Question

Using complex coordinates the rotation T_{a, theta} is defined by T_{a, theta} (z) = e^{i theta}z+a where a in C.

For example, A = T_{i, pi/2} z ->iz+i
B = T_{i, pi/2} z ->iz+1
Describe the rotation BA (A followed by B) What is the center, amount, and direction?

Answer 1

rotation in \[T_{a, \theta}\]\[T_{a, \theta} (z) = e^{i \theta} z+ a a \in C\]

Answer 2

whoa whoa.. that's a in C should be sepearate

Answer 3

\[T_{a, \theta} (z) = e^{i \theta} z+ a, a \in C\]

Answer 4

Example

\[A = T_{i, \frac{\pi}{2}}: \] z ->iz+ i

Answer 5

I would simplify exp(i * pi/2) = cos(90) + i sin(90) = i

Answer 6

I am so sorry my pc is doing popups like there is no tomorrow and I can't get my latex right on here

Answer 7

Bz gives (iz+i)
call that z'
now Az' is iz' + i = i(iz+i) + i

Answer 8

\[B = T_{1, \frac{\pi}{2}} : z \rightarrow iz+1\]

Answer 9

\[A = T_{i, \frac{\pi}{2}} : z \rightarrow iz+i\]

Answer 10

I guess it's A first, but A and B are the same?

Answer 11

so I have to do BA rotation meaning do A first and then B

Answer 12

can I scan the page so e it's easier?

Answer 13

@phi after you're done with helping this friendly madame can you help me on my statistics question?


http://openstudy.com/study#/updates/554bdaf9e4b08c324fbe4398

Answer 14

latex is being too small that's why

Answer 15

I posted the result if both A and B are T(i, pi/2 )
i(iz+i)+i
simplify that

Answer 16

\[i(iz+i)+i = i^2z+i^2+ i = -z-1+i\]

Answer 17

btw, are you sure B isn't (1, pi/2) ?

Answer 18

my rotation should be 90 degrees since we have pi/2 I'm gonna scan it

Answer 19

this part looks wrong
B = T_{i, pi/2} z ->iz+1

is there a typo ? should it be
B = T_{1, pi/2} z ->iz+1

Answer 20

wait a sec I did latex 1 and pi/2 for the B

Answer 21

ok, B is T(1 , pi/2)
and A is T( i, pi/2)
as we know A(z) gives us iz + i
B(z) gives iz+1
put in (iz+i) into B: i(iz+i) + 1 = -z -1+1 = -z
that means BA(z) rotates z 180 degrees (pi radians)

Answer 22

whoa a bit lost here

Answer 23

A is iz+i and B is iz + 1

Answer 24

yes

Answer 25

if you start with z and do A(z) you get back iz+i
that "goes into" B in place of z in the formula iz+1
in other words we now do i(iz+i) + 1

Answer 26

why am I plugging in iz+i into z in B?

I start with z and get back iz+i ?

Answer 27

then when iz+i is in the z portion of B

i(iz+i)+1 = i^2z+i^2+ 1 = -z-1+1 = -z a negative z coordinate?

Answer 28

the direction is negative?

Answer 29

yes, if it helps first do A(z) to get iz+i
rename that "a" (in other words iz+i is some complex number, call it a)
we now do B(a)

Answer 30

how ... do I get iz+i? is it from the original definition?

Answer 31

we are doing BA(z) which means B (A(z) )
A(z) "returns" iz+i
now you are doing B( iz+i)

Answer 32

I guess you don't do computer programming? It is like function calls.

Answer 33

but the idea is "put in z" into A and get back a new complex number
now put that new number into B and get back another complex number.

if we work it out
BA(z) gives back -z which is the same as z*exp(i pi)

Answer 34

put z in a and we get iz+i so if A = iz+i then we put that into the iz+1 ?

Answer 35

yes, though I would write it as
z is the starting number
A(z) is a new complex number
B(A(z)) is the final result.

Answer 36

we start from z? like the z from the T_{a, theta} and then plug it into A which becomes iz+i then we plug in iz+i into iz+1 to get i(iz+i)+1 = i^2z+i^2+1 = -z-1+1 = -z so my direction is negative... it's going to be pi/2 + pi/2 = 2pi/2 = pi which is 180 degrees. where is the center though?

Answer 37

ugh I hate my firefox.. keeps crashing

Answer 38

I would not use pi/2 + pi/2 to find the rotation. I would say -z can be written as
z * exp(i pi) (because exp(i pi) = -1) and thus we rotated the original number by 180 degrees.

if we plot some random z and -z: