How Do the Roots of Complex Numbers Vary in Distribution?

[phospho]

say we had a complex number $w^4$ such that $w^4 = -8 +i8\sqrt{3}$ so $w = 2(cos(\frac{\pi}{6} + \frac{k\pi}{2}) + isin(\frac{\pi}{6} + \frac{k\pi}{2}))$ where k is an integer

in a question I was asked to find the roots of w, as there will be 4 my first assumption is that the roots would be spread throughout the argand diagram, i.e the first root would be $\sqrt{3} + i$ and the second would be $-\sqrt{3} + i$ and so on till I get the 4 roots. However this is not the case, as if I substitute values for k, I get the roots to be $\sqrt{3} + i, -1 + \sqrt{3}i, 1 - \sqrt{3}i, -\sqrt{3} - i$ which is actually correct.

However in the question below, the roots do seem to be spread evenly:


Prove that $cos\frac{\pi}{12} = m$ and $sin\frac{\pi}{12} = n,$ where $m = \frac{\sqrt{3} + 1}{2\sqrt{2}}$ and $n = \frac{\sqrt{3} -1}{2\sqrt{2}}$
Hence find in terms of m and n, in the form a + ib, where a,b are real, the fourth roots of $4(cos(\frac{\pi}{3}) + isin(\frac{\pi}{3}))$

the first root is $z = \sqrt{2}m + i\sqrt{2}n$ second $-\sqrt{2}m + i\sqrt{2}n$ and so on...

why is it in one question the 4 roots are not spread evenly and I have to adjust k to find the roots, while in the other they are spread evenly. How do I spot whether or not they will be spread evenly?

thanks,

 

[tiny-tim]

hi phospho!

the fourth (or nth) roots of any number w⁴ (or wⁿ)will be w times any fourth (or nth) root of 1

ie w, iw, i²w (= -w), and i³w (= -iw)

(so this is wrong …)

Hence find in terms of m and n, in the form a + ib, where a,b are real, the fourth roots of $4(cos(\frac{\pi}{3}) + isin(\frac{\pi}{3}))$

the first root is $z = \sqrt{2}m + i\sqrt{2}n$ second $-\sqrt{2}m + i\sqrt{2}n$

 

[phospho]

hi phospho!

the fourth (or nth) roots of any number w⁴ (or wⁿ)will be w times any fourth (or nth) root of 1

ie w, iw, i²w (= -w), and i³w (= -iw)

(so this is wrong …)

I have the answers in front of me, and they have the same roots that I gave above.

 

[Millennial]

Why don't you simply halve cosine twice? To be more precise, $\cos(\pi/3)=1/2$ and we have the half-angle formula $\cos^2(x)=\frac{\cos(2x)+1}{2}$. Applying it twice should give you $\cos(\pi/12)$, without the need of complex numbers at all.