Suppose$\mathit{A}{\mathit{e}}^{\mathbf{i}\mathbf{a}\mathbf{x}}\mathbf{+}\mathit{B}{\mathit{e}}^{\mathbf{i}\mathbf{b}\mathbf{x}}\mathbf{=}\mathit{C}{\mathit{e}}^{\mathbf{i}\mathbf{c}\mathbf{x}}$, for some nonzero constants A, B, C, a, b, c, and for all x. Prove that a = b = cand A + B = C.

Write the expression for the amplitude equation.

$\mathit{A}{\mathit{e}}^{\mathbf{i}\mathbf{a}\mathbf{x}}\mathbf{+}\mathit{b}{\mathit{e}}^{\mathbf{i}\mathbf{b}\mathbf{x}}\mathbf{=}\mathit{C}{\mathit{e}}^{\mathbf{i}\mathbf{c}\mathbf{x}}$ …… (1)

Write the expression for the Euler equation.

${\mathit{e}}^{\mathbf{i}\mathbf{\theta }}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{\theta }\mathbf{+}\mathit{i}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }$

Substitute the known values in equation (1).

$A(cosax+isinax)+B(cosbx+isinbx)=C(coscx+isincx)$

Write the real part of the equation.

role="math" localid="1657520367793" $A\cos ax+B\cos bx=C\cos cx$

Write the real imaginary of the equation.

$Asinax+Bsinbx=Csincx$

Using boundary conditions at x = 0 , the left-hand side function will be equal to the right-hand side function.

$$\begin{gathered} A{e}^{ia\left(0\right)}+B{e}^{ib\left(0\right)}=C{e}^{ic\left(0\right)} \\ A+B=C \end{gathered}$$

Square and add the equation (2) and (3).

${(A\cos ax+B\cos bx)}^2+{(A\sin ax+B\sin bx)}^2=C^2({\cos }^{2}cx+{\sin }^{2}cx)$

Since, ${\cos }^{2}cx+{\sin }^{2}cx=1$.

On further solving,

$$\begin{gathered} \left\{\begin{array}{c}\left[A^2{\cos }^{2}ax+B^2{\cos }^{2}bx+2AB\left(\cos Ax\right)(\cos bx)\right]+\\ A^2{\sin }^{2}ax+B^2\sin bx+2AB(\sin ax)(\sin bx)\end{array}\right\}=C^2 \\ \left\{\begin{array}{c}{A}^2({\cos }^{2}ax+{\sin }^{2}ax+B^2({\cos }^{2}bx+{\sin }^{2}bx)\\ +2AB(\cos ax\cos bx+\sin ax\sin bx)\end{array}\right\}=C^2 \\ {A}^2+B^2+2AB\cos (a-b)x=C^2 \end{gathered}$$

Substitute the value of $A+B=C$in the above expression.

$$\begin{gathered} A^2+B^2+2AB\cos (a-b)x={(A+B)}^2{A}^2+B^2+2AB\cos (a-b)x=A^2+B^2+2AB \\ 2AB\cos (a-b)x=2AB \\ \cos (a-b)x=1 \end{gathered}$$

Hence,

a = b

Substitute the value of a and b in equation (1).

$$\begin{gathered} e^{iax}(A+B)=C{e}^{icx} \\ {e}^{iax}(A+B)=(A+B)e^{icx} \\ {e}^{iax}=e^{icx} \end{gathered}$$

Solve further as,

a = b

= c

Therefore, it is proved A + B = C and a = b = c.