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Consider the electric fields are$\stackrel{鈫�}{\mathrm{E}}={\stackrel{鈫�}{\mathrm{E}}}_0(\mathrm{x},\mathrm{y}){\mathrm{e}}^{\mathrm{i}(\mathrm{kz}鈭�\mathrm{蝇t})}$.

Consider the electric field vector${\stackrel{鈫�}{\mathrm{E}}}_0={\mathrm{E}}_{\mathrm{x}}\widehat{\mathrm{x}}+{\mathrm{E}}_{\mathrm{y}}\widehat{\mathrm{y}}+{\mathrm{E}}_{\mathrm{z}}\widehat{\mathrm{z}}$.

Consider the magnetic fields are$\stackrel{鈫�}{\mathrm{B}}={\stackrel{鈫�}{\mathrm{B}}}_0(\mathrm{x},\mathrm{y}){\mathrm{e}}^{\mathrm{i}(\mathrm{kz}鈭�\mathrm{蝇t})}$ .

Consider the magnetic fields are ${\stackrel{鈫�}{\mathrm{B}}}_0={\mathrm{B}}_{\mathrm{x}}\widehat{\mathrm{x}}+{\mathrm{B}}_{\mathrm{y}}\widehat{\mathrm{y}}+{\mathrm{B}}_{\mathrm{z}}\widehat{\mathrm{z}}$.

Write the formula electric field in x-axis direction.

${\mathbf{B}}_y=\frac{1}{\mathrm{ik}}\mathbf{(}\frac{\mathbf{鈭�}{\mathbf{B}}_{\mathbf{z}}}{\mathbf{鈭�}\mathbf{y}}\mathbf{+}\frac{\mathbf{i蝇}}{{\mathbf{c}}^{\mathbf{2}}}{\mathbf{E}}_{\mathbf{z}}\mathbf{)}鈫�\left(\mathbf{2}\right)$ 鈥︹�� (1)

Here, $\mathbf{k}$is real, ${\mathbf{B}}_{\mathbf{z}}$is transverse electric wave, $\mathbf{蝇}$ is frequency and ${\mathbf{E}}_{\mathbf{z}}$ is transverse electric wave.

Write the formula of electric field in y-axis direction.

role="math" localid="1658734044435" ${\mathbf{B}}_x=鈭�\frac{蝇}{{\mathbf{c}}^2}\frac{1}{k}{\mathbf{E}}_y鈭�\frac{i}{k}\frac{鈭�{\mathbf{B}}_z}{鈭�\mathbf{x}}鈫�\left(\mathbf{1}\right)$ 鈥︹�� (2)

Here, $\mathrm{蝇}$is frequency, $\mathbf{k}$is real, ${\mathbf{E}}_y$is electric field in y-axis and ${\mathbf{B}}_z$is transverse electric wave.

Write the formula of magnetic field in x-axis direction.

${\mathbf{ikB}}_x鈭�\frac{鈭�{\mathbf{B}}_z}{鈭�\mathbf{x}}$ 鈥︹�� (3)

Here, $\mathbf{k}$ is real, ${\mathbf{B}}_x$ is magnetic field in x-axis and ${\mathbf{B}}_z$ is transverse electric wave.

Write the formula of magnetic field in y-axis direction.

$\frac{鈭�{\mathbf{B}}_z}{鈭�\mathbf{y}}鈭�{\mathbf{ikB}}_y$ 鈥︹�� (4)

Here, $\mathbf{k}$ is real, ${\mathbf{B}}_y$ is magnetic field in y-axis and ${\mathbf{B}}_z$ is transverse electric wave.

From the 3^rd Maxwell equation by components:

$\begin{array}{c}鈭�脳=鈭�\frac{鈭�\stackrel{鈫�}{\mathrm{B}}}{鈭�\mathrm{t}}\\ \frac{鈭�{\mathrm{E}}_{\mathrm{z}}}{鈭�\mathrm{y}}鈭�\frac{鈭�{\mathrm{E}}_{\mathrm{y}}}{鈭�\mathrm{z}}=鈭�\frac{鈭�{\mathrm{B}}_{\mathrm{z}}}{鈭�\mathrm{t}}\\ \frac{鈭�{\mathrm{E}}_{\mathrm{z}}}{鈭�\mathrm{y}}鈭�{\mathrm{E}}_{\mathrm{y}}\mathrm{ik}=\mathrm{i蝇Bz}\end{array}$

Determine the 3^rd Maxwell equation by components in z-axis direction.

$\begin{array}{c}\frac{鈭�{\mathrm{E}}_{\mathrm{z}}}{鈭�\mathrm{z}}鈭�\frac{鈭�{\mathrm{E}}_{\mathrm{z}}}{鈭�\mathrm{x}}=鈭�\frac{鈭�{\mathrm{B}}_{\mathrm{y}}}{鈭�\mathrm{t}}\\ {\mathrm{E}}_{\mathrm{z}}\mathrm{ik}鈭�\frac{鈭�{\mathrm{E}}_{\mathrm{z}}}{鈭�\mathrm{x}}={\mathrm{i蝇B}}_{\mathrm{y}}\end{array}$

Determine the 3^rd Maxwell equation by components in y-axis direction.

$\begin{array}{c}\frac{鈭�{\mathrm{E}}_{\mathrm{y}}}{鈭�\mathrm{x}}鈭�\frac{鈭�{\mathrm{E}}_{\mathrm{z}}}{鈭�\mathrm{y}}=鈭�\frac{鈭�{\mathrm{B}}_{\mathrm{z}}}{鈭�\mathrm{t}}\\ \frac{鈭�{\mathrm{E}}_{\mathrm{y}}}{鈭�\mathrm{x}}鈭�\frac{鈭�{\mathrm{E}}_{\mathrm{z}}}{鈭�\mathrm{y}}=鈭�{\mathrm{i蝇B}}_{\mathrm{z}}\end{array}$

From the 4^th Maxwell equation by components:

$\begin{array}{c}鈭�脳\stackrel{鈫�}{\mathrm{B}}=\frac{1}{{\mathrm{c}}^2}\frac{鈭�\stackrel{鈫�}{\mathrm{E}}}{鈭�\mathrm{t}}\\ \frac{鈭�{\mathrm{B}}_{\mathrm{z}}}{鈭�\mathrm{y}}鈭�\frac{鈭�{\mathrm{B}}_{\mathrm{y}}}{鈭�\mathrm{z}}=鈭�\frac{鈭�{\mathrm{E}}_{\mathrm{z}}}{鈭�\mathrm{t}}\\ \frac{鈭�{\mathrm{B}}_{\mathrm{z}}}{鈭�\mathrm{y}}鈭�{\mathrm{B}}_{\mathrm{y}}\mathrm{ik}=鈭�\frac{\mathrm{i蝇}}{{\mathrm{c}}^2}{\mathrm{E}}_{\mathrm{x}}\end{array}$

Determine the 4^th Maxwell equation by components in z-axis direction.

$\begin{array}{c}\frac{鈭�{\mathrm{B}}_{\mathrm{z}}}{鈭�\mathrm{z}}鈭�\frac{鈭�{\mathrm{B}}_{\mathrm{z}}}{鈭�\mathrm{x}}=\frac{1}{{\mathrm{c}}^2}\frac{鈭�{\mathrm{E}}_{\mathrm{y}}}{鈭�\mathrm{t}}\\ {\mathrm{B}}_{\mathrm{z}}\mathrm{ik}鈭�\frac{鈭�{\mathrm{B}}_{\mathrm{z}}}{鈭�\mathrm{x}}=鈭�\frac{\mathrm{i蝇}}{{\mathrm{c}}^2}{\mathrm{E}}_{\mathrm{y}}\end{array}$

Determine the 4^th Maxwell equation by components in y-axis direction.

