# Square root of complex number

• Nov 28th 2011, 09:56 AM
piglet
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!
• Nov 28th 2011, 11:15 AM
Soroban
Re: Square root of complex number
Hello, piglet!

Quote:

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

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

Did you do any of the arithmetic?

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

The denominator is: . $(9,64\times 10^{-3}) + 2\pi j(10^2\cdot 10^6)(8.854\times 10^{-12})$
. . . . . . . . . . . . . $=\;(9.64 \times 10^{-3}) + 2\pi j (8.854 \times 10^{-4})$
. . . . . . . . . . . . . $=\;(2\cdot10^{-4})(48.2 + 4.427\pi j$

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

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

By the way, what are we solving for?

• Nov 29th 2011, 01:04 AM
piglet
Re: Square root of complex number
Quote:

Originally Posted by Soroban
Hello, piglet!

Did you do any of the arithmetic?

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

The denominator is: . $(9,64\times 10^{-3}) + 2\pi j(10^2\cdot 10^6)(8.854\times 10^{-12})$
. . . . . . . . . . . . . $=\;(9.64 \times 10^{-3}) + 2\pi j (8.854 \times 10^{-4})$
. . . . . . . . . . . . . $=\;(2\cdot10^{-4})(48.2 + 4.427\pi j$

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

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

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.
• Nov 29th 2011, 01:14 AM
CaptainBlack
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 $A+Bj$

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

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

CB