I am attempting to study for an exam and encountered a problem involving an electromagnetic plane wave with a provided electric field and propagation constant, but despite setting up the solution correctly I am unable to understand how to simplify the complex propagation constant with the provided information.
Question and Solution :
An electromagnetic plane wave with an angular frequency of $\omega=4.518 \times 10^{10} \mathrm{rad} / \mathrm{s}$ propagates in the $+z$ direction through a conducting medium. The medium is characterized by $\sigma=0.2 \mathrm{~S} / \mathrm{m}, \varepsilon=\varepsilon_{0}=8.8534 \times 10^{-12} \mathrm{~F} / \mathrm{m}$,and $\mu=\mu_{0}=4 \pi \times 10^{-7} \mathrm{H} / \mathrm{m}$. The associated electric field is represented in phasor form as: $$ \vec{E}=E_{x 0} e^{-\gamma z} \hat{x} $$ where $\gamma$ is the propagation constant given by: $$ \gamma=j \omega \sqrt{\mu_{0} \varepsilon_{0}} \sqrt{1-\frac{j \sigma}{\omega \varepsilon_{0}}} $$ The value of $z$ (in meters) for which the magnitude of the electric field is $0.01 E_{x 0}$ is most nearly:
A. $\quad 0.00273$
B. $\quad 0.0289$
C. $\quad 0.0297$
D. $\quad 0.1259$
Using the equations in the item statement: $$ \begin{array}{l} \frac{1 \angle 90^{\circ}}{2.998 \times 10^{8}} \times 4.518 \times 10^{10} \sqrt{1-j 0.5}=159.3 \angle 76.72^{\circ}=36.59+j 155 \\ e^{-36.59 z}=0.01 \\ z=\frac{\ln (0.01)}{-36.59}=0.1259 \end{array} $$
Essentially my issue comes with the first and second intermediary steps - I understand substituting the three terms with the provided values and simplifying the third term to $$(1-0.5j)^{0.5}$$, but what I do not understand is why the solution has a phasor in the first step instead of a complex j term, and why the $$2.998\cdot10^8$$ term is in the denominator of the first term.
Additionally, I do not understand why this mixed phasor + complex number is able to be represented in an equivalent step as a phasor in the next step, unless the provided solution is skipping a ton of information?
From there I understand the conversion to a complex number from a phasor, then setting up the desired field strength to the electric field strength with the calculated propagation constant, then simplifying for the magnitude of the distance from the field to obtain the desired strength - I am mainly just getting caught up in how the simplification is made as indicated in the solution on the first line.
I think it's just easier to join an angle to a magnitude than multiply by $j$, maybe easier to calculate with. (For the unitiated reader: engineers and physicists often mark the imaginary unit with a $j$ because of $i$'s resemblance of the symbol of alternating current.)
$$\sqrt{\mu_0\varepsilon_0}=\sqrt{8.8534\cdot4\pi\cdot10^{-19}}\approx 3.3354925513\cdot10^{−9} \Rightarrow$$ $$\dfrac{1}{\sqrt{\mu_0\varepsilon_0}}=\dfrac{1}{3.3354925513\cdot10^{−9}}=299805796.182\approx2.998\cdot10^8$$
Thus $$\sqrt{\mu_0\varepsilon_0}\approx\frac{1}{2.998\cdot10^{8}}$$
Both of them are numbers, you can check the physical quantities with dimensional analysis, but both things at the two side of the equals sign represent the same numerical value (apart from the approximations) they are just written differently. Yes, steps were skipped but you can change complex numbers into many forms.
Three main forms i know of, the algebraic form: $$z=a+bi$$ The polar form:$$z=r\cdot e^{i\varphi}$$ where $\varphi$ is the degree of the complex number's vector, in electrical engineering subjects often marked $$z=r\angle\varphi$$ and the trigonometric form: $$r(cos\varphi+i\sin\varphi)$$ They are described formally in this article.
How to switch between algebraic and phase form: $$z=a+bi$$
$$r=\sqrt{a^2+b^2}\hspace{5mm}\varphi=\arctan\left(\frac{b}{a}\right)$$
The trigonometric form uses the same parameters as the phasor one.
Complex division because this is the one people usually struggle with:
$$z=\dfrac{a+bi}{c+di}=\dfrac{a+bi}{c+di}\cdot1=\dfrac{a+bi}{c+di}\cdot\dfrac{c-di}{c-di}=\dfrac{(a+bi)(c-di)}{(c+di)(c-di)}=\dfrac{ac+i(bc-ad)-bdi^2}{c^2-d^2i^2}=$$
$$z=\dfrac{ac+i(bc-ad)+bd}{c^2+d^2}=\color{blue}{\dfrac{ac+bd}{c^2+d^2}}+\color{red}{\dfrac{bc-ad}{c^2+d^2}}i$$