$\begin{array}{c}\frac{鈭�{\mathrm{B}}_{\mathrm{y}}}{鈭�\mathrm{x}}鈭�\frac{鈭�{\mathrm{B}}_{\mathrm{z}}}{鈭�\mathrm{y}}=\frac{1}{{\mathrm{c}}^2}\frac{鈭�{\mathrm{E}}_{\mathrm{z}}}{鈭�\mathrm{t}}\\ \frac{鈭�{\mathrm{B}}_{\mathrm{y}}}{鈭�\mathrm{x}}鈭�\frac{鈭�{\mathrm{B}}_{\mathrm{z}}}{鈭�\mathrm{y}}=鈭�\frac{\mathrm{i蝇}}{{\mathrm{c}}^2}{\mathrm{E}}_{\mathrm{z}}\end{array}$

Now from (4) express ${\mathrm{B}}_{\mathrm{y}}$ and put this into (2), obtaining ${\mathrm{E}}_{\mathrm{x}}$:

Substitute ${\mathrm{E}}_{\mathrm{x}}\mathrm{ik}鈭�\frac{鈭�{\mathrm{E}}_{\mathrm{z}}}{鈭�\mathrm{x}}$for ${\mathrm{B}}_{\mathrm{y}}$ into equation (1).

$\begin{array}{c}{\mathrm{E}}_{\mathrm{x}}\mathrm{ik}鈭�\frac{鈭�{\mathrm{E}}_{\mathrm{z}}}{鈭�\mathrm{x}}=\mathrm{i蝇}\frac{1}{\mathrm{ik}}\left(\frac{鈭�{\mathrm{B}}_{\mathrm{z}}}{鈭�\mathrm{y}}+\frac{\mathrm{i蝇}}{{\mathrm{c}}^2}{\mathrm{E}}_{\mathrm{x}}\right)\\ {\mathrm{E}}_{\mathrm{x}}\left(\mathrm{ik}鈭�\frac{\mathrm{i}}{\mathrm{k}}\frac{{\mathrm{蝇}}^2}{{\mathrm{c}}^2}\right)=\frac{\mathrm{蝇}}{\mathrm{k}}\frac{鈭�{\mathrm{B}}_{\mathrm{z}}}{鈭�\mathrm{y}}+\frac{鈭�{\mathrm{E}}_{\mathrm{z}}}{鈭�\mathrm{x}}\\ {\mathrm{E}}_{\mathrm{x}}\frac{\mathrm{i}}{\mathrm{k}}\left({\mathrm{k}}^2鈭�\frac{{\mathrm{蝇}}^2}{{\mathrm{c}}^2}\right)=\frac{\mathrm{蝇}}{\mathrm{k}}\frac{鈭�{\mathrm{B}}_{\mathrm{z}}}{鈭�\mathrm{y}}+\frac{鈭�{\mathrm{E}}_{\mathrm{z}}}{鈭�\mathrm{x}}\\ {\mathrm{E}}_{\mathrm{x}}=\frac{\mathrm{i}}{{(\mathrm{蝇}/\mathrm{c})}^2鈭�{\mathrm{k}}^2}\left(\mathrm{蝇}\frac{鈭�{\mathrm{B}}_{\mathrm{z}}}{鈭�\mathrm{y}}+\mathrm{k}\frac{鈭�{\mathrm{E}}_{\mathrm{z}}}{鈭�\mathrm{x}}\right)\end{array}$

Now express from (5) the ${\mathrm{B}}_x$ and put it into the (1) to get the ${\mathrm{E}}_{\mathrm{y}}$

Substitute $\frac{鈭�{\mathrm{E}}_{\mathrm{z}}}{鈭�\mathrm{y}}鈭�{\mathrm{E}}_{\mathrm{y}}\mathrm{ik}$ for ${\mathrm{B}}_{\mathrm{x}}$ into equation (2).

