{z-2-3i}{z-6-9i}=\frac{(x-2)+(y-3)i}{(x-6)+(y-9)i}\cdot\frac{(x-6)-(y-9)i}{(x-6)-(y-9)i}=$$

$$=\frac{(x-2)(x-6)+(y-3)(y-9)}{(x-6)^2+(y-9)^2}+\frac{(x-6)(y-3)-(x-2)(y-9)}{(x-6)^2+(y-9)^2}i$$

By the given data, it must be that the real and imaginary parts are identical, and thus

$$(x-2)(x-6)+(y-3)(y-9)=(x-6)(y-3)-(x-2)(y-9)\iff $$

$$\iff x^2-14x+y^2-8y-26=0$$

Complete squares, make some algebraic hokus pokus and get a circle.

$\endgroup$
6
  • $\begingroup$ Hmmmm..... You sure about the working there? I'm getting the same center, but the radius is coming out to be different. $\endgroup$
    – Gummy bears
    Jan 13, 2015 at 16:35
  • $\begingroup$ As sure as someone hating intensively doing boring calculations can be. You check the outcome, the way is shown above. If you want you can add your work into your question so that people can check it. $\endgroup$
    – Timbuc
    Jan 13, 2015 at 16:36
  • $\begingroup$ Well the center of the circle is the same. However, also, in my answer, I am only getting the major arc of the circle as the answer. No such point is noted by your method. Any specific reason for this? $\endgroup$
    – Gummy bears
    Jan 13, 2015 at 16:40
  • $\begingroup$ The arc thing comes from the fact that you have to be, for the general $\;x,y\;$ as in the answer, in the first quadrant. It is not noted in my answer as I don't usually detail all the little things but rather prefer to show general ways as good as I can. $\endgroup$
    – Timbuc
    Jan 13, 2015 at 16:43
  • $\begingroup$ But when writing down the locus, how would I represent only an arc of the circle as the locus? $\endgroup$
    – Gummy bears
    Jan 13, 2015 at 16:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .