# Show that $Z^{\prime }\left( s\right) -2\sqrt{-1}H\left( s\right) Z\left( s\right) -1=0$

Consider the equations

$$2H\left( s\right) y\left( s\right) -x^{\prime }\left( s\right) -x^{\prime \prime }\left( s\right) y\left( s\right) y^{\prime }\left( s\right) +x^{\prime }\left( s\right) y\left( s\right) y^{\prime \prime }\left( s\right) =0$$

$$x^{\prime }\left( s\right) ^{2}+y^{\prime }\left( s\right) ^{2}=1$$

$$Z\left( s\right) =y\left( s\right) y^{\prime }\left( s\right) +\sqrt{-1}% y\left( s\right) x^{\prime }\left( s\right)$$

where $$H,x,y:I\rightarrow %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion$$ are functions real of one variable

Show that $$Z^{\prime }\left( s\right) -2\sqrt{-1}H\left( s\right) Z\left( s\right) -1=0$$, $$s\in I\subset \mathbb{R}$$

I have opened the accounts anyway, but can not show the equality of the differential equation.

• Can you go back a few steps and tell more about the genesis of this problem? The first equation seems to tell something about $\arg(y'+ix)$, the second says that $|y'+ix|=1$ while $Z$ is a multiple of that complex number,... – LutzL Oct 29 '18 at 13:57
• It is a problem of rotational surfaces where $H$ is the mean curvature of the surfaces generated by the curve in $\mathbb{R}^3$ whose coordinates are $x$ and $y$ in the plane $z = 0$ – Mancala Oct 29 '18 at 14:04

From your equations you get $$1=(y'+ix')(y'-ix')\\ 0=x'x''+y'y''$$ So for the function under consideration, $$Z=y(y'+ix')$$, we get \begin{align} \frac{Z'}{Z}&=\frac{y'}{y}+\frac{y''+ix''}{y'+ix'} \\& =\frac{y'}y+(y''+ix'')(y'-ix') =\frac{y'}y+i(x''y'-y''x') \\& =\frac1y\left[y'+i(2Hy-x')\right] \\ \implies Z'& =\frac{Z}{y(y'+ix')}+2iHZ=1+2iHZ \end{align} which is indeed the given differential equation.