Complex numbers (1 Viewer)

[henry08]

How exactly does (x1, y1)(x2, y2) expand out to (c1x2 - y1y2, x1y2 + x2ya)?

How exactly do they 'solve this system', i.e. derive the equations for x' and y', I was just having a mental blank on this earlier.

Thanks for the help.

 

[vds700]

How exactly does (x1, y1)(x2, y2) expand out to (c1x2 - y1y2, x1y2 + x2ya)?

How exactly do they 'solve this system', i.e. derive the equations for x' and y', I was just having a mental blank on this earlier.

Thanks for the help.

For the first one, i guess they are just defining the relationship of the product of 2 complex numbers in the set, i dont think u have to worry about where it comes from, just use it to answer the question.

For the second part, it appears they have just solved the 2 equations simultaneously.

Sorry if I'm wrong.

I must say this is a weird question, where did u get it from?

 

[Trebla]

How exactly does (x1, y1)(x2, y2) expand out to (c1x2 - y1y2, x1y2 + x2ya)?

How exactly do they 'solve this system', i.e. derive the equations for x' and y', I was just having a mental blank on this earlier.

Thanks for the help.

Whenever you say x є R, this means x belongs to the real number set, which is one dimensional (i.e. number line).
If say x and y are both real numbers i.e. x є R and y є R, then on the two dimensional space (i.e. number line) you write (x,y) є R².
In general if you have x_i є R for all integers i = 1,2,...,n, then (x₁, x₂, ....... , x_n) є Rⁿ i.e. the n-dimensional space, often referred to as the hyperplane.

When you plot a complex number x + iy on the Argand diagram, you plot (x,y) which are coordinates of real numbers hence (x,y) є R²

When you add two complex numbers: z₁ + z₂, you get:
x₁ + iy₁ + x₂ + iy₂
= (x₁ + x₂) + i(y₁ + y₂)
So when you plot z₁ + z₂ on the Argand diagram, you are actually plotting (x₁ + x₂, y₁ + y₂)

When you multiply two complex numbers: z₁.z₂, you get
(x₁ + iy₁)(x₂ + iy₂)
= x₁x₂ + i²y₁y₂ + i(x₁y₂ + y₁x₂)
= x₁x₂ - y₁y₂ + i(x₁y₂ + y₁x₂)
When you plot z₁.z₂ on the Argand diagram you plot (x₁x₂ - y₁y₂, x₁y₂ + y₁x₂)

 

[Trebla]

For the second one, you use the formula for multiplying complex numbers as above:
z = x + iy and z^-1 = x' + iy'
z.z^-1 = xx' - yy' + i(xy' + x'y)
BUT the LHS = 1, so you have
1 = xx' - yy' + i(xy' + x'y)
Since 1 is purely real, the imaginary parts must be zero and real parts must be 1:
(1) xx' - yy' = 1
(2) yx' + xy' = 0
To solve for x' and y' you just solve simultaneously as normal (treat the other variables as constants)
(1)y - (2)x
=> - y²y' - x²y' = y
-y'(x² + y²) = y
y' = - y / (x² + y²)
(1)x + (2)y
=> x²x' + y²x' = x
x'(x² + y²) = x
x' = x / (x² + y²)

 

[henry08]

Thanks Trebla, I get it now.

 

[davidbarnes]

= x₁x₂ + i²y₁y₂ + i(x₁y₂ + y₁x₂)
= x₁x₂ - y₁y₂ + i(x₁y₂ + y₁x₂)

how does it go from + i²y₁y₂to - y₁y₂

Does i² = -1?

Thats some complicated BB code as well.

 

[davidbarnes]

For the second one, you use the formula for multiplying complex numbers as above:
z = x + iy and z^-1 = x' + iy'
z.z^-1 = xx' - yy' + i(xy' + x'y)
BUT the LHS = 1, so you have
1 = xx' - yy' + i(xy' + x'y)
Since 1 is purely real, the imaginary parts must be zero and real parts must be 1:
(1) xx' - yy' = 1
(2) yx' + xy' = 0

I understand the above.

To solve for x' and y' you just solve simultaneously as normal (treat the other variables as constants)
(1)y - (2)x
=> - y²y' - x²y' = y
-y'(x² + y²) = y
y' = - y / (x² + y²)
(1)x + (2)y
=> x²x' + y²x' = x
x'(x² + y²) = x
x' = x / (x² + y²)

How do you solve simultaneous questions by 'y - x' and 'x+ y'? Sorry if its real simple.

 

[Trebla]

How do you solve simultaneous questions by 'y - x' and 'x+ y'? Sorry if its real simple.

Its meant to say:
(1) times y - (2) times x
and
(1) times x + (2) times y
The two variables are x' and y'. This first operation eliminates x' so we can express y' on its own. The second operation eliminates y' so we can express x' on its own.

 

[Trebla]

how does it go from + i²y₁y₂to - y₁y₂

Does i² = -1?

Thats some complicated BB code as well.

Yes...that's the defining element of complex numbers...that i² = -1.