Finding Complex Roots: Poles of $ \frac{1}{{z^4}+4} $, {z: |z-1| LE 2}

[ognik]

I think I'm a bit rusty here, started with finding poles for $ \frac{1}{{z^4}+4} $, {z: |z-1| LE 2}

1) Out of interest, is there a complex equivalent of the rational roots test? The above function is obvious, but for a poly that has both real and complex roots?

2) I am using the exponential form to find roots, ie for n roots, $ {z}_{k+1} = r^{-n}e^{i(\frac{\theta}{n} + k\frac{2\pi}{n})}, k=0,\pm1,\pm2...\pm(n-1) $ - is this formula correct? Is there a better way?

3) Using the above, $ (\theta = 0) $, I get the 4 poles to be $ {z}_{k+1} = \sqrt{2}e^{i( k\frac{2\pi}{n})} = \pm\sqrt{2}, \pm\sqrt{2}i $, but the answer given is $ \pm1 \pm i $, can anyone see what I'm doing wrong?

 

[Opalg]

If $z^4+4 = 0$ then $z^4 = -4 = 4e^{i\pi}$. The fourth roots are given by $z = \sqrt2e^{i(2k+1)\pi/4}$.

 

[Fernando Revilla]

Also, it can be useful the exersise 5 b) here.

 

[ognik]

Thanks both.

Fernando, the useful exercises you linked to, uses basically the same formula I had (although in CIS form) - except for the starting value, where they use pi - please tell me why pi and not 0?

 

[Fernando Revilla]

Fernando, the useful exercises you linked to, uses basically the same formula I had (although in CIS form) - except for the starting value, where they use pi - please tell me why pi and not 0?

Take into accont that $4=4(\cos 0+i\sin 0)$ and $-4=4(\cos \pi +i\sin \pi).$

 

[ognik]

Of course, I should have thought of that myself, thanks Fernando.

(May I ask if you wouldn't mind having a look at an earlier post please - http://mathhelpboards.com/analysis-50/find-coefficiant-laurent-series-without-using-residue-15997.html)