Wave impedance is give by the equation:
$$\eta = \sqrt{\frac{i~\omega \mu}{\sigma + i~\omega \varepsilon}}$$
where:
$$i = \sqrt{-1}$$
Now I want to make an approximation of this equation when $\sigma$ is much larger than $\omega\varepsilon$, that is:
$$\sigma >> \omega \varepsilon$$
The book says this should give:
$$\eta \approx \sqrt{\frac{\omega \mu}{\sigma}} \angle 45^\circ$$
I was just curious, how they came to this conclusion...
I could imagine that since $\sigma$ is much larger...it dominates in the denominator... so we can ignore the "$+ i~\omega \varepsilon$" term, yeilding:
$$\eta \approx \sqrt{\frac{i~\omega \mu}{\sigma}}$$
but there's still the "i" term to deal with...
unless i were to multiple the top and bottom by the complex conjugate, to yield:
$$\eta = \sqrt{\frac{i~\omega\mu(\sigma-i~\omega\varepsilon)}{\sigma^2 + i~(\omega\varepsilon)^2}}$$
$$\eta \approx \sqrt{\frac{i~\omega\mu\sigma}{\sigma^2}}$$
its just the same thing again...
Start from $\sqrt{\frac{\mathrm{i} \omega \mu}{\sigma}} = \sqrt{\frac{\omega \mu}{\sigma}} \sqrt{\mathrm{i}}$. $\sqrt{\mathrm{i}}$ is either $\angle 45^\circ$ or $\angle {-135^\circ}$. Since $\sigma + \mathrm{i} \omega \varepsilon$ has a small positive argument, and $\mathrm{i} \omega \mu$ has argument $\angle 90^\circ$, we know we should retain the positive argument.