Find f(z) if possible from u(x,y) and v(x,y)

[IHazAName]

Homework Statement

a. u(x,y)=(x+2)²-y², v(x,y)=2(x+2)y
b. u(x,y)=cosxcoshy, v(x,y)=-sinxsinhy
c. u(x,y)=(x²+y²+y)/(x²+y²), v(x,y)=x/(x²+y²)
d. u(x,y)=e^ycosx, v(x,y)=e^ysinx

The Attempt at a Solution

a. f(x+iy)=(x+2)²-y²+i2(x+2)y
=(x+2+iy)(x+2+iy)
f(z)=(z+2)²

b. f(x-iy)=cosxcoshy-isinxsinhy
Am I allowed to answer this question as f(\bar{z})=cos\bar{z}?

c.f(x-iy)=((x²+y²+y)/(x²+y²))-i(x/(x²+y²))
=((x-iy-i)(x+iy))/((x-iy)(x+iy))
=(x-iy-i)/(x-iy)
Same as b... f(\bar{z})=(\bar{z}-i)/\bar{z}?

d. du/dx != dv/dy, fails Cauchy-Reimann?

Edit: How do I go from x-iy to x+iy? This escapes me. :(

 

[tiny-tim]

Welcome to PF!

Hi IHazAName! Welcome to PF!

Is the question to find an f such that f = u + iv ?

a and b look ok

(for b, just replace y by -y )

I haven't checked c

For d, isn't cauchy-riemann only relevant to differentiability? Try again

 

[HallsofIvy]

Homework Statement

a. u(x,y)=(x+2)²-y², v(x,y)=2(x+2)y
b. u(x,y)=cosxcoshy, v(x,y)=-sinxsinhy
c. u(x,y)=(x²+y²+y)/(x²+y²), v(x,y)=x/(x²+y²)
d. u(x,y)=e^ycosx, v(x,y)=e^ysinx

The Attempt at a Solution

a. f(x+iy)=(x+2)²-y²+i2(x+2)y
=(x+2+iy)(x+2+iy)
f(z)=(z+2)²

b. f(x-iy)=cosxcoshy-isinxsinhy
Am I allowed to answer this question as f(\bar{z})=cos\bar{z}?

If f(\bar{z})= cos(\bar{z}), then f(z)= cos(z).

c.f(x-iy)=((x²+y²+y)/(x²+y²))-i(x/(x²+y²))
=((x-iy-i)(x+iy))/((x-iy)(x+iy))
=(x-iy-i)/(x-iy)
Same as b... f(\bar{z})=(\bar{z}-i)/\bar{z}?

And, as in b, if f(\bar{z})= (\bar{z}- i)/\bar{z} then f(z)= (z- i)/z.

d. du/dx != dv/dy, fails Cauchy-Reimann?

The problem asked you to find a function- it did not say the function had to be analytic.

Edit: How do I go from x-iy to x+iy? This escapes me. :(

If z= x- iy, then \bar{z}= x+ iy.

 

[IHazAName]

Thanks guys!

If f(\bar{z})= cos(\bar{z}), then f(z)= cos(z). And, as in b, if f(\bar{z})= (\bar{z}- i)/\bar{z} then f(z)= (z- i)/z. The problem asked you to find a function- it did not say the function had to be analytic.

If z= x- iy, then \bar{z}= x+ iy.

...Oops. I forgot to mention that...these are supposed to be analytic functions as that is what we're learning. :(

So, just to make extra sure: for f(x-iy)=cos(x)cosh(y)-isin(x)sinh(y) that is still f(z)=cos(z)? Even though using x+iy grants f(x+iy)=cos(x)cosh(y)+isin(x)sinh(y)

I guess my confusion with that in our notes we only solve it one way...where we're given a f(z)...say f(z)=z and plug in x+iy so that f(z) becomes f(x+iy)=x+iy, and solve for u(x,y) and v(x,y). In these problems I was just trying to go in the reverse direction, so getting a result that needed x-iy was confusing to me.