Question:
Given $f(z) = 3z^2 + 9z^3 -z$.
1. Find $f^\prime(z)$
2. Find $f(z)$ when $z = 3 + 2i$
3. Use Cauchy-Riemann to find if $f(z)$ is differentiable at $3 + 2i$
My Attepmt:
1. $f^\prime(z) = 6z + 27z^2 - 1$
2.
$$
\begin{align}
f(3 + 2i)
&= 3(3+2i)^2 + 9(3+2i)^3 - (3 + 2i)\\
&= 3(9-4+6i)+9(9-8i+54i-36) - (3+2i)\\
&= (15 + 18i) + (81 - 72i + 486i - 324) -(3 + 2i)\\
&= -231 + 430i\\
\end{align}
$$
3. For this, I substituted $z = x + iy$ in $f(z)$, then expanded the brackets. After that, I grouped the like terms together and expressed $f(z) = u(x,y) + iv(x,y)$. I got this as a result of my calculations:
$$
\begin{align}
u(x,y) &= 3x^2 - 3y^2 + 9x^3 -9y^3 - 27xy^2\\
v(x,y) &= 6xiy + 2x^2iy - iy
\end{align}
$$
Then, I calculated partial derivatives $u_x, v_y, u_y$ and $v_x$. For Cauchy-Riemann, $u_x = v_y$ and $u_y = -v_x$. I substituted x = 3 and y = 2 in the partial derivative and found that it fails C.R. equations. Hence, NO, IT IS NOT DIFERENTIABLE AT $3 + 2i$
Can someone please verify this for me ?
Wrong. This is a polynomial, so it is differentiable, and the C-R equations should be true.
EDIT: In this case I get $$\eqalign{ u &= 9\,{x}^{3}-27\,x{y}^{2}+3\,{x}^{2}-3\,{y}^{2}-x, \cr v &= 27\,{x}^{2}y-9\,{y}^{ 3}+6\,xy-y\cr u_x &= 27 x^2 - 27 y^2 + 6 x - 1 = v_y\cr u_y &= -54 x y - 6 y = - v_x}$$