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$f\in C[[x+iy]]$ means $f(x,y)=g(x+iy),$ where $g\in C[[z]]$. Then $$f_x=g'(x+iy),\; f_y=ig'(x+iy),$$ therefore $f_y=if_x$ and this is Cauchy-Riemann: indeed, if $f=u+iv$ then $f_x=u_x+iv_x$ and $f_y=u_y+iv_y$, so $f_y=if_x$ is equivalent to
$u_x=v_y$, $u_y=-v_x$.

To prove the converse, make the linear change of the variables: $z=x+iy,\; \overline{z}=x-iy$. A simple calculation shows that $f_z=(1/2)(f_x-if_y),\; f_{\overline{z}}=(1/2)(f_x+if_y).$ Then the Cauchy-Riemann conditions in the new variables become $f_{\overline{z}}=0$. Assuming that this is satisfied, set $h=f_z$. Then $$h_{\overline{z}}=f_{z\overline{z}}=f_{\overline{z}z}=0,$$ so $h$ does not depend on $\overline{z}$ that is $h\in C[[z]]$. Taking an antiderivative of $h$ in $C[[z]$ we obtain the $g$ as above.

Remark. This is how old textbooks like Whittaker Watson explain the Cauchy-Riemann conditions. Pure algebra.

$f\in C[[x+iy]]$ means $f(x,y)=g(x+iy),$ where $g\in C[[z]]$. Then $$f_x=g'(x+iy),\; f_y=ig'(x+iy),$$ therefore $f_y=if_x$ and this is Cauchy-Riemann: indeed, if $f=u+iv$ then $f_x=u_x+iv_x$ and $f_y=u_y+iv_y$, so $f_y=if_x$ is equivalent to
$u_x=v_y$, $u_y=-v_x$.

To prove the converse, make the linear change of the variables: $z=x+iy,\; \overline{z}=x-iy$. A simple calculation shows that $f_z=(1/2)(f_x-if_y),\; f_{\overline{z}}=(1/2)(f_x+if_y).$ Then the Cauchy-Riemann conditions in the new variables become $f_{\overline{z}}=0$. Assuming that this is satisfied, set $h=f_z$. Then $$h_{\overline{z}}=f_{z\overline{z}}=f_{\overline{z}z}=0,$$ so $h$ does not depend on $\overline{z}$ that is $h\in C[[z]]$. Taking an antiderivative of $h$ in $C[[z]]$ we obtain the $g$ as above.

Remark. This is how old textbooks like Whittaker-Watson explain the Cauchy-Riemann conditions. Pure algebra.

$f\in C[[x+iy]]$ means $f(x,y)=g(x+iy),$ where $g\in C[[z]]$. Then $$f_x=g'(x+iy),\; f_y=ig'(x+iy),$$ therefore $f_y=if_x$ and this is Cauchy-Riemann: indeed, if $f=u+iv$ then $f_x=u_x+iv_x$ and $f_y=u_y+iv_y$, so $f_y=if_x$ is equivalent to
$u_x=v_y$, $u_y=-v_x$.

To prove the converse, make the linear change of the variables: $z=x+iy,\; \overline{z}=x-iy$. A simple calculation shows that $f_z=(1/2)(f_x-if_y),\; f_{\overline{z}}=(1/2)(f_x+if_y).$ Then the Cauchy-Riemann conditions in the new variables become $f_{\overline{z}}=0$. Assuming that this is satisfied, set $g=f_z$. Then $$g_{\overline{z}}=f_{z\overline{z}}=f_{\overline{z}z}=0,$$ so $g$ does not depend on $\overline{z}$ that is $g\in C[[z]]$.

Remark. This is how old textbooks like Whittaker Watson explain the Cauchy-Riemann conditions. Pure algebra.

$f\in C[[x+iy]]$ means $f(x,y)=g(x+iy),$ where $g\in C[[z]]$. Then $$f_x=g'(x+iy),\; f_y=ig'(x+iy),$$ therefore $f_y=if_x$ and this is Cauchy-Riemann: indeed, if $f=u+iv$ then $f_x=u_x+iv_x$ and $f_y=u_y+iv_y$, so $f_y=if_x$ is equivalent to
$u_x=v_y$, $u_y=-v_x$.

To prove the converse, make the linear change of the variables: $z=x+iy,\; \overline{z}=x-iy$. A simple calculation shows that $f_z=(1/2)(f_x-if_y),\; f_{\overline{z}}=(1/2)(f_x+if_y).$ Then the Cauchy-Riemann conditions in the new variables become $f_{\overline{z}}=0$. Assuming that this is satisfied, set $h=f_z$. Then $$h_{\overline{z}}=f_{z\overline{z}}=f_{\overline{z}z}=0,$$ so $h$ does not depend on $\overline{z}$ that is $h\in C[[z]]$. Taking an antiderivative of $h$ in $C[[z]$ we obtain the $g$ as above.

Remark. This is how old textbooks like Whittaker Watson explain the Cauchy-Riemann conditions. Pure algebra.

$f\in C[[x+iy]]$ means $f(x,y)=g(x+iy),$ where $g\in C[[z]]$. Then $$f_x=g'(x+iy),\; f_y=ig'(x+iy),$$ therefore $f_y=if_x$ and this is Cauchy-Riemann: indeed, if $f=u+iv$ then $f_x=u_x+iv_x$ and $f_y=u_y+iv_y$, so $f_y=if_x$ is equivalent to
$u_x=v_y$, $u_y=-v_x$.

To prove the converse, make the linear change of the variables: $z=x+iy,\; \overline{z}=x-iy$. Then the Cauchy-Riemann conditions in new variables become $f_{\overline{z}}=0$. Assuming that this is satisfied, set $g=f_z$. Then $$g_{\overline{z}}=f_{z\overline{z}}=f_{\overline{z}z}=0,$$ so $g$ does not depend on $\overline{z}$ that is $g\in C[[z]]$.

Remark. This is how old textbooks like Whittaker Watson explain the Cauchy-Riemann conditions. Pure algebra.

$f\in C[[x+iy]]$ means $f(x,y)=g(x+iy),$ where $g\in C[[z]]$. Then $$f_x=g'(x+iy),\; f_y=ig'(x+iy),$$ therefore $f_y=if_x$ and this is Cauchy-Riemann: indeed, if $f=u+iv$ then $f_x=u_x+iv_x$ and $f_y=u_y+iv_y$, so $f_y=if_x$ is equivalent to
$u_x=v_y$, $u_y=-v_x$.

To prove the converse, make the linear change of the variables: $z=x+iy,\; \overline{z}=x-iy$. A simple calculation shows that $f_z=(1/2)(f_x-if_y),\; f_{\overline{z}}=(1/2)(f_x+if_y).$ Then the Cauchy-Riemann conditions in the new variables become $f_{\overline{z}}=0$. Assuming that this is satisfied, set $g=f_z$. Then $$g_{\overline{z}}=f_{z\overline{z}}=f_{\overline{z}z}=0,$$ so $g$ does not depend on $\overline{z}$ that is $g\in C[[z]]$.

Remark. This is how old textbooks like Whittaker Watson explain the Cauchy-Riemann conditions. Pure algebra.