$\begin{array}{c}\frac{鈭�{\mathrm{E}}_{\mathrm{z}}}{鈭�\mathrm{y}}鈭�{\mathrm{E}}_{\mathrm{y}}\mathrm{ik}=鈭�\frac{{\mathrm{蝇}}^2}{{\mathrm{c}}^2}\frac{1}{\mathrm{k}}{\mathrm{E}}_{\mathrm{y}}+\frac{\mathrm{蝇}}{\mathrm{k}}\frac{鈭�{\mathrm{B}}_{\mathrm{z}}}{鈭�\mathrm{x}}\\ {\mathrm{E}}_{\mathrm{y}}\frac{\mathrm{i}}{\mathrm{k}}\left(\frac{{\mathrm{蝇}}^2}{{\mathrm{c}}^2}鈭�{\mathrm{k}}^2\right)=\frac{\mathrm{蝇}}{\mathrm{k}}\frac{鈭�{\mathrm{B}}_{\mathrm{z}}}{鈭�\mathrm{x}}鈭�\frac{鈭�{\mathrm{E}}_{\mathrm{z}}}{鈭�\mathrm{y}}\\ {\mathrm{E}}_{\mathrm{y}}=\frac{\mathrm{i}}{(\mathrm{蝇}/{\mathrm{c}}^2)鈭�{\mathrm{k}}^2}\left(\mathrm{k}\frac{鈭�{\mathrm{E}}_{\mathrm{z}}}{鈭�\mathrm{y}}鈭�\mathrm{蝇}\frac{鈭�{\mathrm{B}}_{\mathrm{z}}}{鈭�\mathrm{x}}\right)\end{array}$

Now to get the other two equations put the result (7) into (4):

Substitute ${\mathrm{B}}_{\mathrm{y}}$ for $\frac{鈭�{\mathrm{B}}_{\mathrm{z}}}{鈭�\mathrm{y}}鈭�{\mathrm{ikB}}_{\mathrm{y}}$ into equation (3).

$\begin{array}{c}{\mathrm{B}}_{\mathrm{y}}=\frac{1}{\mathrm{ik}}\frac{鈭�{\mathrm{B}}_{\mathrm{z}}}{鈭�\mathrm{y}}鈭�\frac{1}{\mathrm{ik}}\frac{\mathrm{蝇}}{{\mathrm{c}}^2}\frac{1}{{(\mathrm{蝇}/\mathrm{c})}^2鈭�{\mathrm{k}}^2}\left(\mathrm{蝇}\frac{鈭�{\mathrm{B}}_{\mathrm{z}}}{鈭�\mathrm{y}}+\mathrm{k}\frac{鈭�{\mathrm{E}}_{\mathrm{z}}}{鈭�\mathrm{x}}\right)\\ =\frac{1}{\mathrm{ik}}\left[\frac{鈭�{\mathrm{B}}_{\mathrm{z}}}{鈭�\mathrm{y}}\left(1鈭�\frac{{(\mathrm{蝇}/\mathrm{c})}^2}{{(\mathrm{蝇}/\mathrm{c})}^2鈭�{\mathrm{k}}^2}\right)鈭�\frac{\mathrm{k蝇}}{{\mathrm{c}}^2}\frac{1}{{(\mathrm{蝇}/\mathrm{c})}^2鈭�{\mathrm{k}}^2}\frac{鈭�{\mathrm{E}}_{\mathrm{z}}}{鈭�\mathrm{x}}\right]\\ {\mathrm{B}}_{\mathrm{y}}=\frac{\mathrm{i}}{{(\mathrm{蝇}/\mathrm{c})}^2鈭�{\mathrm{k}}^2}\left(\mathrm{k}\frac{鈭�{\mathrm{B}}_{\mathrm{z}}}{鈭�\mathrm{y}}+\frac{\mathrm{蝇}}{{\mathrm{c}}^2}\frac{鈭�{\mathrm{E}}_{\mathrm{z}}}{鈭�\mathrm{x}}\right)\end{array}$

Now put the result (8) into (5) to finally reproduce all the eq. (9.180).

