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Thanks

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- Thread starter Heirot
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- #1

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Thanks

- #2

nicksauce

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Office_Shredder

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That would be f(z) = Re(z)*sin(Im(z)) + icos(Re(z))

Where if x+iy=z, Re(z):= x, Im(z):= y

Where if x+iy=z, Re(z):= x, Im(z):= y

- #5

lurflurf

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x=Re(z)

y=Im(z)

so

u(x,y)+iv(x,y)=u(Re(z),Im(z))+iv(Re(z),Im(z))=f(z)

That said

Re(z) and Im(z) are not considered proper functions of z as they break z apart and do not treat it as whole variable

So we are led to the concept of an analystic function that maps z as a whole

The Cauchy-Riemann Equations allow us to check if a function is analystic

for a function in

u+iv form the condition is written

Dx(u)=Dy(v)

Dy(u)=-Dx(v)

where Dx and Dy are partial derivatives with respect to x and y

so for your example

Dx(x sin(y))=Dy(cos(x))

sin(y)=0

Dy(x sin(y))=-Dx(cos(x))

x cos(y)=sin(x)

so we se your function is not analytic and cannot be written as f(z) in a proper way

we could write it as f(z,z*) (where z* is the conjugate of z and z*z=|z|^2)

when written in this form the condition is

Dz*(f)=0 where Dz* is the partial derivative with respect to z*

- #6

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Yes, that's it! Thank you!

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