# Let $f(z)$ be holomorphic on C such that the Re $f(z)=2x+y$ and $f(i)=1$ find the imaginary part of $f$

I know I'm supposed to use the Cauchy Riemann equations with partial derivatives. Then take the anti-derivative, but I'm not sure exactly how this works. The real part of $f$ is $u(x,y)$ and the imaginary part is $v(x,y)$ $u_x=2$ and $u_y=1$ so $v_x=-1$ and $v_y=2$. When I integrate the partial derivatives do I integrate $v_x$ with respect to $x$ or $y$? Either way, what do I do with that result?

If I understand correctly, I integrate $v_y$ and get $v=2y+\phi(x)$ for some $\phi(x)$. Taking the derivative with respect to $x$ gives $v_x=2+\phi'(x)$ and since $v_x=-1$, $\phi'(x)=-3$ thus $\phi(x)=-3x+c$. From there I don't really have a clue what to do.

• $v_x$ is the derivative with respect to x so you should integrate w.r. to x and add a function of y. Then the second equation gives you this function. Mar 29, 2017 at 19:52
• So $v=-x+\phi(y)$, $v_y=\phi'(y)=2$, then $\phi(y)=2y+c$ and $v=-x+2y+c$?. Then since $f(i)=1$, $c=-1$ so $v=-x+2y-1$? Mar 29, 2017 at 20:00

Well we have f(z):=u(x,y) + iv(x,y)

We have u(x,y) = 2x + y

CR =>

$u_x$ = $v_y = 2$

$u_y$ = $-v_x = 1$

1.) v = $\int v_y dy$ = $\int 2 dy$ = 2y + C(x)

2.) v = $\int v_x dx$ = $\int -1 dx$ = -x + C(y)

=>

v = 2y - x + C

To find the constant, I'll leave to you.

When you are given a function $u$ defined as a harmonic polynomial in two variables and you are looking for an holomorphic function $f$ such that $\Re{f} = u$, then you may find $f$ easily by saying that $f(z) = 2u\left( \frac{z}{2}, \frac{z}{2i}\right)- u(0,0)$ up to an imaginary constant.

Suppose $f$ exists. Then $f$ is a complex polynomial in $z$ and for all real $x,y$ you must have $$u(x,y) = \frac{f(x+iy) + \overline{f}(x-iy)}{2}$$ Therefore all coefficients of these polynomials must be equal to ane another. But then, you may see $u$ as a polynomial in two complex variables and hence deduce that $$2u\left(\frac{z}{2}, \frac{z}{2i}\right) = f(z) + \overline{f(0)}$$ and since $u(0,0) = \Re(f(0))$, the function $z \mapsto 2 u\left(\frac{z}{2}, \frac{z}{2i}\right) - u(0,0)$ is a good candidate to solve the problem. The computations that we did above can be done backwards so you may conclude

In your exercise apply, the formula and get $f(z) = (2-i)z - 2i$.