Substitute ${\mathrm{B}}_{\mathrm{x}}$ for ${\mathrm{ikB}}_{\mathrm{x}}鈭�\frac{鈭�{\mathrm{B}}_{\mathrm{z}}}{鈭�\mathrm{x}}$ into equation (4).

$\begin{array}{c}{\mathrm{B}}_{\mathrm{x}}=鈭�\frac{\mathrm{i}}{\mathrm{k}}\left[\frac{鈭�{\mathrm{B}}_{\mathrm{z}}}{鈭�\mathrm{x}}(1鈭�\frac{{(\mathrm{蝇}/\mathrm{c})}^2}{{(\mathrm{蝇}/\mathrm{c})}^2鈭�{\mathrm{k}}^2})+\frac{\mathrm{蝇k}}{{\mathrm{c}}^2}\frac{1}{{(\mathrm{蝇}/\mathrm{c})}^2鈭�{\mathrm{k}}^2}\frac{鈭�{\mathrm{E}}_{\mathrm{z}}}{鈭�\mathrm{y}}\right]\\ {\mathrm{B}}_{\mathrm{x}}=\frac{\mathrm{i}}{{(\mathrm{蝇}/\mathrm{c})}^2鈭�{\mathrm{k}}^2}\left(\mathrm{k}\frac{鈭�{\mathrm{B}}_{\mathrm{z}}}{鈭�\mathrm{x}}鈭�\frac{\mathrm{蝇}}{{\mathrm{c}}^2}\frac{鈭�{\mathrm{E}}_{\mathrm{z}}}{鈭�\mathrm{y}}\right)\end{array}$

With this all of the Eq. 9.180 have been reproduced. Nothing fancy, just a lot of algebra.

The 1^st Maxwell equation is:

$\begin{array}{c}鈭�鈰�\stackrel{鈫�}{\mathrm{E}}=\frac{鈭�{\mathrm{E}}_{\mathrm{x}}}{鈭�\mathrm{x}}+\frac{鈭�{\mathrm{E}}_{\mathrm{y}}}{鈭�\mathrm{y}}+\frac{鈭�{\mathrm{E}}_{\mathrm{z}}}{鈭�\mathrm{z}}\\ \frac{鈭�{\mathrm{E}}_{\mathrm{x}}}{鈭�\mathrm{x}}+\frac{鈭�{\mathrm{E}}_{\mathrm{y}}}{鈭�\mathrm{y}}+\frac{鈭�{\mathrm{E}}_{\mathrm{y}}}{鈭�\mathrm{y}}+{\mathrm{ikE}}_{\mathrm{z}}=0鈫�\left(7\right),\left(8\right)\\ \frac{\mathrm{i}}{{(\mathrm{蝇}/\mathrm{c})}^2鈭�{\mathrm{k}}^2}\mathrm{k}\left(\frac{{鈭�}^2{\mathrm{E}}_{\mathrm{x}}}{鈭�{\mathrm{x}}^2}+\frac{{鈭�}^2{\mathrm{E}}_{\mathrm{z}}}{鈭�{\mathrm{y}}^2}\right)+{\mathrm{ikE}}_{\mathrm{z}}=0\text{鈥�}/{(\mathrm{蝇}/\mathrm{c})}^2鈭�{\mathrm{k}}^2\\ \left[\frac{{鈭�}^2}{鈭�{\mathrm{x}}^2}+\frac{{鈭�}^2}{鈭�{\mathrm{y}}^2}+{(\mathrm{蝇}/\mathrm{c})}^2鈭�{\mathrm{k}}^2\right]{\mathrm{E}}_{\mathrm{z}}=0\end{array}$

