What is the range of $r$ here?

Question

I am new to complex numbers.

Let $C_1=\arg(z)=\frac\pi6$ and $C_2=\vert z-i2\sqrt3\vert=r$ be two curves which intersect at two points.

Then I want to find the value (may be range) of $r$. How to find that?

I assumed $z=a+ib$ then $r=\sqrt{a^2 + (b-2\sqrt{3})^2}$ Or $r^2={a^2 + (b-2\sqrt{3})^2}$ which is the locus of a circle, centered at $(0, 2\sqrt{3})$ and of radius $r$.

score 5 · Accepted Answer · edited Dec 11 '17 at 07:42

5

Note that:

$C_1=\arg(z)=\frac{\pi}{6}$ is the Ray "$l$"

$C_2=\vert z- i2\sqrt{3} \vert =r$ is a circle centered in $(0,2\sqrt{3})$

Here a sketch of what is going on:

Since $$d=\sqrt{(2\sqrt 3)^2-(\sqrt 3)^2}=\sqrt{(12-9)}=3$$

The condition for $r$ in order to have 2 intersection points is: $$3<r<2\sqrt3$$ as the ray is not defined at the origin.

edited Dec 11 '17 at 07:42

answered Dec 10 '17 at 07:53

user

154,566

What is $d$ ? ? – Fghj Dec 10 '17 at 07:57
d is the distance between the point and the line – user Dec 10 '17 at 08:12
My solution is uncorrect since Arg(z) define a ray and not a line! – user Dec 10 '17 at 08:28
I've updated the solution! – user Dec 10 '17 at 08:38
Where is your image – Fghj Dec 10 '17 at 16:58
@RaviPrakash I've fixed for the image! – user Dec 10 '17 at 17:35
Can $r=6$? I think it should be – Fghj Dec 11 '17 at 04:27
@Ravi Prakash No it eas my initial fault. Infact Arg(z) is well defined in [0,2\pi) with positive value of $\rho$, thus thr Ray is not a straight line. – user Dec 11 '17 at 07:15

Parcly Taxel · Answer 2 · 2017-12-10T07:39:52.393

2

There is a neat geometric approach that treats the portion of $C_1$ with $0<|z|\le2\sqrt3$ as one side of a regular hexagon whose centre is $2\sqrt3i$, also the centre of the circle $C_2$. For the two curves to intersect in precisely two points, the exclusive upper bound for $r$ is attained when $C_2$ circumscribes the hexagon, i.e. has a radius of $2\sqrt3$. The exclusive lower bound is attained when $C_2$ is inscribed in the hexagon, i.e. has a radius of 3. Thus the admissible range of $r$ is $3<r<2\sqrt3$.

edited Dec 10 '17 at 07:39

answered Dec 10 '17 at 07:25

Parcly Taxel

103,344

1

Science Princess, can you give me the intersection points? – Fghj Dec 10 '17 at 07:31
@RaviPrakash The intersection points were not asked for! But if you want, their magnitudes are $\sqrt3\pm\sqrt{r^2-9}$ and their phases are $\pi/6$. – Parcly Taxel Dec 10 '17 at 07:36
Science Princess, so the region of $C_1$ is the area bounded by $O$ , $(0,2√3)$ and $(3,Y)$ where Y is the ordinate of second intersection of both the curves – Fghj Dec 10 '17 at 07:42
@RaviPrakash Area is irrelevant here, just care about the intersection points! – Parcly Taxel Dec 10 '17 at 07:44
Okay, I am not asking the area – Fghj Dec 10 '17 at 07:46

Deepak · Answer 3 · 2017-12-10T07:37:40.783

Easiest to solve geometrically with a quick Argand diagram sketch. The first locus is a ray in the first quadrant making an angle with the horizontal real axis of $\frac{\pi}{6}$ with a "hole" at the origin (since the origin is excluded from $C_1$). The second locus $C_2$ is a circle of undetermined radius but centered at $(0,2\sqrt 3)$ on the vertical imaginary axis.

For a particular value of $r$ (which is geometrically the radius of the circle of the second locus), the circle will be just tangent to the ray of the first locus. Using trigonometry, you should be able to determine this minimal value of $r$ to be $2\sqrt 3 \cos \frac{\pi}{6} = 3$. Any value of $r$ smaller than this will give a circle that doesn't contact the ray at all.

For a certain higher value of $r$, the circle will just pass through the origin (which is not on $C_1$ - it's a "hole" at the origin along the ray). For higher values of $r$, the circle will intersect the ray once in the first quadrant and not pass through the origin. This maximal value of $r$ is simply $2\sqrt 3$.

The bounds are both exclusive. So the required range is $3 < r < 2\sqrt 3$.

What is the range of $r$ here?

3 Answers3