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?
$\endgroup$ 97 Answers
$\begingroup$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 $k=0,1,2,3,4.$
$\endgroup$ $\begingroup$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 $\theta$.
$\endgroup$ 2 $\begingroup$Hint: express them in polar form.
$\endgroup$ $\begingroup$Hint: First, find a real one. Then use polar form.
$\endgroup$ $\begingroup$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!]
$\endgroup$ 2 $\begingroup$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: $32=32\cdot 1$.
$\endgroup$ $\begingroup$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.
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