# Recover the holomorphic function from one of its components using Cauchy-Riemann equations

I got two similar questions:

1. Find the holomorphic function $f(x+iy)$ if $\Re(f(x+iy))=x(3-2y)\text{ and }f(i)=2i$
2. Find the holomorphic function $f(x+iy)$ if $\Im(f(x+iy))=3(x-1)^2y-y^3\text{ and }f(0)=1$

I tried both using this source but got problem at the end. My steps for the first one :

$$\frac {\partial u}{\partial x} = 3-2y = \frac {\partial v}{\partial y}$$ $$\frac {\partial u}{\partial y} = -2x =- \frac {\partial v}{\partial x}$$

Now integrate both $$\int 3-2y\ dy = 3y-y^2 + g(x) = v$$ $$\int 2x\ dx = x^2 + h(y) = v$$

And now I should compare but mine v's are not equal. What am I missing, what to do with g(x) and h(y) and how to use $f(i)=2i$

• Presumably, that's two different questions - find an $f$ that matches (a) then find a different $f$ that matches (b)? Not: "Find a single $f$ that matches both (a) and (b)." Aug 27, 2015 at 14:22
• Yes you are right, sorry, I'll edit that.
– Mike
Aug 27, 2015 at 14:22
• Also, note that $\int 2xdy = 2xy + h(x)$, not $x^2+h(y)$. Aug 27, 2015 at 14:23
• Thanks, should be dx there. I'm new here, and don't know how to put text into neat lines.
– Mike
Aug 27, 2015 at 14:29
• Also, $\int (3-2y)dy = 3y-y^2$, not $3y-2y^2$. Aug 27, 2015 at 14:37

Integrate first one $$\int 3-2y\ dy = 3y-y^2 + g(x) = v$$ $$\frac{dv}{dx}=g'(x)$$ compare it to $\frac{du}{dx}$ from above $$\frac{du}{dx}=g'(x)=-2x$$ $$g(x)=\int-2xdx=x^2 +c$$ $$v(x,y)=3y-y^2+x^2 +c$$ $$f(x+iy)=f(i)=2i \quad=> \quad x=0, y=1$$ $$f(i)=i(3-1+c)=2i\quad=>\quad c=0$$ Finally $$v(x,y)=3y-y^2+x^2$$ $$f(x,y)=x(3−2y)+i(3y-y^2+x^2)$$