Question

Find the analytic function f(z) = u+iv in terms of z if
u-v = e^x (cos y - sin y)

Answer 1

i know that if f is analytic,
then
\(u_x = v_y \\ u_y = - v_x\)

Answer 2

cauchy hehe

Answer 3

yep

Answer 4

so how do i start ?

Answer 5

There are many definitions of analytic functions. But in terms of complex analysis like this, we're talking about the Cauchy-Riemann equations that you mentioned.

Answer 6

just apply the thm
f(z) = u+iv
f(z)=u(x,y) + i v(x,y)
f'(z)=ux+ivx
....

Answer 7

so i differentiate both sides of
u-v = e^x (cos y - sin y)

once with x , once with y
?

Answer 8

to get
ux , uy, vx, vy

Answer 9

yep

Answer 10

why didn't i think of that earlier :P

Answer 11

thanks
i'll solve it and let you know my answer to verify!

Answer 12

2 heads are better than 1 @hartnn :[

Answer 13

Since \(u-v=e^x\cos y - e^x\sin y\), my first attempt would be to try two candidates for \((u,v)\) : \((e^x\cos y,e^x\sin y)\) and \((-e^x\sin y, -e^x\cos y)\). it would take one minute to verify if this is true.

Answer 14

does
\(\Large v_x = e^x \sin y \)

Answer 15

wow, what a guess reemii !

Answer 16

(not sure it deserves a wow..)
I thought you were discarding this bet to go for the general method.

Answer 17

how do you find it with ux,uy,vx,vy? im curious

Answer 18

i will upload my work in a minute

Answer 19

can any of you check whether my work is correct ??
any mistakes ?

Answer 20

@BSwan @Miracrown @reemii

Answer 21

are you guys also getting
\(\Large f(z)=e^x cosy+i e^x siny \)

Answer 22

looking it over still
good so far...

Answer 23

in my opinion, once you obtain \(v_x\) with the addition of the two equation (end of page 1), you don't need more play with the equations.
integrate wrt. x the function \(v_x\), and since \(-u_y=v_x\), integrate wrt \(y\) the function \(-v_x\).

Answer 24

perfect :)
only add C :P hehe

Answer 25

and it's correct I think. \(\checkmark\)

Answer 26

Yes, it all looks good. Just one part concerns me
But I want to check this is correct against your work

Answer 27

Thanks you so much!
i wish i could give all of you medals :)

Answer 28

you can give all of us thanks, that's all what we need!

Answer 29

awesoem

Answer 30

thanks <3

Answer 31

integral of f ' (z) w.r.t. dx = integral (e^x cos(y) + i e^x sin(y) ) dx
implies
f(z) = e^x cos(y) + i * e^x sin(y) + F(y) where F(y) is some function of y, possibly complex.
We should also examine f ' (z) w.r.t. y
f ' (z) w.r.t. y = u_y + i v_y

Answer 32

^ you got a point!

Answer 33

= - v_x + i u_x by (A) (the cauchy riemann equations)

Answer 34

f(z) = integral of f ' (z) w.r.t. dy = integral ( - e^x sin(y) + i * e^x cos(y) ) dy

Answer 35

= e^x cos(y) + i e^x sin(y) + G(x)

Answer 36

therefore G(x) = F(y) this can only happen if they are both 0.
so it works out in the end, but maybe it's not obvious to show

Answer 37

infact it is incorrect, if i don't verify it...!

Answer 38

I think that the method @hartnn used is general and works always.

Answer 39

Well, I mean, if it's not obvious, yes

Answer 40

this is in real case , in complex just take a constants

Answer 41

So we agree up to that point then. Then the issue becomes representing x, y in terms of z,z-bar

This method will work, but I mean you run into not so nice functions easily where you won't be able to get closed form solutions pretty quickly
But yes "it works"

Answer 42

how do i express my answer in terms of z ?

Answer 43

:)
when u verify u would had a problem with i incase of G(X) or F(Y")
function
so be carfull what to choose

Answer 44

so
we have to do this
which is a bit messy imho

Answer 45

x = (z+z-bar)/2
z-bar = complex conjugate
y = (z-z-bar)/2

over (2i)** sorry for y
So
You substitute these into the equation for f(z)
But we also need to know how to take sin and cos of complex numbers
it is extended to complex arguments like so

Answer 46

cos(z) = 1/2(e^(iz) + e^(-iz))

and sin(z) =
forgot the sign
thinking

Answer 47

sin(z) = i (e^(iz) - e^(-iz) ) / 2
Just let me look it up
I am not sure

Answer 48

ux = e^x cos y
uy = -ex sin y

i can find 'u' from here ?

Answer 49

sin(z) = (e^(iz) - e^(-iz) ) / (2i)
gotta keep the i in the denominator
sorry I have no complex analysis textbook with me here
so
Let's just start with the cos(y)
cos(y) = cos( (z-z-bar) / (2i) )
= (e^(i* (z-z-bar) / (2i) ) + e^(-i (z-z-bar) / (2i) ) ) / 2

Answer 50

= (e^((z-z-bar)/2) + e^(-(z-z-bar)/2))/2

yes hartnn

Answer 51

then u = e^x cos y

Answer 52

you will arrive at the same f(z) this way

Answer 53

right ??

Answer 54

yes

Answer 55

so f(z) will just be z ?

Answer 56

because v is also e^x sin y

Answer 57

So let's multiply this by e^x = e^( (z+z-bar)/2)
e^x*cos(y) = (1/2) * (e^z + e^z-bar)

Answer 58

uhm, I'm not sure. I hope not :(

Answer 59

No, I don't think so
so?
z = x + i y not u + i v

Answer 60

oh yes :(

Answer 61

this is tough to type anyway... just checking it

Answer 62

yes so e^x cos(y) = (1/2) * (e^z+e^z-bar)
now for the other term

Answer 63

this question was not supposed to be so difficult...

Answer 64

sin(y) = sin ( (z-z-bar)/(2i)) = 1/(2i) * (e^(i * (z-z-bar)/(2i) ) - e^(-i *(z-z-bar)/(2i) ) )

Answer 65

:) now multiply by e^((z+z-bar)/2)) again

Answer 66

i got your method,
will try to find it.
thanks, no need to further explain

Answer 67

alright =) I am sorry that it's so tedious. Not much I can do about that

Answer 68

e^z
should be final answer
f(z) = e^z

Answer 69

Now that I worked it out. I suppose it's shown somewhere that Euler's identity over the complex numbers takes the form we have here, so if you knew that you could immediately write down e^z from e^x cos(y) + i e^x sin(y). But I did not know that

Answer 70

oh wow! saves lot of work!

*deleting all those extra work* :P

Answer 71

Of course, did you expect different ;)
just kidding

Answer 72

If only in life, hard work translated always to the correct answer

Answer 73

it does! sooner or later :)

Answer 74

Optimism :)

Answer 75

as always ;)