Find a regular function $f$ satisfying Re $(f) =u$

Question

$$u = 2x^3 - 6xy^2$$

How does one do this?

I can find a function $f(0) = i$

Not sure what this is asking me to do...

Answer 1

Let's say $z=x+iy$ and $f(z)=u(x,y)+iv(x,y)$, where $u,v$ are real functions. You need to make sure the Cauchy-Riemann equations are satisfied:

$\begin{cases}\dfrac{\partial u}{\partial x}=\dfrac{\partial v}{\partial y}\\\dfrac{\partial u}{\partial y}=-\dfrac{\partial v}{\partial x}\end{cases}$

Since $u=2x^3-6xy^2\Rightarrow\dfrac{\partial u}{\partial x}=6x^2-6y^2=\dfrac{\partial v}{\partial y}\Rightarrow v=\displaystyle\int(6x^2-6y^2)\,\text{d}y\;+\phi(x)=6x^2y-2y^3+\phi(x)$. We are adding a function $\phi(x)$ because when doing $\dfrac{\partial v}{\partial y}$ we would get rid of $\phi(x)$, since it doesn't depend on $y$, so just in case there was a part of $v$ not depending on $y$, we should add it.

Now you have to make sure $\dfrac{\partial u}{\partial y}=-\dfrac{\partial v}{\partial x}$: Try it for yourself first!

$\dfrac{\partial u}{\partial y}=-12xy$ and $-\dfrac{\partial v}{\partial x}=-12xy-\phi'(x)$, so we get $\phi'(x)=0\Rightarrow\phi(x)=C$, with $C$ a real constant. So $v=6x^2y-2y^3+C$.

Therefore

$f(z)=f(x+iy)=2x^3-6xy^2+i(6x^2y-2y^3+C)=2(x^3-3xy^2+3ix^2y-iy^3)+iC=2(x+iy)^3+iC=2z^3+iC$

If we also impose $f(0)=i$ then

$f(0)=20^3+iC=i\Rightarrow C=1$

So the final answer is

$f(z)=2z^3+i$

Answer 2

If $f=u+iv$, then $u_x=v_y$. So $$ u=2x^3-6xy^2\quad\Longrightarrow\quad u_x=6x^2-6y^2=v_y \quad\Longrightarrow\quad v=6x^2y-2y^3+h(x) $$ But $v_x=-u_y$ and hence $h'(x)=0$, and thus $h(x)=c$. So $$ f(x+iy)=2x^3-6xy^2+i(6x^2y-2y^3+c) $$ Since $f(0)=i$, then c=1, and finally $$ f(x+iy)=2x^3-6xy^2+i(6x^2y-2y^3+1) $$