Context
I am working on the series solution of a second-order, homogeneous, linear, ordinary differential equation. Irrespective of the two roots of the indical equation, one of the two linearly independent solutions has the form $$ |z|^{z_1}\left(1 + \sum_{n=1}^\infty b_n\,z^n\right). $$ I will need to differentiate this solution to plug it into the governing differential equation. Thus, I need to answer the following question.
Question
Given $z\in \mathbb{C}$ and $z_1\in \mathbb{C}$, and $f(z) = |z|^{z_1}$, if $f'{(z)}$ exists, what is $f'{(z)}$?
Solution
In this solution, I use the following nation:
$$ z = x+ i\,y\qquad\text{and}\qquad z_1 = x_1+ i\,y_1.$$
Additionally, from the fact that
\begin{align*}
f(z)
&= |z|^{z_1}
\\
&= |x+ i\,y|^{x_1+ i\,y_1}
\\
&= \left[\sqrt{x^2+ y^2}\right]^{x_1+ i\,y_1}
\\
&= \left[\sqrt{x^2+ y^2}\right]^{x_1 }\,\left[\sqrt{x^2+ y^2}\right]^{ i\,y_1}
\\
&= \left[\sqrt{x^2+ y^2}\right]^{x_1 }\,\left[e^{\log{\sqrt{x^2+ y^2}}}\right]^{ i\,y_1}
\\
&= \left[\sqrt{x^2+ y^2}\right]^{x_1 }\,\left[e^{ i\,y_1\,\log{\sqrt{x^2+ y^2}}}\right]
\\
&=
\left[\sqrt{x^2+ y^2}\right]^{x_1 }\,\left[
\cos{ \left(y_1\,\log{\sqrt{x^2+ y^2}}\right) }
+i\,
\sin{ \left( y_1\,\log{\sqrt{x^2+ y^2}} \right)}
\right]
\\
&=
\cos{ \left(y_1\,\log{\sqrt{x^2+ y^2}}\right) }\left[ x^2+ y^2 \right]^{\frac{x_1}{2} }
+
i\,
\sin{ \left( y_1\,\log{\sqrt{x^2+ y^2}} \right)}
\left[ x^2+ y^2 \right]^{\frac{x_1}{2} }
\end{align*}
I write the real and imaginary parts of $f$, respectively, as
\begin{align*}
u(x,y)
&=
\cos{ \left(y_1\,\log{\sqrt{x^2+ y^2}}\right) }\left[ x^2+ y^2 \right]^{\frac{x_1}{2} },
\\
v(x,y)
&=
\sin{ \left( y_1\,\log{\sqrt{x^2+ y^2}} \right)}
\left[ x^2+ y^2 \right]^{\frac{x_1}{2} }
.
\end{align*}
I know that if, and only if, the Cauchy-Riemann conditions are satisfied that the derivative exists.
First Cauchy-Riemann condition
I need to check if \begin{align*} \frac{\partial u}{\partial x} &= \frac{\partial v}{\partial y}. \end{align*} Now, \begin{align*} \frac{\partial u}{\partial x} &= \frac{\partial \cos{ \left(y_1\,\log{\sqrt{x^2+ y^2}}\right) }\left[ x^2+ y^2 \right]^{\frac{x_1}{2} }}{\partial x} \\ &= - y_1\,\frac{\frac{2\,x}{2\,\sqrt{x^2+ y^2}} }{\sqrt{x^2+ y^2}} \sin{ \left(y_1\,\log{\sqrt{x^2+ y^2}}\right) }\left[ x^2+ y^2 \right]^{\frac{x_1}{2} } + \cos{ \left(y_1\,\log{\sqrt{x^2+ y^2}}\right) }\frac{x_1\,2\,x}{2}\,\left[ x^2+ y^2 \right]^{\frac{x_1}{2}-1 } \\ &= x\left[ - y_1\, \sin{ \left(y_1\,\log{\sqrt{x^2+ y^2}}\right) } + x_1\, \cos{ \left(y_1\,\log{\sqrt{x^2+ y^2}}\right) } \right] \left[ x^2+ y^2 \right]^{\frac{x_1}{2}-1 } . \end{align*} Further, \begin{align*} \frac{\partial v}{\partial y} &= \frac{\partial \sin{ \left( y_1\,\log{\sqrt{x^2+ y^2}} \right)} \left[ x^2+ y^2 \right]^{\frac{x_1}{2} }}{\partial y} \\ &= y_1\,\frac{\frac{2\,y}{2\sqrt{x^2+ y^2}}}{ \sqrt{x^2+ y^2}} \cos{ \left( y_1\,\log{\sqrt{x^2+ y^2}} \right)} \left[ x^2+ y^2 \right]^{\frac{x_1}{2} } + \frac{2\,y\,x_1}{2} \sin{ \left( y_1\,\log{\sqrt{x^2+ y^2}} \right)} \left[ x^2+ y^2 \right]^{\frac{x_1}{2}-1 } \\ &= y\left[ y_1\, \cos{ \left( y_1\,\log{\sqrt{x^2+ y^2}} \right)} + x_1\, \sin{ \left( y_1\,\log{\sqrt{x^2+ y^2}} \right)} \right]\left[ x^2+ y^2 \right]^{\frac{x_1}{2}-1 } \end{align*}
I have found that \begin{align*} \frac{\partial u}{\partial x} \neq \frac{\partial v}{\partial y}. \end{align*}
My conclusion is that given $z\in \mathbb{C}$ and $z_1\in \mathbb{C}$, and $f(z) = |z|^{z_1}$, $f'{(z)}$ does not exist. Yet, I have doubts and seek alternative solutions from the community.