Problem 6.6

Show that, if $$f$$ is holomorphic and nonzero in $$G$$, then $$ln|f(x,y)|$$ is harmonic in G.

Assume $$f=u+iv$$ is holomorphic and nonzero in $$G$$. Since $$f$$ is holomorphic, its real and imaginary parts are harmonic and therefore infinitely differentiable by Proposition 6.4 and Corollary 6.9. The first derivative of $$ln|f(x,y)|$$ with respect to $$x$$ is \begin{aligned} \\ (ln|f(x,y)|)_x &= [ln(\sqrt{u^{2}+v^{2}})]_x\\ &= \frac{1}{2}[ln(u^{2}+v^{2})]_x\\ &=\frac{2uu_x+2vv_x}{2(u^{2}+v^{2})}\\ &=\frac{uu_x+vv_x}{u^{2}+v^{2}}\end{aligned} And the second derivative with respect to $$x$$ is \begin{aligned} (ln|f(x,y)|)_{xx}&= (\frac{uu_x+vv_x}{u^{2}+v^{2}})_x\\ &= \frac{(uu_x + uu_{xx} +v_xv_x + vv_{xx})(u^{2}+v^{2})-(uu_x + vv_x)(2uu_x+2vv_x)}{(u^{2}+v^{2})^{2}}\end{aligned} Similarly the first and second derivatives of $$ln|f(x,y)|$$ with respect to $$y$$ are \begin{aligned} (ln|f(x,y)|)_y &= [ln(\sqrt{u^{2}+v^{2}})]_y\\ &= \frac{1}{2}[ln(u^{2}+v^{2})]_y\\ &=\frac{2uu_y+2vv_y}{2(u^{2}+v^{2})}\\ &=\frac{uu_y+vv_y}{u^{2}+v^{2}}\end{aligned} and \begin{aligned} (ln|f(x,y)|)_{yy}&= (\frac{uu_y+vv_y}{u^{2}+v^{2}})_y\\ &= \frac{(uu_y + uu_{yy} +v_yv_y + vv_{yy})(u^{2}+v^{2})-(uu_y + vv_y)(2uu_y+2vv_y)}{(u^{2}+v^{2})^{2}}\end{aligned} Since $$f$$ is nonzero and $$u$$ and $$v$$ are harmonic in $$G$$, the second partials of $$ln|f(x,y)|$$ are defined on $$G$$. Now consider the Laplace equation, \begin{aligned} (ln|f(x,y)|)_{xx}+(ln|f(x,y)|)_{yy} &=\frac{1}{(u^{2}+v^{2})^{2}}[(uu_x + uu_{xx} +v_xv_x + vv_{xx})(u^{2}+v^{2})-(uu_x + vv_x)(2uu_x+2vv_x)\\ &+ (uu_y + uu_{yy} +v_yv_y + vv_{yy})(u^{2}+v^{2})-(uu_y + vv_y)(2uu_y+2vv_y)]\end{aligned} We will show that the numerator expression will be equal to zero and thus the Laplace equation will equal zero.\begin{aligned} (uu_x + uu_{xx} +v_xv_x + vv_{xx})(u^{2}+v^{2}) &-(uu_x + vv_x)(2uu_x+2vv_x)+(uu_y + uu_{yy} +v_yv_y + vv_{yy})(u^{2}+v^{2})\\ &-(uu_y + vv_y)(2uu_y+2vv_y)\\ &=u^{2}u^{2}_x+u^{3}u_{xx}+u^{2}v^{2}_x+u^{2}vv_{xx}+v^{2}u^{2}_x+v^{2}uu_{xx}+v^{2}v^{2}_x+v^{3}v_{xx}\\ &-2u^{2}u^{2}_x-2uvv_xu_x-2uvu_xv_x-2v^{2}v^{2}_x+u^{2}u^{2}_y+u^{3}u_{yy}+u^{2}v^{2}_y+u^{2}vv_{yy}+\\ &v^{2}u^{2}_y+v^{2}uu_{yy}+v^{2}v^{2}_y+v^{3}v_{yy}-2u^{2}u^{2}_y-2uvv_yu_y-2uvu_yv_y-2v^{2}v^{2}_y\\ &-2u^{2}u^{2}_y-2uvv_yu_y-2uvu_yv_y-2v^{2}v^{2}_y\end{aligned} But since $$u$$ and $$v$$ are harmonic, $$u_{xx}+u_{yy}=0$$ and $$v_{xx}+v_{yy}=0$$. Using this fact and combining like terms gives us, \begin{aligned} -u^{2}u^{2}_x+u^{2}v^{2}_x+v^{2}u^{2}_x-v^{2}v^{2}_x-4uvu_xv_x-u^{2}u^{2}_y+u^{2}v^{2}_y+v^{2}u^{2}_y-v^{2}v^{2}_y-4uvu_yv_y\end{aligned} And now we can use the Cauchy-Riemann equations to get \begin{aligned} -u^{2}u^{2}_x+u^{2}v^{2}_x+v^{2}u^{2}_x-v^{2}v^{2}_x-4uvv_yv_x-u^{2}v^{2}_x+u^{2}u^{2}_x+v^{2}v^{2}_x-v^{2}u^{2}_x+4uv_xv_y\end{aligned} Which simplifies to 0 and thus $$(ln|f(x,y)|)_{xx}+(ln|f(x,y)|)_{yy}=0$$. Since $$ln|f(x,y)|$$ has continuous second partials on G and the Laplace equation is satisfied, $$ln|f(x,y)|$$ is harmonic on $$G$$.