Question

Bi-linear Transformation anyone ?

Find the image of the circle x^2+y^2=1,under the transformation w=(5-4z)/(4z-2).

x^2+y^2 =1, gave me |z| =1, so i got |w| =1/2, but then what ?

Answer 1

I think it will be straight line in w plane

Answer 2

show me what have you done so far...

Answer 3

|z| =1

|w|=|(5-4z)|/|(4z-2)|

lol, i guess i can't do this ?? >> |w| = (5-4|z|)/(4|z|-2) = 1/2
?

Answer 4

if u want i can give the answer direct cuz i have sum imp work

Answer 5

you can help me later instead :)

Answer 6

bt for shall i give u the answer?
will explain later

Answer 7

ok, no problem :)

Answer 8

i think i can't do this
|w| = (5-4|z|)/(4|z|-2) = 1/2

|5-4z| is not 5=4|z|
i will have to start over...

Answer 9

z = a+ib

Answer 10

z= x+iy
w= u+iv
??
should i plug those in the transformation ?

Answer 11

and then compare real and imaginary co-efficients ?

Answer 12

yes

|w| = (5-4|z|)/(4|z|-2) = 1/2

w^2 = 1/4

=>

|5-4(u+iv)|^2 = 1/4|4(u+iv)-2|^2

Answer 13

but i can't do this, right ?
|w| = (5-4|z|)/(4|z|-2) = 1/2

because
|5-4z| is not 5-4|z|
or can i ?

Answer 14

so |w| is NOT 1/2
right ?

Answer 15

oh yes sorry, i think we need to solve z first

Answer 16

u+iv=(5-4x-4iy)/(4x+4iy-2)

multiply by conjugate of denominator ? :O

Answer 17

w=(5-4z)/(4z-2)

z = (2w+5)/(4w+4)

Answer 18

how does that help ?

Answer 19

now we can set the absolute value of z to 1 and get a relation in w :

|z| = 1

|(2w+5)/(4w+4)| = 1

Answer 20

:D

Answer 21

square both sides, right ?

Answer 22

after plugging in w =u +iv

Answer 23

yes just algebra..

Answer 24

will try and let you know my answer...please verify it...

Answer 25

yeah @ikram002p will verify... me also started complex analysis recently....

Answer 26

eshh im not expert xD

Answer 27

i am getting
4u+7=0

Answer 28

Let's not forget the simple Pole at \(z = 1/2\). Where did that pole come from? You should be able to trace that behavior.

A little long division shows \(w = -1 + \dfrac{7}{4\cdot z - 2}\). That's interesting.

Answer 29

http://www.wolframalpha.com/input/?i=simplify+%7C%282%28u%2Biv%29%2B5%29%7C%5E2+%3D+%7C%284%28u%2Biv%29%2B4%29%7C%5E2

Answer 30

how will that help though,
pole at z= 1/2
x+iy =1/2
x= 1/2, y= 0

i am getting a straight line 4u+7= 0 in w plane, so it has no pole....so did i do something wrong ?

Answer 31

i am not squaring both sides, i am taking the magnitude

Answer 32

4 |u+1+iv|=|5+2u+2iv|
4 (u+1)^2+4v^2=(5+2u)^2+4v^2

what is wrong there ?

Answer 33

http://www.wolframalpha.com/input/?i=4+%7Cu%2B1%2Biv%7C%3D%7C5%2B2u%2B2iv%7C

http://www.wolframalpha.com/input/?i=4+%28u%2B1%29%5E2%2B4v%5E2%3D%285%2B2u%29%5E2%2B4v%5E2

Answer 34

Also, i should have got straight lines,
in the transformation,
w = (az+b)/(cz+d)
if |a| = |c|, then circles transform into straight lines

Answer 35

My bad. I was mapping the whole unit disk. You're mapping just the Circle.

Answer 36

when you pull out 4 from the mag, you should get 16 right ?

Answer 37

** circles in w plane, map to straight line in z plane

is the reverse also true ?

Answer 38

|4a| = 4|a|

Answer 39

oops, i got my mistake

Answer 40

4 |u+1+iv|=|5+2u+2iv|
\( 4 \sqrt {(u+1)^2+4v^2}=\sqrt{(5+2u)^2+4v^2} \)

Answer 41

haha yes :)

Answer 42

so, u^2+u+v^2 = 3/4 is the final answer ?

which means the converse is not true!
if circles in w plane map to straight line in z plane, then it is not true that
circles in z plane map to straight line in w plane!

Answer 43

wish i was good at maths T,T

Answer 44

welcome to the club zeke :)

Answer 45

it looks good to me ^