Next do the similar thing with the 2^nd Maxwell equation:

$\begin{array}{c}鈭�鈰�\stackrel{鈫�}{\mathrm{B}}=\frac{鈭�{\mathrm{B}}_{\mathrm{x}}}{鈭�\mathrm{x}}+\frac{鈭�{\mathrm{B}}_{\mathrm{y}}}{鈭�\mathrm{y}}+\frac{鈭�{\mathrm{B}}_{\mathrm{z}}}{鈭�\mathrm{z}}=\frac{鈭�{\mathrm{B}}_{\mathrm{x}}}{鈭�\mathrm{x}}+\frac{鈭�{\mathrm{B}}_{\mathrm{y}}}{鈭�\mathrm{y}}+{\mathrm{ikB}}_{\mathrm{z}}=0鈫�\left(9\right),\left(10\right)\\ \frac{1}{{(\mathrm{蝇}/\mathrm{c})}^2鈭�{\mathrm{k}}^2}\mathrm{k}\left(\frac{{鈭�}^2{\mathrm{B}}_{\mathrm{z}}}{鈭�{\mathrm{x}}^2}+\frac{{鈭�}^2{\mathrm{B}}_{\mathrm{z}}}{鈭�{\mathrm{y}}^2}\right)+{\mathrm{ikB}}_{\mathrm{z}}=0\text{鈥�}/鈰�{(\mathrm{蝇}/\mathrm{c})}^2鈭�{\mathrm{k}}^2\\ \left[\frac{{鈭�}^2}{鈭�{\mathrm{x}}^2}+\frac{{鈭�}^2}{鈭�{\mathrm{y}}^2}+{(\mathrm{蝇}/\mathrm{c})}^2鈭�{\mathrm{k}}^2\right]{\mathrm{B}}_{\mathrm{z}}=0\end{array}$

In both cases the crucial thing was the cancellation of the mixed second derivatives.

Finally use the results of 9.180 with (i) and (iv) of 9.179 to obtain the same equations:

(i)

$\begin{array}{c}\frac{鈭�{\mathrm{E}}_{\mathrm{y}}}{鈭�\mathrm{x}}鈭�\frac{鈭�{\mathrm{E}}_{\mathrm{x}}}{鈭�\mathrm{y}}={\mathrm{i蝇B}}_{\mathrm{z}}\\ \frac{鈭�{\mathrm{E}}_{\mathrm{y}}}{鈭�\mathrm{x}}=\frac{\mathrm{i}}{{(\mathrm{蝇}/\mathrm{c})}^2鈭�{\mathrm{k}}^2}\left(\mathrm{k}\frac{{鈭�}^2{\mathrm{E}}_{\mathrm{z}}}{鈭�\mathrm{x}鈭�\mathrm{y}}鈭�\mathrm{蝇}\frac{{鈭�}^2{\mathrm{B}}_{\mathrm{z}}}{鈭�{\mathrm{x}}^2}\right)\\ \frac{鈭�{\mathrm{E}}_{\mathrm{y}}}{鈭�\mathrm{x}}=\frac{\mathrm{i}}{{(\mathrm{蝇}/\mathrm{c})}^2鈭�{\mathrm{k}}^2}\left(\mathrm{k}\frac{{鈭�}^2{\mathrm{E}}_{\mathrm{z}}}{鈭�\mathrm{x}鈭�\mathrm{y}}+\mathrm{蝇}\frac{{鈭�}^2{\mathrm{B}}_{\mathrm{z}}}{鈭�{\mathrm{x}}^2}\right)\\ \frac{鈭�{\mathrm{E}}_{\mathrm{y}}}{鈭�\mathrm{x}}=\frac{\mathrm{i}}{{(\mathrm{蝇}/\mathrm{c})}^2鈭�{\mathrm{k}}^2}\mathrm{蝇}\left(\frac{{鈭�}^2{\mathrm{B}}_{\mathrm{z}}}{鈭�{\mathrm{x}}^2}+\frac{{鈭�}^2{\mathrm{B}}_{\mathrm{z}}}{鈭�{\mathrm{y}}^2}\right)鈭�{\mathrm{i蝇B}}_{\mathrm{z}}=0\text{鈥�}/鈰�{(\mathrm{蝇}/\mathrm{c})}^2鈭�{\mathrm{k}}^2\end{array}$

