M4.6 Some identities on Analytic and harmonic functions

01. If $u$ and $v$ are harmonic functions show that $\left(\frac{\partial u}{\partial y}-\frac{\partial v}{\partial x}\right)+i\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)$ is analytic.
Ans:
Let $P=\left(\frac{\partial u}{\partial y}-\frac{\partial v}{\partial x}\right)$ and $Q=\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)$.
Now we have to show that $P+iQ$ is analytic, i.e., we have to show that $P$ and $Q$ will satisfies CR-equations
$\frac{\partial P}{\partial x}-\frac{\partial Q}{\partial y}=\frac{\partial }{\partial x}\left(\frac{\partial u}{\partial y}-\frac{\partial v}{\partial x}\right)-\frac{\partial }{\partial y}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)$
$\Longrightarrow \frac{\partial P}{\partial x}-\frac{\partial Q}{\partial y}=\frac{\partial^2 u}{\partial x \partial y}-\frac{\partial^2 v}{\partial x^2}-\frac{\partial^2 u}{\partial y \partial x}-\frac{\partial^2 v}{\partial y^2}=-\left(\frac{\partial^2 v}{\partial x^2}+\frac{\partial^2 v}{\partial y^2}\right)=0$
[because $\frac{\partial^2 v}{\partial x \partial y}=\frac{\partial^2 v}{\partial y \partial x}$ and since $u$ and $v$ are harmonic, $u$ and $v$ satisfies Laplace equations]. Hence $\frac{\partial P}{\partial x}-\frac{\partial Q}{\partial y}=0$$\Longrightarrow \frac{\partial P}{\partial x}=\frac{\partial Q}{\partial y}$
Similarly, consider
$\frac{\partial P}{\partial y}+\frac{\partial Q}{\partial x}=\frac{\partial }{\partial y}\left(\frac{\partial u}{\partial y}-\frac{\partial v}{\partial x}\right)+\frac{\partial }{\partial x}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)$
$\Longrightarrow \frac{\partial P}{\partial y}+\frac{\partial Q}{\partial x}=\frac{\partial^2 u}{\partial y^2}-\frac{\partial^2 v}{\partial y \partial x}+\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 v}{\partial x \partial y}=\left(\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}\right)=0$
[because $\frac{\partial^2 u}{\partial x \partial y}=\frac{\partial^2 u}{\partial y \partial x}$ and since $u$ and $v$ are harmonic, $u$ and $v$ satisfies Laplace equations]. Hence $\frac{\partial Q}{\partial x}+\frac{\partial Q}{\partial y}=0$$\Longrightarrow \frac{\partial Q}{\partial x}=-\frac{\partial Q}{\partial y}$
Hence $P+iQ$ is analytic.

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02. If $f(z)$ is analytic, show that $\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right)|f(z)|^2=4|f'(z)|^2$ is analytic.
Ans:
Let $f(z)=u+iv$ be analytic. $\Longrightarrow |f(z)|=\sqrt{u^2+v^2}\Longrightarrow |f(z)|^2=u^2+v^2=\phi$<\br> Hence, we need to show that $\phi_{xx}+\phi_{yy}=4|f'(z)|^2$.<\br> $$\phi=u^2+v^2\Longrightarrow \phi_x=2uu_x+2vv_x \Longrightarrow \phi_{xx}=2[uu_{xx}+u_x^2+vv_{xx}+v_x^2]\rightarrow (1)$$ $$\phi=u^2+v^2\Longrightarrow \phi_y=2uu_y+2vv_y \Longrightarrow \phi_{yy}=2[uu_{yy}+u_y^2+vv_{yy}+v_y^2]\rightarrow (2)$$ Now add (1) and (2)
$$\phi_{xx}+\phi_{yy}=2[u(u_{xx}+u_{yy})+v(v_{xx}+v_{yy})+u_x^2+v_x^2+u_y^2+v_y^2]\rightarrow (3)$$ Since $f(z)$ is analytic, $u$ and $v$ are harmonic $\Longrightarrow u_{xx}+u_{yy}=0,v_{xx}+v_{yy}=0$, also using CR-equations in (3), we get.
$$\phi_{xx}+\phi_{yy}=2[2u_x^2+2v_x^2]=4[u_x^2+v_x^2]\rightarrow (4)$$ But $f'(z)=u_x+iv_x \Longrightarrow |f'(z)|=u_x^2+v_x^2$, hence we can conclude that $$\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right)|f(z)|^2=4|f'(z)|^2$$

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03. If $f(z)$ is a regular analytic, show that $\left(\frac{\partial}{\partial x}|f(z)|\right)^2+\left(\frac{\partial}{\partial y}|f(z)|\right)^2=|f'(z)|^2$
Ans:
Let $f(z)=u+iv$ be analytic. $\Longrightarrow |f(z)|=\sqrt{u^2+v^2}=\phi\Longrightarrow |f(z)|^2=u^2+v^2=\phi^2$<\br> Consider $\phi^2=u^2+v^2$ $$\Longrightarrow 2\phi \phi_x=2uu_x+2vv_x\Longrightarrow \phi \phi_x=uu_x+vv_x \rightarrow (1)$$ $$\text{similarly} 2\phi \phi_y=2uu_y+2vv_y\Longrightarrow \phi \phi_y=uu_y+vv_y\rightarrow (2)$$ Squaring and adding both (1) and (2), we will get $$\phi^2 (\phi_x^2+\phi_y^2)=(uu_x+vv_x)^2+(uu_y+vv_y)^2$$ $$=(u^2 u_x^2+v^2 v_x^2+2uvu_x v_x)+(u^2u_y^2+v^2v_y^2+2uvu_y v_y)$$ Since $f(z)$ is analytic, $u$ and $v$ satisfies CR-equations. $$\Longrightarrow \phi^2 (\phi_x^2+\phi_y^2)=u^2(u_x^2+v_x^2)+v^2(u_x^2+v_x^2)=(u^2+v^2)(u_x^2+v_x^2)=\phi^2 (u_x^2+v_x^2)$$ Hence, we get $$\phi_x^2+\phi_y^2=u_x^2+v_x^2$$ Also we have $f(z)=u+iv\Longrightarrow f'(z)=u_x+iv_x \Longrightarrow |f'(z)|=\sqrt{u_x^2+v_x^2}$, hence above equation will become $$\phi_x^2+\phi_y^2=|f'(z)|^2$$