Express rational function $f: \mathbb{C} \to \mathbb{C}$ in the form $u(x,y) + iv(x,y)$.

Question

Simple question. I have the function $f(z) = \frac{4z+1}{z^3+z}$ defined on $\mathbb{C}\setminus\{0,i,-i\}$.

I want to show the function is analytic on its domain of definition. To do so, I want to show that the Cauchy-Riemann equations hold everywhere, plus some other sufficient conditions for differentiability.

To do this, I want to find $u$ and $v$ such that $$ f(z) = \frac{4z+1}{z^3+z} = u(x,y) + iv(x,y) $$ for $z = x+iy$. I've tried fiddling around with it but haven't gotten anywhere. Any ideas?

Alternate approaches to show that $f$ is analytic on its domain are welcome.

Answer 1

If you replace $z$ with $x+yi$, you get$$\frac{4 x^4+x^3+4 x^2-3 y^2 x+x-4 y^4+4 y^2}{\left(x^3-3xy^2+x\right)^2+\left(-y^3+3 x^2 y+y\right)^2}+\frac{-8 y x^3-3 y x^2-8 y^3 x+y^3-y}{\left(x^3-3xy^2+x\right)^2+\left(-y^3+3 x^2 y+y\right)^2}i.$$

Answer 2

decompose $f$

after you're done with that you'll get

$f(z) = \frac{3}{z^2+1} +\frac1z = 3\frac{\partial}{\partial z}(\arctan(z)) + \frac{\partial}{\partial z} \ln z = 3\frac{\partial}{\partial z}(\arctan(x+iy)) + \frac{\partial}{\partial z} \ln (x+iy)$

and you have $\frac{\partial}{\partial z} = \frac12(\frac{\partial}{\partial x} - i\frac{\partial}{\partial y})$

Answer 3

Or you could just note that $$f(z) = \frac{4z+1}{z(z-i)(z+i)}=\frac{p(z)}{q(z)}$$ where $$p(z)=4z+1$$ and $$q(z)=z(z-i)(z+i)$$

Since $p$ and $q$ are analytic on $\mathbb C$, being polynomials, their ratio is analytic eveyrwhere $q\neq 0$, which is exactly $\mathbb C\setminus\{0,i,-i\}$.