Complex numbers loci.

Original post by Victor Victus

When illustrating the loci of arg((z-z1)/(z-z2)) = theta, where z1 and z2 are constants respectively and theta is an angles less than pi radians, the negative angles of arg(z-z1) and arg(z-z2) respectively are never considered. That is, Instead of two circles subtended by the same arc length (a binocular shape), it's just a single circle subtended by one arc. Why is this?

Excuse inappropriate mathematical language :colondollar:

.

Ok, correct me if I am wrong but you mean something of the kind eg:

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

, ps.

z_{1}

and

z_{2}

are complex numbers.

Which produces something similar to:

This is because, if you remember back to your circle theorems, the angle subtended at the centre of the circle is twice the angle at the circumference.

So we can then deduce that 4i to 3 is the diameter of the circle loci because 2*90 = 180:

Tidying up a bit leaves:

This is just useful if we want to find the centre of the circle and the radius etc.

Reply 2

OP

Original post by Joshmeid

Ok, correct me if I am wrong but you mean something of the kind eg:

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

, ps.

z_{1}

and

z_{2}

are complex numbers.

Which produces something similar to:

This is because, if you remember back to your circle theorems, the angle subtended at the centre of the circle is twice the angle at the circumference.

So we can then deduce that 4i to 3 is the diameter of the circle loci because 2*90 = 180:

Tidying up a bit leaves:

This is just useful if we want to find the centre of the circle and the radius etc.

Thanks for your time. It's duly appreciated. I can comprehend the principles your explaining. However, that isn't my problem. Considering the special case you've illustrated, why is it not feasible to construct a full circle (in the context of your special case only). More generically, Why isn't the loci symmetric about the segment (in this case, the segment is the line joining 4i and 3) so as to produce to identical sectors of circles with symmetry about the segment?

Reply 3

We know the argument is the angle made between the complex number and the tangent to the positive real axis from the complex number.

so

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

tells us that the angle the first argument makes with the real axis is bigger than the second argument to produce a positive angle.

So this is why only above the circle is the loci. Anything below the circle would produce a negative angle as arg(z - 3) would be smaller than arg(z - 4i).

Posted from TSR Mobile

Reply 4

OP

Original post by Joshmeid

We know the argument is the angle made between the complex number and the tangent to the positive real axis from the complex number.

so

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

tells us that the angle the first argument makes with the real axis is bigger than the second argument to produce a positive angle.

So this is why only above the circle is the loci. Anything below the circle would produce a negative angle as arg(z - 3) would be smaller than arg(z - 4i).

Posted from TSR Mobile

I think i've got a way of understanding it now, thanks to your helpful insight!

+rep (ya know you want it :wink:

)

Reply 5

OP

Original post by Joshmeid

We know the argument is the angle made between the complex number and the tangent to the positive real axis from the complex number.

so

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

tells us that the angle the first argument makes with the real axis is bigger than the second argument to produce a positive angle.

So this is why only above the circle is the loci. Anything below the circle would produce a negative angle as arg(z - 3) would be smaller than arg(z - 4i).

Posted from TSR Mobile

Wait! Wait! false alarm, sorry :frown:

. surely arg(z-3) can be negative but be of smaller magnitude than arg(z-4i)??

Reply 6

ImageUploadedByStudent Room1371411117.847111.jpg40.3KB

Original post by Victor Victus

Wait! Wait! false alarm, sorry :frown:

. surely arg(z-3) can be negative but be of smaller magnitude than arg(z-4i)??

Do you mean something like this:

Posted from TSR Mobile

(edited 10 years ago)

Reply 7

OP

Original post by Joshmeid

Do you mean something like this:

ImageUploadedByStudent Room1371411117.847111.jpg

Posted from TSR Mobile

Yeah, that's it. I would've thought that because the diagramatic representation uses the magnitude of the arguments, it subtends the same angle. However, after observing that the y components can either both be negative or positive to intercept (it's a half line), then the subtended angle is multiplied by -1 to accommodate the negative y component case. This will yield a negative angle?

(edited 10 years ago)

Reply 8

Changed the image to relate more to your question.

Reply 9

OP

Oh, so when the angle lies below the line joining 3 and 4i, then the angle at 3 becomes greater in magnitude than the angle at 4i. Therefore, the subtended angle will become negative! (by applying my observations in the previous post)?!?!

Reply 10

Ok, think of how the argument works.

If we have a complex number for example

z = 3 - 4i

.

then we construct a triangle and find the angle between the complex number and the positive x axis using TOA from GCSE.

In this case,

\tan{\theta} = \frac{4}{3}

so

\theta = 53.1^{\circ}

(3 sig.fig.)

Now because z is under the x axis, that angle is negative.

so

arg(z - 3 + 4i) = -53.1^{\circ}

.

In image i've just shown in the last post,

|arg(z - 3)| < |arg(z - 4i)|

but they are negative, so

arg(z - 3) > arg(z - 4i)

because

arg(z - 3)

is less negative than

arg(z - 4i)

.

(edited 10 years ago)

Reply 11

OP

Eh, yeah. What i said works. It's all down to you, though! Thanks! I'd rep you again but i can't! It's very altruistic of you to contribute your time!

Reply 12