Question:
If $ \Im[f'(z)]=6x(2y-1)$ and $f(0)=3-2i,f(1)=6-5i$
then, $f(1+i)=?$
my attempt:
i used property
${f'(z)=u_{1} (z,0)-i\ u_{2}(z,0)}$ where
$ u_{1} (x,y)=\dfrac{\partial u}{\partial x}$ and
$u_{2} (x,y)=\dfrac{\partial u}{\partial y}$
i used Cauchy -Reimann equations also
but failed to get right answer ...please provide right way to approach ,thank you.
Let $f(z) = u(x,y) + iv(x,y)$, where $z = x+iy$. Then $f'(z) = u_x + iv_x$, and $u,v$ must satisfy the Cauchy-Riemann equations: $$u_x = v_y$$ $$u_y = -v_x$$ We are given that $v_x = 6x(2y-1) = -u_y$. Integrating, we find $v = 3x^2(2y-1) + \phi(y)$ and $u = 6x(y-y^2) + \psi(x)$, where $\phi(y), \psi(x)$ are arbitrary functions. Differentiating, we find: $$u_x = 6(y-y^2)+\psi'(x)$$ $$v_y = 6x^2+\phi'(y)$$ It is now apparent that if $u_x = v_y$, then $$\psi'(x) -6x^2=\phi'(y) + 6(y^2-y)$$ It follows that the right and left hand sides must be constant. We find that $$\psi(x) = 2x^3 + C_0x+C_1$$ $$\phi(y) = 3y^2-2y^3 + C_0y + C_2$$ for constants $C_0,C_1, C_2$. So our $f(z)$ now looks like: $$f(z) = 6x(y-y^2) + 2x^3 + C_0x + C_1 +i(3x^2(2y-1) + 3y^2 - 2y^3 + C_0y + C_2)$$ We can identify the constants $C_0,C_1$ using the given values of $f$ at $0,1$. Since $f(0) = 3-2i$, $C_1 = 3$ and $C_2 = -2$. Since $f(1) = 6-5i$, $C_0 = 1$. Hence, $$f(z) = 6x(y-y^2)+2x^3+x+3 + i(3x^2(2y-1)+3y^2-2y^3 + y -2)$$ Then $f(1+i)$ is: $$f(1+i) = 6+3i$$