2

I want to rewrite $(1-i)^ \frac 1 3$ using de Moivre's formula.

I defined $z := 1 - i$, then $r_z = \sqrt{2}$ and

$1 = \sqrt2\cos\theta$ and $-1 = \sqrt2 \sin\theta \Rightarrow \theta = -\frac \pi 4$

So: $$1 - i = \sqrt{2}(\cos(\frac \pi 4) + i\sin(\frac \pi 4))$$ $$(1-i)^\frac 1 3 = 2^ \frac 1 6 (\cos(\frac \pi {12} + i\sin(\frac \pi {12}))$$

Am I correct with this derivation?

Lucian
  • 1,785
  • 6
  • 12

3 Answers3

1

You have only take the principal root, indeed we have (there is also a typo for the angle in the $\sin$ term):

$$1 - i = \sqrt{2}\left(\cos\left(\frac \pi 4+2n\pi\right) + i\sin\left(-\frac \pi 4+2n\pi\right)\right)$$

and therefore the three possible roots are

$$2^ \frac 1 6 \left(\cos\left(\frac \pi {12}\right) + i\sin\left(-\frac \pi {12}\right)\right)$$ $$2^ \frac 1 6 \left(\cos\left(\frac \pi {12}+\frac 2 3 n\right) + i\sin\left(-\frac \pi {12}+\frac 2 3 n\right)\right)$$ $$2^ \frac 1 6 \left(\cos\left(\frac \pi {12}+\frac 4 3 n\right) + i\sin\left(-\frac \pi {12}+\frac 4 3 n\right)\right)$$

with $n=0,1,2$.

user
  • 154,566
1

$$(1-i)^{1/3}=2^{1/6}e^{-i\pi/12} e^{2in\pi/3},n=0,1,2$$ So you have got the root for $n=0$ only, there are two more for $n=1,2$.

This is like $(1)^{1/3}=(e^{2in\pi/3}), n=0,1,2.$ These roots are well known as $1,\omega,\omega^2$, respectively.

Z Ahmed
  • 43,235
1
  • Your final step is invalid (discarding solutions!) because it misapplies De Moivre's theorem, which does not generally hold for non-integer powers like $\frac13.$ So, for example, \begin{align}1^\frac12&=\left(\cos(2\pi)+i\sin(2\pi)\right)^{\frac12}\\&\neq\cos(\frac12\times2\pi)+i\sin(\frac12\times2\pi)\\&=-1.\end{align} This is because in the complex world, raising a number to a non-integer power generally outputs multiple values.
  • In your exercise, the goal is to determine $(1-i)^\frac13,$ i.e., the third roots of unity of $1-i.$ This process (abbreviating $(\cos x+i\sin x)$ as $\text{cis }x)$ invokes not De Moivre's theorem but the multi-valued definition of $e^z:$ \begin{align}(1-i)^\frac13&=\left[\sqrt2 \text{ cis} \left(-\frac\pi4\right)\right]^\frac13\\&=2^\frac16\text{ cis} \left(\frac{-\frac\pi4+2k\pi}3\right)\\&=2^\frac16\text{ cis} \left(\frac{8k-1}{12}\pi\right)\\&=2^\frac16\text{ cis} \left(\frac{-\pi}{12}\right),\,2^\frac16\text{ cis} \left(\frac{7\pi}{12}\right)\text{ or }\,2^\frac16\text{ cis} \left(\frac{5\pi}{4}\right).\end{align}
ryang
  • 38,879
  • 14
  • 81
  • 179