[number23]

im not quite sure about this topic, like why does the locus of lz-5-3il=3 represent a circle?

[SBarns]

(Original post by number23)
im not quite sure about this topic, like why does the locus of lz-5-3il=3 represent a circle?

If |something| = 5, then the distance of all the 'somethings' from a central point is 5 units. This is the same as the definition of a circle, as if you take any point on it's edge, it's the same distance from the centre of the circle.

You could think of this as a complex number being rotated around the axis, the only think changing is it's argument, the modulus is always the same.

[number23]

(Original post by SBarns)
If |something| = 5, then the distance of all the 'somethings' from a central point is 5 units. This is the same as the definition of a circle, as if you take any point on it's edge, it's the same distance from the centre of the circle.

You could think of this as a complex number being rotated around the axis, the only think changing is it's argument, the modulus is always the same.

ok so the length of all the point from the origin is constant, but the angle changes? also when its like z-5i, why do you move upwards rather than downwards? thanks

[Venomilys]

You must revise!

A magnitude in the complexplane represents a cirlce.

|z - 5 - 3i| = 3
|z - (5+ 3i)| = 3

Look at it this way:
z is any number, so
z = x + iy

-->
|x + iy - 5 - 3i| = 3
|(x -5) + i(y - 3)| = 3

Take the magnitude, so square the imaginary and real components and square the RHS:

(x -5)^2 + (y-3)^2 = 9.

This represents a cirlce in the cartesian plane.

[SBarns]

(Original post by number23)
ok so the length of all the point from the origin is constant, but the angle changes? also when its like z-5i, why do you move upwards rather than downwards? thanks

In C1, if you had f(x-5), we shifted up the x axis by 5 units, not down - it's counterintuitive.

So, say we had a locus represented by f(z) = k. If we had (z-(a+bi)) = k, we have a circle of radius k with a centre that is shifted a units right/left and b units up/down. In the example you gave: z-5i = z - (0+5i), so we have a translation 5 units upwards.

It's difficult to explain without drawing diagrams/graphs.

[Jodin]

|z| where z = x + yi is defined as (x^2 + y^2)^0.5

So if |z - c - di| = r, we have |(x-c) + (y-d)i| = r, or rather (x-c)^2 + (y-d)^2 = r^2

So yes, the circle of radius r centred at (c,d)

Edit: beaten to it.

[number23]

(Original post by Ilyas)
You must revise!

lol, thanks.

so the modulus of a complex number, equalled to a constant, represents a circle and the centre is given by the transformation of z? and the radius=constant^0.5

thanks everyone

[Dangerous Theory]

(Original post by Ilyas)
You must revise!

A magnitude in the complexplane represents a cirlce.

|z - 5 - 3i| = 3
|z - (5+ 3i)| = 3

Look at it this way:
z is any number, so
z = x + iy

-->
|x + iy - 5 - 3i| = 3
|(x -5) + i(y - 3)| = 3

Take the magnitude, so square the imaginary and real components and square the RHS:

(x -5)^2 + (y-3)^2 = 9.

This represents a cirlce in the cartesian plane.

Mate, that is an excellent representation. I've been struggling with the concept of loci too, but your expansion in the form of a cartesian circle really puts things into perspective. Amazing how things suddenly piece together, thanks!

[Venomilys]

(Original post by number23)
lol, thanks.

so the modulus of a complex number, equalled to a constant, represents a circle Yes

and the centre is given by the transformation of z? Yes. You should always bracket it so you get the centre straight away. Just picture it as a movement from the origin.

and the radius=constant^0.5 Not sure what you mean here. When given the complex form, |something| = 3, the radius is three. So the radius of |something| = n always has radius n with complex numbers.

In red.

[Venomilys]

(Original post by Dangerous Theory)
Mate, that is an excellent representation. I've been struggling with the concept of loci too, but your expansion in the form of a cartesian circle really puts things into perspective. Amazing how things suddenly piece together, thanks!

No problem. I think I just upped the grade boundaries, sorry everyone!

[number23]

(Original post by Ilyas)
In red.

so the constant equals the raduis, thanks

[aysha.19]

can i get some help on this question :

Showing your working, find the two square roots of the complex number 1 − (2√6)i. Give your answers in the form x + iy, where x and y are exact.

[notnek]

(Original post by aysha.19)
can i get some help on this question :

Showing your working, find the two square roots of the complex number 1 − (2√6)i. Give your answers in the form x + iy, where x and y are exact.

A complex number which is a square root of will satify:



Write as then expand and compare real and imaginary parts.