Therefore, the value of use the results of 9.180) with (i) to obtain the same equations is $\left[\frac{{鈭�}^2}{鈭�{\mathrm{x}}^2}+\frac{{鈭�}^2}{鈭�{\mathrm{y}}^2}+{(\mathrm{蝇}/\mathrm{c})}^2鈭�{\mathrm{k}}^2\right]{\mathrm{B}}_2=0$.

(iv)

$\begin{array}{c}\frac{鈭�{\mathrm{B}}_{\mathrm{y}}}{鈭�\mathrm{x}}鈭�\frac{鈭�{\mathrm{B}}_{\mathrm{x}}}{鈭�\mathrm{y}}=鈭�\frac{\mathrm{i蝇}}{{\mathrm{c}}^2}{\mathrm{E}}_{\mathrm{z}}\\ \frac{鈭�{\mathrm{B}}_{\mathrm{y}}}{鈭�\mathrm{x}}=\frac{\mathrm{i}}{{(\mathrm{蝇}/\mathrm{c})}^2鈭�{\mathrm{k}}^2}\left(\mathrm{k}\frac{{鈭�}^2{\mathrm{B}}_{\mathrm{z}}}{鈭�\mathrm{x}鈭�\mathrm{y}}+\frac{\mathrm{蝇}}{{\mathrm{c}}^2}\frac{{鈭�}^2{\mathrm{E}}_{\mathrm{z}}}{鈭�{\mathrm{x}}^2}\right)\\ \frac{鈭�{\mathrm{B}}_{\mathrm{x}}}{鈭�\mathrm{y}}=\frac{\mathrm{i}}{{(\mathrm{蝇}/\mathrm{c})}^2鈭�{\mathrm{k}}^2}\left(\mathrm{k}\frac{{鈭�}^2{\mathrm{B}}_{\mathrm{z}}}{鈭�\mathrm{x}鈭�\mathrm{y}}鈭�\frac{\mathrm{蝇}}{{\mathrm{c}}^2}\frac{{鈭�}^2{\mathrm{E}}_{\mathrm{z}}}{鈭�{\mathrm{x}}^2}\right)\\ \frac{\mathrm{i}}{{(\mathrm{蝇}/\mathrm{c})}^2鈭�{\mathrm{k}}^2}\frac{\mathrm{蝇}}{{\mathrm{c}}^2}\left(\frac{{鈭�}^2{\mathrm{E}}_{\mathrm{z}}}{鈭�{\mathrm{x}}^2}+\frac{{鈭�}^2{\mathrm{E}}_{\mathrm{z}}}{鈭�{\mathrm{y}}^2}\right)+\mathrm{i}\frac{\mathrm{蝇}}{{\mathrm{c}}^2}{\mathrm{E}}_{\mathrm{z}}\text{鈥�}/鈰�{(\mathrm{蝇}/\mathrm{c})}^2鈭�{\mathrm{k}}^2\end{array}$

Therefore, the value of use the results of 9.180) with (iv) to obtain the same equations is $\left[\frac{{鈭�}^2}{鈭�{\mathrm{x}}^2}+\frac{{鈭�}^2}{鈭�{\mathrm{y}}^2}+{(\mathrm{蝇}/\mathrm{c})}^2鈭�{\mathrm{k}}^2\right]{\mathrm{E}}_2=0$.