FP2 complex numbers

I don't really understand using complex numbers to represent a locus of points on an Argand diagram...

eg. |z-5-3i|=3
Does it mean the magnitude of the complex number z minus another complex number (5+3i) equals 3? Or does it mean that the absolute value of the difference of the complex numbers is 3?
(Or are they the same thing?)

Furthermore, when you're finding the Cartesian equation of this locus, what exactly are you solving for? And why are we using an x and y axis instead of a real and imaginary one?

Also :
|(x-5)+i(y-3)| = 3
And so the the moduli is removed by squaring :
(x-5)^2 + (y-3)^2 = 3^2

What happened to the i?

thanks

Reply 1

baffled_mathman

8

It's a circle, so we need the equation of a circle.

(x-a)² + (y-b)² = r² where (a,b) is the circle's centre and r is its radius.

We know that the centre is at (5,3) and the radius is 3. So our equation is (x-5)²+(y-3)²=9

Reply 2

nuodai

17

Original post by confuzzled92

eg. |z-5-3i|=3
Does it mean the magnitude of the complex number z minus another complex number (5+3i) equals 3? Or does it mean that the absolute value of the difference of the complex numbers is 3?
(Or are they the same thing?) The absolute value and modulus of a complex number are the same thing. You can think of complex numbers as vectors, where

x+iy

corresponds to the vector

\begin{pmatrix} x \\ y \end{pmatrix}

. With vectors, you should know that, say,

\mathbf{a} - \mathbf{b}

gives you the vector pointing from the point with position vector

\mathbf{b}

to the point with position vector

\mathbf{a}

, and hence

|\mathbf{a} - \mathbf{b}|

gives you the distance between the two points. It's exactly the same with complex numbers; if

w,z

are two complex numbers then

|w-z|

tells you the distance between them on an Argand diagram. So here you have

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

, which tells you that the corresponding locus is the set of all points at distance 3 from the point

5+3i

; in other words, it's a circle about

5+3i

of radius 3.

Original post by confuzzled92

Furthermore, when you're finding the Cartesian equation of this locus, what exactly are you solving for? And why are we using an x and y axis instead of a real and imaginary one? The Argand diagram (with real and imaginary axes) corresponds to the two-dimensional plane (with x- and y-axes), and so when you find the Cartesian equation of the locus, you stop thinking in terms of

z

and start thinking in terms of

(x,y)

.

Original post by confuzzled92

Also :
|(x-5)+i(y-3)| = 3
And so the the moduli is removed by squaring :
(x-5)^2 + (y-3)^2 = 3^2

What happened to the i? You should know the identity

|z|^2 = z \bar z

, and so if

z=a+ib

then

\bar{z} = a-ib

; multiplying the two thus gives

a^2+b^2

.

Reply 3

ztibor

10

Original post by confuzzled92

I don't really understand using complex numbers to represent a locus of points on an Argand diagram...

eg. |z-5-3i|=3
Does it mean the magnitude of the complex number z minus another complex number (5+3i) equals 3? Or does it mean that the absolute value of the difference of the complex numbers is 3?
(Or are they the same thing?)

Furthermore, when you're finding the Cartesian equation of this locus, what exactly are you solving for? And why are we using an x and y axis instead of a real and imaginary one?

Also :
|(x-5)+i(y-3)| = 3
And so the the moduli is removed by squaring :
(x-5)^2 + (y-3)^2 = 3^2

What happened to the i?

thanks

You can not remove the moduli by squaring so. THe square of a complex number is another complex number. You would like to get |z|^2 and tried to wrote |z^2|
|z^2|=|((x-5)+i(y-3))^2|=|(x-5)^2-(y-3)^2+2i(x-5)(y-3)|

|z^2| \neq |z|^2

|z|^2=z\cdot \bar{z}=(x-5)^2+(y-3)^2

in your example.

Reply 4

nuodai

17

Original post by ztibor

You can not remove the moduli by squaring so. THe square of a complex number is another complex number. You would like to get |z|^2 and tried to wrote |z^2|
|z^2|=|((x-5)+i(y-3))^2|=|(x-5)^2-(y-3)^2+2i(x-5)(y-3)|

|z^2| \neq |z|^2

|z|^2=z\cdot \bar{z}=(x-5)^2+(y-3)^2

in your example.

As it happens, in fact,

|z^2|=|z|^2

, but I don't think the OP ever did try to find

|z^2|

anyway.

FP2 complex numbers

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you