I'm trying to find function $v(x,y)$ such that the pair $(u,v)$ satisfies the Cauchy-Riemann equations for the following functions $u(x,y)$:
a) $u = \log(x^2+y^2)$
$$
u_x = v_y \Rightarrow \frac{2x}{x^2+y^2} = v_y \Rightarrow v = \frac{2xy}{x^2+y^2}?
$$
b) $u = \sin x \cosh y$
$$
u_x = \cos x \cosh y = v_y \Rightarrow v = \sinh y \cos x + C
$$
c) $u = \frac{x}{x^2+y^2}$
$u_x = v_y$, but I am getting a mess with integration. The reason is that is there a way to do this by integration, or is the way I have started this seem correct? Thanks!
Best Answer
Here's how to find the corresponding imaginary parts by educated guessing. Maybe not the most systematic method, but maybe it improves your intuition about how these things behave.
For the first, remember that $e^{a + ib} = e^a(\cos b + i\sin b)$ (assuming $a,b \in \mathbb{R}$). In other words, $e^z$ sort of maps from polar coordinates to cartesian coordinates. $\log(z)$ thus does the reverse mapping, i.e. $\log(a+ib) = \log(|z|) + i\arg(a+ib) + i2\pi n$. This motivates the assumption that your first $u$ is the real part of $$ f(z) = \log(z^2) $$ Which it actually is, since $\Re(f(x+iy)) = \log(|(x+iy)^2|) = \log(|x+iy|^2) = log(x^2+y^2)$. The corresponding $v$ is thus $v(x,y) = \Im(f(x+iy))$, i.e. $$ v(x,y) = \arg(z^2) + 2\pi n= 2\arg(z) + 2\pi n= 2\left(arctan\left(\frac{y}{x}\right) + \pi \tilde{n}\right) $$ Your integration yields the wrong result because you forgot to take the denominator (which depends on $y$!) into account when finding the antiderivative.
For the second, use that $\sin(x+iy)$ = $\sin(x)\cos(iy) + \cos(x)\sin(iy)$ = $\sin(x)\cosh(y) +i\cos(x)\sinh(y)$ . From that, you get that $$ f(z) = \sin(z) $$ and $$ v(x,y) = \cos(x)\sinh(y) $$ which is the same as your integration yields.
For the third, observe that $x^2+y^2 = z\bar{z}$ if $z = x+iy$. Thus, one guess for $f(z)$ could be $\frac{z}{z\bar{z}}$. That's no good, however, because $z \to \bar{z}$ is not holomorphic. But since the real parts of $z$ and $\bar{z}$ are identical, you can also try $$ f(z) = \frac{\bar{z}}{z\bar{z}} = \frac{1}{z} $$ And voilá, since $$ \frac{1}{z} = \frac{1}{x+iy} = \frac{x-iy}{(x+iy)(i-iy)} = \frac{x - iy}{x^2 + y^2} $$ you indeed have $\Re(f(x+iy)) = \frac{x}{x^2+y^2}$, and thus $$ v(x,y) = \frac{-y}{x^2+y^2} $$