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FP2 complex numbers locus of points

number23

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

Reply 1

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? :s-smilie:

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.

Reply 2

number23

Original post by SBarns

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

Reply 3

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.

(edited 11 years ago)

Reply 4

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.

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

Reply 7

Dangerous Theory

Original post by Ilyas

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!

Reply 8

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.

Reply 9

Venomilys

Original post by Dangerous Theory

No problem. :smile:

I think I just upped the grade boundaries, sorry everyone! :tongue:

Reply 10

number23

Original post by Ilyas

In red.

so the constant equals the raduis, thanks :smile:

Reply 11

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.

Reply 12

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.