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Prof. A.F.Guimarães Mathematical Physics 1 – Problems 2a

Let us now calculate the following products, &* &+ and &* &̅+ . So:

Problem 1 Verify the followings rules for scalar and vector products. [ Solution: Let ! and

(

!

&* &+ = 68 − 79 + 3(69 + 87) &* &̅+ = 68 + 79 + 3(87 − 69 ) (2.2)

! ∙ # ) = $%(&̅* &+ ) = $% (&* &̅+ ), × # ] = ,-(&̅* &+ ) = −,-(&* &̅+ ).

By the use of (2.2), we have #

be the complex numbers given by:

$%(&* &+ ) + $%(&* &̅+ ) = 268 = 2($% &* $% &+ ) 1 ∴ $% &* $% &+ = [$%(&* &+ ) + $%(&* &̅+ )] 2 (2.3)

! = 0* (cos 1 + 3 sin 1) # = 0+ (cos 4 + 3 sin 4 ) (1.1)

Let us calculate &̅* &+ ,

Problem 3 Prove that, if &(3 − 1) = −&̅(3 + 1), then arg & is either < ?< or − > . > Solution: Let us choose & = 6 + 37, then,

&̅* &+ = 0* 0+ (cos 1 + 3 sin 1 )(cos 4 − 3 sin 4 ) &̅* &+ = 0* 0+ [cos 4 cos 1 + sin 4 sin 1 +3 (sin 4 cos 1 − sin 4 cos 1 )] ∴ &̅* &+ = 0* 0+ [cos(4 − 1) + 3 sin(4 − 1)] (1.2)

(6 + 37)(3 − 1) = −(6 − 37)(3 + 1) 63 − 6 − 7 − 37 = −63 − 6 − 7 + 37 23(6 − 7) = 0 ∴ 6 = 7 (3.1)

Where | ! | = 0* and | # | = 0*. Another product (&* &̅+ ) is given by: &* &̅+ = 0* 0+ [cos(4 − 1) − 3 sin(4 − 1)] (1.3)

The argument of & is given by:

7 6 ∴ arg & = tg A* 1 (3.2)

The scalar product is given by: (

!

#)

∙

arg & = tg A*

= |&* ||&+ | cos(4 − 1) (1.4)




By the use of the result from (1.2) or (1.3) we can write, (

!

∙

#)

= $%(&̅* &+ ) = $%(&* &̅+ ) (1.5)

!

!

×

×

#]

#]

= |&* ||&+ | sin(4 − 1) (1.6)

= ,-(&̅* &+ ) = −,-(&* &̅+ ) (1.7)

& = 6* + 36+ + E(7* + 37+ − 6* − 36+ ) ∴ & = 6* + E(7* − 6* ) + 3 [6+ + E(7+ − 6+ )] (4.1)

Problem 2 *

.

If 6 and 7 are two complex numbers and E is a real parameter, then the expression & = 6 + E(7 − 6) represents a curve in complex plane. Set & = F + 3G and determine the parametric equations F = F(E) e G = G(E) of this curve. What kind of curve is it? Solution The numbers 6 and 7 are given by 6 = 6* + 36+ and 7 = 7* + 37+ . Here, 7* − 6* ≠ 0 and 7+ − 6+ ≠ 0. So we can write:

Hence, [

>

Problem 4

And the product of vectors, [

?