This is a problem from a book that I'm using to study complex analysis. I'm a little insecure with what I have to prove here, because, I don't know what it means $g_w$ for example . I'm a little confused... sorry for asking this stupid things
a question about the derivate with respect to z , of a composition
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0It seems to me we should be reading $g_w,g_{\bar{w}}$ as the derivative of $g$ AT $w=f(z), \bar{w}=\bar{f(z)}$, by analogy with the real chain rule, rather than "with respect to" those variables. – 2012-08-07
1 Answers
In exercise $2.8$ of the same book they define $f_z=\frac{1}{2}(f_x-if_y)\\f_{\bar{z}}=\frac{1}{2}(f_x+if_y)$ Let $f=u(x,y)+iv(x,y)$, $w=u+iv$, and $g=s(u,v)+it(u,v)$. Then we have $\begin{align}(g\circ f)_z&=\frac{1}{2}((g\circ f)_x-i(g\circ f)_y)\\&=\frac{1}{2}(s_uu_x+s_vv_x+it_uu_x+it_vv_x-i(s_uu_y+s_vv_y+it_uu_y+it_vv_y))\end{align}$
Similarly we have $\begin{align}g_wf_z&=\frac{1}{2}(g_u-ig_v)\frac{1}{2}(f_x-if_y)\\&=\frac{1}{2}(s_u+it_u-i(s_v+it_v))\frac{1}{2}(u_x+iv_x-i(u_y+iv_y))\\&=\frac{1}{4}(s_u+t_v-is_v+it_u)(u_x+v_y-iu_y+iv_x)\end{align}$
and
$\begin{align}g_{\bar{w}}\bar{f}_z&=\frac{1}{2}(g_u+ig_v)\frac{1}{2}(\bar{f}_x-i\bar{f}_y)\\&=\frac{1}{2}(s_u+it_u+i(s_v+it_v))\frac{1}{2}(u_x-iv_x-i(u_y-iv_y))\\&=\frac{1}{4}(s_u-t_v+is_v+it_u)(u_x-v_y-iu_y-iv_x)\end{align}$
Expanding and adding you will see that $(g\circ f)_z=g_wf_z+g_{\bar{w}}\bar{f}_z$. The second equality is shown similarly.