Hard Root of Complex Number (1 Viewer)

[mezza333]

Hi i was wondering if someone could find the sqaure roots of (12-6i)
I usually use Terry Lee's quick method, but this is a bit harder.
Thanks for the help

 

[Shadowdude]

And Terry Lee's method is...?

 

[mezza333]

It's in his book on page 31. He equates the terms really quickly but it generally works for roots that are fairly clean. i think this one is a little more complex :O have any idea on the answer?

 

[Sup3rDry]

Hmm idk his method is weird. Do you know how to do it the algebraic way?

 

[slyhunter]

Which method? He has 2.

 

[mezza333]

yea i can do it algebraically but i just want a worked solution to see if it's right

 

[mezza333]

Which method? He has 2.

Method 2 but i dont think that works for this question

 

[slyhunter]

Then use Method 1? I've always used that method, never 2.

 

[mezza333]

can u post a latex solution to it please?

 

[Sup3rDry]

He factorises it somehow :O

 

[slyhunter]

I'm getting a very complicated solution, can you post what you've done seeing as how you've done it algebraically?

 

[tohriffic]

Derp wrong question. Soz bro.

 

[slyhunter]

12-6i isn't as easy it seems

 

[tohriffic]

12-6i isn't as easy it seems

It's too complex!

 

[mezza333]

I'm getting a very complicated solution,

I'm getting a really complicated answer as well. i think it will just be a messy result

 

[slyhunter]

Will post my solution soon

 

[slyhunter]

Let where are real.

upon squaring both sides.

Equate real parts,

Equate imaginary parts,

From

Substituting that into









But so,





The square roots of are

 

[marksgm]

nub

 

[mezza333]

Thanks mate you're a boss

 

[lolcakes52]

Isn't it just plus or minus the square root of the modulus times cis(theta) where theta is half the argument? That seems like a much more simplistic way considering the argument is so clean.