The problem is specific as an example from hw. But it is more the concept/process I could use clarification on. Given a complex number


Find all possible roots. I know the method is to change into the exponential form, solving for magnitude (r) and theta. Which I did and got $e^{i4\pi/3}$. Do I then multiply theta by $4$, or $1/4$ and then add $\pi/2$ ? By either multiplying or dividing I get the same 4 roots, only with different starting points. But, if all I want are the roots, does it matter what the starting point is?


In general, we can write


where the complex logarithm is the multi-valued function and can be written as


for all integer values of $n$.

For the specific problem in the OP, $z=\frac{-2}{1-i\sqrt 3}=\frac{2e^{-i\pi}}{2e^{-i\pi/3}}=e^{-i2\pi/3}$ and $c=1/4$. Therefore, we have

$$\begin{align} \left(\frac{-2}{1-i\sqrt 3}\right)^{1/4}&=e^{\frac14\log(1)+i\frac14(-2\pi/3+2n\pi)}\\\\ &e^{i(-\pi/6+n\pi/2)}\\\\ &=(i)^ne^{-i\pi/6}\\\\ &= \begin{cases} ie^{-i\pi/6}&,n=1\\\\ -1e^{-i\pi/6}&,n=2\\\\ -ie^{-i\pi/6}&,n=3\\\\ e^{-i\pi/6}&,n=4 \end{cases} \end{align}$$

  • $\begingroup$ maybe it's just a different way of looking at the same thing or I am missing something. However, once we find the angle and our magnitude, would we not want to keep it on the unit circle by just adding pi/2 4 consecutive times. Your method seems to keep the same angle, but change the magnitude r. $\endgroup$ – NotSoSN Feb 5 '16 at 4:11
  • 1
    $\begingroup$ Actually, the magnitude never changed. Since the complex logarithm is multi-valued, and has imaginary part equal to $\arg(z)+2n\pi$, it is only the "angle" (i.e., $\arg(z)+2n\pi$) that changes. $\endgroup$ – Mark Viola Feb 5 '16 at 4:25
  • $\begingroup$ Ok, to make sure I'm tracking 100%. For the final 4 angles, while theta stays at -pi/6, the axis from which that angle is being taken is changing. Such that, your n=1 is the same as subtracting the angle pi/6 from the imaginary axis and is at pi/3 (from the real axis)? $\endgroup$ – NotSoSN Feb 5 '16 at 4:34
  • $\begingroup$ Yes! Well done. For $n=1$ we have $ie^{-i\pi/6}=e^{i(\pi/2-\pi/6)}=e^{i\pi/3}$ $\endgroup$ – Mark Viola Feb 5 '16 at 5:13
  • $\begingroup$ One last question, to circle back to my op. For your soln we take c=1/4, which makes since as that is the nth of the given imaginary number. However, if we were to have c=4, we would end up with the same roots, only the order would change (n=1 becomes n=3, etc). So, if we have the same roots regardless of using c=1/n or c=n, does it matter which we use? $\endgroup$ – NotSoSN Feb 5 '16 at 5:27

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