Cauchy-riemann using polar co-ordinates

[randybryan]

I've been given a function U(x,y) where f(z) = U(x,y) + iV(x,y)

and asked to check if it is harmonic and then work out what V(x,y) is

U(x,y) = $$\frac{y}{(1 - x)^{2} + y^{2}}$$

To check if it is harmonic I can see if d²U/dx² + d²U/dy² = 0

I've tried differentiating and it's fairly arduous, so I'm thinking it might be easier to use polar co-ordinates (As a lot of the other questions simplify when doing this). Can anyone think of a suitable substitution to make the differentiation easier?

the answer in the back of the book says that f(z) = -i / (1 - z)

 

[matematikawan]

I suggest you consider the logarithm of the function.
log u = log y + log((1-x)²+y²)

$$\frac{1}{u}\frac{\partial u}{\partial y}=\frac{1}{y}+\frac{2y}{(1-x)^2+y^2}$$

$$\frac{\partial u}{\partial y}=\frac{1}{(1-x)^2+y^2}+2u^2$$

 

[randybryan]

Thanks, but I still don't know where to go from here

 

[matematikawan]

So its not helping after all.
I give a try for the second derivatives and obtain

$$u_{yy}=\frac{2u^2}{y}+8u^3$$ and $$u_{xx}=\frac{-2u^2}{y}+\frac{8(1-x)^2u^3}{y^2}$$

so that
$$u_{xx}+u_{yy}=\frac{8u^2}{y}\neq 0$$ ?

 

[anirudh dutta]

TRY this
for a function to be harmonic its laplacian should be zero
so $$\nabla$$²U = 0

( i worked it out and it is true)

THEN use the Cauchy Rienman conditions to determine v

Finally the f(z)=U+iV