Let $f$ be an analytic function such that $\Re (f'(z))=2y$ and $f(1+i)=2$. What can one say about the imaginary part of $f(z)$?
Options given in the book:
(1) $ -2xy $
(2) $x^2-y^2$
(3) $2xy$
(4) $y^2-x^2$
My efforts
Using the formula that $f'(z)=u_x +iv_x=v_y-iu_y$
So we get that $u_x=2y$ and $v_y=2y$
$u_x=2y$, so $u(x,y)=2xy + $ some function of $y$ ( say $p(y)$)
$v_y=2y$, so $v(x,y)=y^2 +$ some function of $x$ ( say $q(x)$ )
I still have not used the condition that $f(1+i)=2$
$f(1+i)=u(1,1)+iv(1,1)$
$2= 2 + p(1)+ 1 + q(1)$.
I don't know how to proceed from here.
Option (1) and (3) are incorrect. It is clear.
Let $g(x)=-2iz$. Then, if $x,y\in\mathbb R$,\begin{align}\operatorname{Re}\bigl(g(x+yi)\bigr)&=\operatorname{Re}(2y-2xi)\\&=2y\\&=\operatorname{Re}\bigl(f'(x+yi)\bigr).\end{align}Therefore,$$(\forall z\in\mathbb{C}):\operatorname{Re}(f'(z)-g(z)\bigr)=0.$$So, since $f'-g$ is analytic, $f'-g$ is constant, and, more precisely, $f'-g=ik$, with $k\in\mathbb R$. So, $f'(z)=ki-2iz$, and so $f(z)=kiz-iz^2+k'$, for some $k'\in\mathbb C$. Since $f(1+i)=2$, $k'=(1-i)k$. What this means is that$$f(z)=kiz-iz^2+(1-i)k,$$or$$f(x+yi)=-ky+2xy+k+i(kx-x^2+y^2-k).$$The only possibility consistent with this is the option (4).