Question

Show that $\mathbf{\Phi }\mathbf{\left(}\mathbf{\Phi }\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{{\mathbf{im}}_{\mathbf{l}}\mathbf{\varphi }}\mathbf{=}\mathbf{\Phi }\mathbf{(}\mathbf{\Phi }\mathbf{+}\mathbf{2}\mathbf{\pi }\mathbf{)}$(that is, show thatis periodic with period 2) if and only ifis restricted to the values 0, ±1, ±2,….. (Hint: Euler’s formula states that${\mathbf{e}}^{\mathbf{i\varphi }}\mathbf{=}\mathbf{cos\varphi }\mathbf{=}\mathbf{isin\varphi }$)

Short answer

We indeed prove that it is a periodic function.

Step-by-step solution

Important Concepts

We know from Euler’s formula

${\mathbf{e}}^{\mathbf{i\varphi }}\mathbf{=}\mathbf{cos\varphi }\mathbf{=}\mathbf{isin\varphi }$………………..(1)

Application

We rewrite are the function as

$\mathrm{\Phi }\left(\mathrm{\Phi }\right)={\mathrm{e}}^{{\mathrm{im}}_{\mathrm{l}}\mathrm{\varphi }}$

Rewrite the function at

$\mathrm{\Phi }\left(\mathrm{\Phi }+2\mathrm{\pi }\right)={\mathrm{e}}^{{\mathrm{im}}_{\mathrm{l}}\left(\mathrm{\varphi }+2\mathrm{\pi }\right)}$

Equate the two functions

role="math" localid="1663955560288" $$\begin{gathered} {\mathrm{cosm}}_{\mathrm{l}}\mathrm{\Phi cos}\left({\mathrm{m}}_{\mathrm{l}}\left(\mathrm{\varphi }+2\mathrm{\pi }\right)\right)+\mathrm{isin}\left({\mathrm{m}}_{\mathrm{l}}\left(\mathrm{\varphi }+2\mathrm{\pi }\right)\right) \\ {\mathrm{sinm}}_{\mathrm{l}}\mathrm{\Phi }=\sin \left({\mathrm{m}}_{\mathrm{l}}\left(\mathrm{\varphi }+2\mathrm{\pi }\right)\right) \end{gathered}$$

Rearranging

$$\begin{gathered} {\mathrm{cosm}}_{\mathrm{l}}{\mathrm{\Phi cosm}}_{\mathrm{l}}\mathrm{\varphi }+2{\mathrm{m}}_{\mathrm{l}}\mathrm{\pi } \\ {\mathrm{sinm}}_{\mathrm{l}}\mathrm{\Phi }={\mathrm{sinm}}_{\mathrm{l}}\mathrm{\varphi }+2{\mathrm{m}}_{\mathrm{l}}\mathrm{\pi } \end{gathered}$$

We know that $\cos {\varphi }_1=\cos {\varphi }_2$if and only if${\mathrm{\varphi }}_2={\mathrm{\varphi }}_1=2\mathrm{n\pi },\mathrm{n}\in \mathrm{Z}$,

Thus in comparison, we get

${\mathrm{m}}_{\mathrm{l}}\mathrm{\varphi }\pm \mathrm{n}\left(2\mathrm{\pi }\right)={\mathrm{m}}_{\mathrm{l}}\mathrm{\varphi }\pm 2{\mathrm{\pi m}}_{\mathrm{l}}$

Solving we get

${\mathrm{m}}_{\mathrm{l}}=\pm \mathrm{n}$

Hence, the function is periodic with${\mathrm{m}}_{\mathrm{l}}=\pm \mathrm{n}$