The integer function $f(z) = u(x,y) + iv(x, y)$ is such that $u(x, y) = e^{−x}\sin(y -1) + y$. What's the value of $f'(i)$?

Question

The integer function $f(z) = u(x,y) + iv(x, y)$ is such that $u(x, y) = e^{−x}\sin(y -1) + y$. What's the value of $f'(i)$?

The solution is $-2i$.

I did:

\begin{align}\frac{\partial u}{\partial x} &= -e^{-x} \sin(y-1)\\\frac{\partial u}{\partial y} &= e^{-x}\cos(y-1) + 1\end{align}

\begin{align}f'(i)&=\frac{\partial u}{\partial x}(0,i) - \frac{\partial u}{\partial y} (0,i) \\ &= -\text{sin}(i-1) - (\cos(i-1) +1) \\ &= -\text{sin}(i-1) - \cos(i-1)-1\end{align}

I also try $\cos z=((e^{iz}+e(-iz))/2)$ and the same to $\sin z$ with $z=i-1$ but I didn't get there

Could someone help me please?

Answer 1

As $u(x, y) = e^{−x} \sin(y -1) + y$, we have \begin{align}u_x&=-e^{-x}\sin(y-1)\\u_y&=e^{-x}\cos(y-1)+1\end{align} and since $v_x=-u_y$ by Cauchy-Riemann, \begin{align}f'(0+1i)&=u_x(0,1)+i\cdot(-u_y(0,1))\\&=-\sin(1-1)-i\cdot(\cos(1-1)+1)\\f'(i)&=-2i.\end{align} This, of course, assumes differentiability of $f$.