Square root of complex number

I'm at a loss how to go about solving equations like the one below, where there are j terms under the square root sign.

http://i43.tinypic.com/fx84yr.png

I've tried getting the complex conjugate of what's under the square root sign but i'll still have a j term.

I'm convinced it's a straightforward process and that I'm missing some simple trick.

Clues/Hints/Ideas/Whatever welcome!

Re: Square root of complex number

Hello, piglet!

Quote:

I'm at a loss how to go about solving equations like the one below.

$\displaystyle \eta \,=\,\sqrt{\frac{j\omega\mu}{\sigma + j\omega\epsilon}} \,=\,\sqrt{\frac{j\!\cdot\!2\pi(100 \times 10^6)(100)(4\pi \times 10^{-7})}{0.00964 + j\!\cdot\!2\pi(100 \times 10^6)(8.854 \times 10^{-12})}} \,=\,2664e^{j30}\Omega$

Did you do any of the arithmetic?

The numerator is: .$\displaystyle j\!\cdot\!2\pi (10^8)(10^2)(4\pi\times 10^{-7}) \;=\;8\pi^2j \times 10^3$

The denominator is: .$\displaystyle (9,64\times 10^{-3}) + 2\pi j(10^2\cdot 10^6)(8.854\times 10^{-12})$

. . . . . . . . . . . . . $\displaystyle =\;(9.64 \times 10^{-3}) + 2\pi j (8.854 \times 10^{-4}) $

. . . . . . . . . . . . . $\displaystyle =\;(2\cdot10^{-4})(48.2 + 4.427\pi j $

The fraction becomes: .$\displaystyle \frac{8\pi^2j \times 10^3} {(2\cdot 10^{-4})(48.2 + 4.427\pi j)} $

. . . . . . . . . . . . . . . $\displaystyle =\;\frac{4\pi^2j \times 10^7}{48.2 + 4.427\pi j} $

By the way, what are we **solving** for?

Re: Square root of complex number

Quote:

Originally Posted by

**Soroban** Hello, piglet!

Did you do any of the arithmetic?

The numerator is: .$\displaystyle j\!\cdot\!2\pi (10^8)(10^2)(4\pi\times 10^{-7}) \;=\;8\pi^2j \times 10^3$

The denominator is: .$\displaystyle (9,64\times 10^{-3}) + 2\pi j(10^2\cdot 10^6)(8.854\times 10^{-12})$

. . . . . . . . . . . . . $\displaystyle =\;(9.64 \times 10^{-3}) + 2\pi j (8.854 \times 10^{-4}) $

. . . . . . . . . . . . . $\displaystyle =\;(2\cdot10^{-4})(48.2 + 4.427\pi j $

The fraction becomes: .$\displaystyle \frac{8\pi^2j \times 10^3} {(2\cdot 10^{-4})(48.2 + 4.427\pi j)} $

. . . . . . . . . . . . . . . $\displaystyle =\;\frac{4\pi^2j \times 10^7}{48.2 + 4.427\pi j} $

By the way, what are we **solving** for?

Yes I got it down that far but wondering how to solve that fraction when it's under a square root and then convert it to an polar exponential like the picture in my first post. Any ideas?

By the way, we're solving for Eta, the intrinsic impedance of a medium that has a magnetic field passing through it.

Re: Square root of complex number

Quote:

Originally Posted by

**piglet** Yes I got it down that far but wondering how to solve that fraction when it's under a square root and then convert it to an polar exponential like the picture in my first post. Any ideas?

By the way, we're solving for Eta, the intrinsic impedance of a medium that has a magnetic field passing through it.

You multiply the top and bottom of the fraction by the complex conjugate of the denominator, which reduces the fraction to something of the form $\displaystyle A+Bj$

Now convert that into polar form and the square root is easy-peasea from there.

$\displaystyle \sqrt{A+Bj}=\sqrt{\rho e^{i(\theta+2n\pi)}}=\sqrt{\rho}\ e^{i(\theta/2+n\pi)}};\ n=0, 1$

CB