1. ## Cauchy-Riemann (Complex Analysis)

Let $\displaystyle f=u + iv$ be analytic. In each of the following, find $\displaystyle v$ given $\displaystyle u$.

(1) $\displaystyle u= x^2 - y^2$

(2) $\displaystyle u = \frac{x}{x^2 + y^2}$ (Figured this one out)

(3) $\displaystyle u = 2x^2 +2x +1 -2y^2$

(4) $\displaystyle u = cosh y sin x$

(5) $\displaystyle u = cosh x cos y$

If someone could show me step by step how to do one or two of these problems I would appreciate it. I'm not sure where I am going with these since I don't have the answers. Thanks!

These are Cauchy-Riemann equation problems.

2. Let's do the first part. The C-R equations state that $\displaystyle u_{x}=v_{y}\mbox{ and }u_{y}=-v_{x}$
Given that $\displaystyle u=x^2-y^2$ we can use the C-R equations to find that $\displaystyle v_{y}=2x\mbox{ and }v_{x}=2y.$
For here it is fairly obvious what v is...

3. The method is the same each time, so if you can do (2) you should have been able to do the others.

Use the C–R equations. Here's how to do (1). If $\displaystyle u=x^2-y^2$ then $\displaystyle \tfrac{\partial v}{\partial y} = \tfrac{\partial u}{\partial x} = 2x$. Integrate with respect to y to see that $\displaystyle v = 2xy + f(x)$ for some function f(x). (This function f(x) takes the place of the constant of integration when integrating $\displaystyle \tfrac{\partial v}{\partial y}$, because x counts as a constant in the partial derivative.)

Now differentiate v with respect to x, to get $\displaystyle \tfrac{\partial v}{\partial x} = 2y + f'(x)$. But $\displaystyle \tfrac{\partial v}{\partial x} = -\tfrac{\partial u}{\partial y} = 2y$, which tells us that $\displaystyle f'(x) = 0$. So f(x) is constant.

Conclusion: $\displaystyle v = 2xy\ +$ const.