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This was the last question on my Year 11 Complex Numbers/Matrices Exam

Name all 5 possible values for $z$ in the equation $z^5=32$

I could only figure out $2$. How would I go about figuring this on paper?

VikeStep
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  • Which values were you able to name? – Mark Bennet Sep 11 '13 at 07:33
  • I could only find 1 value which was $2$ – VikeStep Sep 11 '13 at 07:33
  • Bam bam bam. Those were some fast answers... – dfeuer Sep 11 '13 at 07:34
  • What all the folks with answers are hinting about is: have you learned DeMoivre's Theorem for roots of complex numbers? – colormegone Sep 11 '13 at 07:34
  • the only demoivres theorem I know of is $(r \times cis\theta)^n$=$(r^n \times cis(n \times \theta))$ – VikeStep Sep 11 '13 at 07:37
  • I would think of it in terms of $32 = (re^{i\theta})^5 = r^5e^{5i\theta}$. – dfeuer Sep 11 '13 at 07:39
  • all of these solutions are using $(r{e}^{i\theta})^5$. where are yous getting this from because it isn't familiar to me – VikeStep Sep 11 '13 at 07:41
  • Are you familiar with the polar form of a complex number ? The modulus of the polar form is exactly the "$r$" i the above expression, and the argument is exactly the "$\theta$" in the above expression. It's just a more convenient expression of the polar form. If you haven't seen it, just use it, and take it for granted, and i'll bet you will see it in class very soon! – New_to_this Sep 11 '13 at 07:47
  • ok thanks, and yes we have been doing polar form, but only in the form $r*cis\theta$. I'll just take it for granted – VikeStep Sep 11 '13 at 08:43

7 Answers7

2

We have $z^5=2^5$ or $$\left(\frac{z}{2}\right)^5=1$$ then we have the solutions are $$z=2*\large{(e^\frac{2ki\pi}{5})}$$ where $k=0,1,2,3,4.$

Mikasa
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Shobhit
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Try $z_k = 2 e^{i \frac{2 k\pi}{5}}$.

To see where this comes from write $(r e^{i \theta} )^5 = r^5 e^{i 5\theta} = 32$. Taking absolute values gives $r = 2$, then you need to solve $e^{i 5\theta} = 1$. This requires $5 \theta = 2 k\pi$, and there are only 5 distinct solutions (modulo $2 \pi$) for $\theta$.

copper.hat
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  • This is the best explanation I guess, as someone mentioned below: 0,1,2,3,4 are the values for $k$, so that means that $\theta=0,\frac{2\pi}{5},\frac{4\pi}{5},\frac{6\pi}{5},\frac{8\pi}{5}$. Is that correct? – VikeStep Sep 11 '13 at 08:50
  • @VikeStep: Yes, that is correct. (Well, you can pick any 5 'non-equivalent' values, but it could be customary to pick the ones you mentioned, or possible the values corresponding to $0, \pm 1, \pm 2$.) – copper.hat Sep 11 '13 at 14:59
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Hint: express them in polar form.

Robert Israel
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Hint: First, find a real one. Then use polar form.

dfeuer
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Hint: $32 = 2^5$, so the values of $z$ will be of the form $2w$ where $w^5 = 1$. So ask yourself: what are the complex 5th roots of 1? [And that's standard bookwork!]

Peter Smith
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  • We haven't been taught the complex 5th roots of 1. Are there 4 because I already found one? – VikeStep Sep 11 '13 at 07:39
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    Then you'd better teach yourself about the complex $n$-th roots of 1 -- the question evidently presumes you know this absolutely basic stuff! – Peter Smith Sep 11 '13 at 08:15
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You have to consider this:

$$ \begin{align*} z^{5}=r^{5}e^{i\theta\cdot 5} = 32 \end{align*} $$

So in other words, you have to convert 32 to exponential form, and solve for $r$ and $\theta$. Hint: $32=32\cdot 1$.

New_to_this
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Here is a cheat answer (really) which uses a different method, but is instructive to know about. Let $x=\frac z2$ so that we have $0=x^5-1=(x-1)(x^4+x^3+x^2+1)$ and aim to factorise the second factor into quadratics of the form $x^2+px+1$, $x^2+qx+1$.

We write $a=\frac {p+q}2, b=\frac {p-q}2$ so our factorisation becomes $$(x^2+(a+b)x+1)(x^2+(a-b)x+1)=x^4+2ax^3+(2+a^2-b^2)x^2+2ax+1$$

So we need $2a=1$ and $b^2=\frac 54$. Our factors become $x^2+\frac {1\pm \sqrt 5}2x+1$ which leave us solving $$z^2+(1\pm \sqrt 5)z+4=0$$

And these two quadratic equations can be solved for $z$ in the usual way.

This is a cheat, really, because it involves knowing what to do - though it is possible to negotiate this method via intelligent guesswork. It involves pairing the complex conjugate solutions to the original equation to find the real quadratic factors which are known to exist. I noted it because it relates to constructing a regular pentagon, which can be done by solving quadratic equations in this way and hence by ruler and compasses.

Mark Bennet
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