*Notes to a video lecture on http://www.unizor.com*

__Vectors+ 11__

2D Vector · Complex scalar

2D Vector · Complex scalar

We know that a two-dimensional vector multiplied by a real scalar changes its length proportionally, but the direction remains the same for a positive multiplier or changes to opposite for a negative multiplier.

Let's examine how vectors on a two-dimensional plane are affected if multiplied by a non-zero complex scalar.

Our first task is to define a multiplication operation of a vector by a complex number.

To accomplish this, let's recall that any complex number

*, where both*

**a+i·b***and*

**a***are real numbers and*

**b***, can be represented by a vector in a two-dimensional Euclidean space (that is, on the coordinate plane) with abscissa*

**i²=−1***and ordinate*

**a***.*

**b**This representation establishes one-to-one correspondence between vectors on a plane and complex numbers.

Using this representation, let's define an operation of multiplication of a vector on a coordinate plane {

*} by a complex number*

**a,b***as follows:*

**z=x+i·y**1. Find a complex number that corresponds to our vector. So, if a vector has coordinates {

*}, consider a complex number*

**a,b***.*

**a+i·b**2. Multiply this complex number by a multiplier

*using the rules of multiplication of complex numbers. This means*

**z=x+i·y**

**(a+i·b)·z = (a+i·b)·(x+i·y) =**

= a·x+a·i·y+i·b·x+i·b·i·y =

= (a·x−b·y) + i·(a·y+b·x)= a·x+a·i·y+i·b·x+i·b·i·y =

= (a·x−b·y) + i·(a·y+b·x)

3. Find the vector that corresponds to a result of multiplication of two complex number in a previous step. This vector should have abscissa

*and ordinate*

**a·x−b·y***.*

**a·y+b·x**4. Declare the vector in step 3 as a result of an operation of multiplication of the original vector {

*} by a complex multiplier*

**a,b***. So, the result of multiplication of vector {*

**z=x+i·y***} by a complex multiplier*

**a,b***is a vector {*

**z=x+i·y***}*

**a·x−b·y,a·y+b·x**Now we have to examine the geometric aspect of this operation.

For this, let's represent a multiplier

*as*

**z=x+i·y**

**z = √x²+y²·(x/√x²+y² + i·y/√x²+y²)**Two numbers,

*and*

**x/√x²+y²***, are both in the range from*

**y/√x²+y²***−1*to

*1*and a sum of their squares equals to

*1*.

Find an angle

*such that*

**φ***cos(φ)=x/√x²+y²*and

*sin(φ)=y/√x²+y²*.

Now the multiplier looks like

*[*

**z = |z|·***cos(φ) + i·sin(φ)*]

where

**|z| = √x²+y²**Using this representation, the product of a vector {

*} by multiplier*

**a,b***looks like*

**z=x+i·y**{

*} = {*

**a·x−b·y,a·y+b·x***}*

**a',b'**where

*[*

**a' = |z|·***] and*

**a·**cos(φ)**−b·**sin(φ)*[*

**b' = |z|·***]*

**a·**sin(φ)**+b·**cos(φ)The geometric meaning of the transformation from vector {

*} to vector {*

**a,b***} is a rotation of the vector by angle*

**a',b'***.*

**φ**Here is why.

The length of a vector {

*} is*

**a,b***.*

**L=√a²+b²**If the angle this vector makes with an X-axis is

*, the abscissa and ordinate of our vector can be expressed as*

**α**

**a = L·**cos(α)

**b = L·**sin(α)Using this representation, let's express the coordinates of a vector {

*} obtained as a result of multiplication of the original vector {*

**a',b'***} by a complex*

**a,b***[*

**z=x+i·y=|z|·***cos(φ) + i·sin(φ)*]

in terms of

*and*

**L***α*.

*[*

**a'=|z|·***]*

**L·**cos(α)**·**cos(φ)**−L·**sin(α)**·**sin(φ)and

*[*

**b'=|z|·***]*

**L·**cos(α)**·**sin(φ)**+L·**sin(α)**·**cos(φ)Recall from Trigonometry:

*cos(α+φ)=cos(α)·cos(φ)−sin(α)·sin(φ)*

*sin(α+φ)=cos(α)·sin(φ)+sin(α)·cos(φ)*

Now you see that

**a'=|z|·L·**cos(α+φ)

**b'=|z|·L·**sin(α+φ)So, the multiplication of a vector {

*} by a complex number*

**a=L·**cos(α),**b=L·**sin(α)*[*

**z=x+i·y=|z|·***cos(φ)+i·sin(φ)*],

**|z|***φ*.

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