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This problem would be less confusing to me if one was not subtracted from $z^6$.
Here is the problem:
If $z=cos(\frac{\pi}{3})+jsin(\frac{\pi}{3})$, show that $z$ is a root of $z^6-1$.
My approach to this problem involved using de Moivre's theorem to convert $z$ to it's complex polar form. I think converting it to complex polar form first makes it easy to identify the root:

$$z=cos(\frac{\pi}{3})+jsin(\frac{\pi}{3})$$ $$a=cos(\frac{\pi}{3}), b=sin(\frac{\pi}{3})$$ $$r=\sqrt{cos(\frac{\pi}{3})^2+sin(\frac{\pi}{3})^2}$$ $$r=1$$ $$\theta=Tan^1(\frac{b}{a})$$ $$\theta=Tan^-1(\frac{sin(\frac{\pi}{3})}{cos(\frac{\pi}{3})})$$ $$\theta=\frac{\pi}{6}$$ $$\therefore z=1\angle{\frac{\pi}{6}}$$


This part is the confusing bit. Because $1$ is subtracted from $z^6$, I decided to expand the whole thing and then show that $z$ is a root of $z^6-1$.
$$(cos(\frac{\pi}{3})+jsin(\frac{\pi}{3}))^6-1$$ $$(\frac{1}{2}+\frac{\sqrt{3}}{2}j)^6-1$$
Noting that $j^6=-1$...
$$(\frac{1}{2}-\frac{\sqrt{3}}{2})^6-1$$
The equation becomes... $$(-\frac{13}{32})-1$$
Which equates to... $$-\frac{45}{32}$$
Convert to complex polar form...
$$z^6=-\frac{45}{32}\angle{0rad}$$ $$z^6=(-\frac{45}{32}\angle{0rad})^6$$ $$z^6=(-\frac{45}{32})^6\angle{(0)}^6$$ $$z^6=(-\frac{45}{32})^6\angle{6(0+2\pi n)}$$ $$z^6=(-\frac{45}{32})^6\angle{(12\pi-11\pi n)}$$ $$z^6=(-\frac{45}{32})^6\angle{(\pi n)}$$
Noting that $z$ is raised to the first power, make $n=1$ $$z^6=(-\frac{45}{32})^6\angle{\pi (1)}$$
I get... $$z^6=(-\frac{45}{32})^6\angle{\pi}$$
Clearly the complex polar form I got is not equal to that of $z$. Can someone lead me to the right path?

uhoh
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AugieJavax98
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3 Answers3

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Alt. hint (not using the polar form): $\;z=\dfrac{1}{2}+ i\,\dfrac{\sqrt{3}}{2}\,$, so $\,|z|=1\,$, and:

$$z^2 = \left(\dfrac{1}{2}+ i\,\dfrac{\sqrt{3}}{2}\right)^2= \dfrac{1}{4}+ 2i \cdot \dfrac{1}{2}\cdot\dfrac{\sqrt{3}}{2} + i^2 \dfrac{3}{4}=-\dfrac{1}{2}+ i\,\dfrac{\sqrt{3}}{2} = - \overline{z}$$

Multiplying with $\,z\,$ gives$\,z^3 = - z \cdot \overline z = -|z|^2 = -1\,$.

dxiv
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  • I've been working mostly with the polar form to solve such problems. This gives me another way to attack such problems. Thanks! – AugieJavax98 May 07 '18 at 16:30
  • @AugieJavax98 In this case, using the polar form and De Moivre's formula is the natural way to go (as pointed out in other answers), but it's still good to be aware that there is more than one way to solve it. – dxiv May 07 '18 at 16:52
2

HINT

Note that according to JuliusL33t and rtybase suggestions by De Moivre's formula

  • $z=\cos(\frac{\pi}{3})+i\sin(\frac{\pi}{3})=e^{i\frac{\pi}{3}}\implies z^6=\cos(\frac{6\pi}{3})+i\sin(\frac{6\pi}{3})$

or also by exponential form

  • $z=\cos(\frac{\pi}{3})+i\sin(\frac{\pi}{3})=e^{i\frac{\pi}{3}}\implies z^6=e^{6\cdot i\frac{\pi}{3}}$
user
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2

Not really sure what our OP is trying to do here; what the intended method might be. But I'm pretty sure that

$\theta = \text{Tan}^{-1} \left ( \dfrac{\sin(\pi/3)}{\cos (\pi/3)} \right ) = \text{Tan}^{-1} ( \tan (\pi/3)) = \dfrac{\pi}{3} \ne \dfrac{\pi}{6}, \tag 1$

so it seems like the first wrong turn was made at this point. And I can't see the logical connection 'twixt

$\left ( \dfrac{1}{2} + j \dfrac{\sqrt 3}{2} \right)^6 \tag 2$

and

$\left (\dfrac{1}{2} - \dfrac{\sqrt3}{2} \right )^6; \tag 3$

so the OP's remarks from this point on are pretty much difficult for me to decipher.

Nevertheless, we do have de Moivre's formula,

$(\cos \theta + j\sin \theta)^n = \cos (n\theta) + j \sin(n\theta); \tag 4$

and if we set

$z = \cos \dfrac{\pi}{3} + j\sin \dfrac{\pi}{3}, \tag 5$

then

$z^6 = \left ( \cos \dfrac{\pi}{3} + j\sin \dfrac{\pi}{3} \right)^6 = \cos \left (6 \dfrac{\pi} 3 \right ) + j \sin \left (6 \dfrac{\pi}{3} \right) = \cos 2 \pi + j \sin 2\pi = \cos 2 \pi = 1, \tag 6$

whence

$z^6 - 1 = 0. \tag 7$

Robert Lewis
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