Question

Prove that if the function${\mathit{e}}^{\mathbf{i}\sqrt{\mathbf{D}}\mathbf{\phi }}$is to meet itself smoothly when$\mathit{\phi }$changes by $\mathbf{2}\mathbf{\pi }$, $\sqrt{\mathbf{D}}$ must be an integer.

Short answer

For the function $e^{\mathrm{i}\sqrt{\mathrm{D}}\mathrm{\phi }}$to have a period of $2\mathrm{\pi }$, $\sqrt{\mathrm{D}}$ must be an integer.

Step-by-step solution

A concept:

The functions which repeat itself after regular interval of time is called a Cyclic function and that interval in which it is repeating itself is called the period of that cyclic function.

As you know that, by Euler’s equation,

${\mathbf{e}}^{\mathbf{ix}}\mathbf{=}\mathbf{cos}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}\mathbf{i}\mathbf{}\mathbf{sin}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$

Value of the function at φ=0 and φ=2π :

Let, the function be

$$\begin{gathered} \mathrm{F}\left(\mathrm{\phi }\right)={\mathrm{e}}^{\mathrm{i}\sqrt{\mathrm{D}}\mathrm{\phi }} \\ =\cos \left(\sqrt{\mathrm{D}}\mathrm{\phi }\right)+\mathrm{isin}\left(\sqrt{\mathrm{D}}\mathrm{\phi }\right) \end{gathered}$$

Where,$D=-\frac{1}{\Phi }\frac{{\partial }^2\Phi }{\partial {\phi }^2}$ and $\phi$is the Azimuthal Angle.

Now at $\phi =0$:

$$\begin{gathered} \mathrm{F}\left(0\right)={\mathrm{e}}^{\mathrm{i}\sqrt{\mathrm{D}}0} \\ =\cos \left(\sqrt{\mathrm{D}}0\right)+\mathrm{isin}\left(\sqrt{\mathrm{D}}0\right) \\ =1 \end{gathered}$$

Again, if the function meets itself at $\phi =2\mathrm{\pi }$,

$$\begin{gathered} \mathrm{F}\left(2\mathrm{\pi }\right)=\mathrm{F}\left(0\right) \\ =1 \end{gathered}$$

Value of  :

If, $\mathrm{F}\left(2\mathrm{\pi }\right)=1$

$$\begin{gathered} {\mathrm{e}}^{\mathrm{i}\sqrt{\mathrm{D}}2\mathrm{\pi }}=1 \\ \cos \left(\sqrt{\mathrm{D}}2\mathrm{\pi }\right)+\mathrm{isin}\left(\sqrt{\mathrm{D}}2\mathrm{\pi }\right)=1 \end{gathered}$$ ….. (1)

In equation (1), if the real part is 1 the imaginary part should be zero and for that to hold, must be an integer$\sqrt{\mathrm{D}}$