Complex Numbers Help (1 Viewer)

someth1ng · Oct 21, 2011

I've tried this question...

$-2\sqrt{3}-2i$

Is one root this:
$\frac{\sqrt{6}-\sqrt{2}}{2}+\frac{-\sqrt{6}-\sqrt{2}}{2}i$

If not, can someone give me a run through?

nightweaver066 · Oct 21, 2011

Pretty long question..
The square roots are:
$\pm\sqrt{2-\sqrt{3}} \mp \frac{1}{\sqrt{2-\sqrt{3}}}i$

I'm not going to run through all the steps as it is troublesome.

From letting $-2\sqrt{3} -2i = (x + iy)^2$ and after equating coefficients, you end up with:
$x^2 -y^2 = -2\sqrt{3}$
$2xy = -2$

Putting these two together by eliminating y, you end up with:
$x^4 + 2\sqrt{3}x^2 -1 = 0$

Applying the quadratic formula, you simplify to this:
$x^2 = -\sqrt{3}\pm2$

As you need to square root this again, it must not be < 0, so you ignore $x^2 = -\sqrt{3}-2$

$x = \pm\sqrt{2-\sqrt{3}}$

So $y = \mp\frac{1}{\sqrt{2-\sqrt{3}}}$

D94 · Oct 22, 2011

Why not leave it in/answer in mod-arg form?

$\\z^{2} = -2\sqrt{3} - 2i \\z^{2} = 4cis(\frac{7\pi}{6} + 2\pi n) \\z = \left (4cis(\frac{7\pi}{6} + 2\pi n) \right )^{\frac{1}{2}} \\z = 2cis\left (\frac{\frac{7\pi}{6} + 2\pi n}{2} \right ) \\z_{0} = 2cis\left (\frac{7\pi}{12} \right ) \\ \\z_{1} = 2cis\left (\frac{19\pi}{12} \right ) = 2cis\left (\frac{-5\pi}{12} \right ) \\ \\\therefore\ Two\ square\ roots\ are\ 2cis\left (\frac{7\pi}{12} \right ) \textsc{and}\ 2cis\left (\frac{-5\pi}{12} \right )$

someth1ng · Oct 22, 2011

That seemed difficult - I just started this roots stuff...

I got the 2cis(-5pi/12) one - getting the other one now.

EDIT: Got the 2cis (7pi/12)...I was getting it right after all...LOL

D94 · Oct 22, 2011

someth1ng said:
That seemed difficult - I just started this roots stuff...

I got the 2cis(-5pi/12) one - getting the other one now.

EDIT: Got the 2cis (7pi/12)...I was getting it right after all...LOL

Nice. Patel gives a decent explanation of complex roots (and complex numbers in general) so just follow the book for at least complex numbers.

someth1ng · Oct 22, 2011

D94 said:
Nice. Patel gives a decent explanation of complex roots (and complex numbers in general) so just follow the book for at least complex numbers.

I haven't actually got the textbook yet - just the digital version - they didn't give me the answers to that specific exercise...I'm trying to keep ahead of everything.
So you're allowed to leave in mod-arg form?

D94 · Oct 22, 2011

someth1ng said:
I haven't actually got the textbook yet - just the digital version - they didn't give me the answers to that specific exercise...I'm trying to keep ahead of everything.
So you're allowed to leave in mod-arg form?

Oh right, makes sense.

Depends on the question. Sometimes it requires mod-arg, sometimes it requires x+iy. Because that question was a bit obscure, the other method of finding the roots probably wouldn't have suited it. (ie. equate real and imaginary parts to find x and y). But from what I've seen in the past papers, they'll ask for a mod-arg form, and then you need to evaluate the complex number raised to a high power, eg. 10, and then simplify that into x+iy.

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Complex Numbers Help (1 Viewer)

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