0
$\begingroup$

I have attempted to solve it by converting to polar form first using $z=r\cdot e^{(i\cdot\theta)}$ and $i=e^{i\cdot(pi/2 \cdot ( 2\cdot n \cdot pi))}$.

Then I was taught to compare like terms independently for the magnitude r and angle $\theta$.

The issue I'm having is not knowing how to handle the "$1 -$ " term in front of i when I am matching terms. The similar example problem we solved in class was $z^4=i$ was much easier to solve since $i$ was by itself so comparing $r$ and the $\theta$ terms was simple. Any advice is much appreciated!

$\endgroup$
3
  • $\begingroup$ Just for fun (and as a trig-less alternative), note that squaring gives $\,z^6=(1-i)^2$ $=-2i$ $=\big(i \sqrt[3]{2}\big)^3\,$. $\endgroup$
    – dxiv
    Commented Feb 14, 2018 at 6:56
  • $\begingroup$ I initially used that approach and came up with answers for my three roots but they were not matching Wolfram Alpha so I asked my professor via e-mail if my approach was incorrect or if I might have made an error and he stated "Your approach does not work. $z^3=1-i$ and $z^6=-2i$ are not equivalent but two different problems. " $\endgroup$
    – J. Paredes
    Commented Feb 14, 2018 at 12:57
  • $\begingroup$ Squaring does indeed introduce new roots, so at the end you need to doublecheck which $3$ of the $6$ roots do in fact satisfy the original equation. Solving it this way, however, has the advantage of requiring nothing more than square roots. $\endgroup$
    – dxiv
    Commented Feb 14, 2018 at 18:01

2 Answers 2

Reset to default
0
$\begingroup$

Find the norm and the angle to write out the polar form of $1-i$.

$$1-i =\sqrt{2} \exp((-\frac{\pi}{4} + 2n \pi)i) $$

Now, the problem reduces to $$z^3=\sqrt{2} \exp((-\frac{\pi}{4} + 2n \pi)i)$$

$\endgroup$
-1
$\begingroup$

You are correct in your first step of converting $z^3=1-i$ to polar form.

$r= \sqrt{1^2+(-1)^2} = \sqrt{2}$, and $\theta= \arctan(\frac{-1}{1})=\frac{7}{4}\pi$

Thus, $z^3 = \sqrt{2} \cdot e^{i(\frac{7}{4}\pi)}$.

But remember that you can use de Moivre's Theorem in that form while taking the cube root of $(1-i)$ is just the cube root of the radius times and the exponent (for simplicity since I don't know LaTeX) is just:

$i[(\frac{7}{4} + 2n)\frac{\pi}{3}]=i[\frac{7}{12}\pi + 2n\frac{\pi}{3}]$, where $n=0,1,2$.

Thus, for this problem, we get three answers:

1) $z = 2^{\frac{1}{6}} \cdot e^{i(\frac{7}{12}\pi)}$;

2) $z = 2^{\frac{1}{6}} \cdot e^{i(\frac{7}{12}\pi + \frac{2}{3}\pi)} = 2^{\frac{1}{6}} \cdot e^{i(\frac{5}{4}\pi)}$;

3) $z = 2^{\frac{1}{6}} \cdot e^{i(\frac{7}{12}\pi + \frac{4}{3}\pi)} = 2^{\frac{1}{6}} \cdot e^{i(\frac{23}{12}\pi)}$.

$\endgroup$
4
  • $\begingroup$ Please, use MathJax (i.e. LaTeX commands) for mathematical notations. $\endgroup$
    – Taroccoesbrocco
    Commented Feb 14, 2018 at 7:12
  • $\begingroup$ A minima, one should also compute $\cos(\pi/12)$ and $\sin(\pi/12)$... $\endgroup$
    – Did
    Commented Feb 14, 2018 at 8:21
  • $\begingroup$ Thank you! These three roots match Wolfram Alpha's results. So my biggest mistake was not converting $1-i$ into polar form correctly since I was just converting the i into polar form but not the 1. Our professor never showed any examples where he used the two formulas to calculate r and $\theta$ like you did, he simply used one example where we just had i so he converted it to polar by drawing it on a polar plot. So you did not end up using Moivre's Theorem correct? You simply converted to polar then took the cube root of both sides while keeping in mind that we needed the $2n \pi$ term. $\endgroup$
    – J. Paredes
    Commented Feb 14, 2018 at 13:05
  • $\begingroup$ To whoever formatted my answer thank you. I will try to learn MathJax over the next few days. $\endgroup$
    – Kaleb Jen
    Commented Feb 14, 2018 at 15:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .