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I want to get some perspective on complex exponentiation and the best possible approach to solve them.

$ z^{3/4}=-1 ;z \in \mathbb{C}$

Let $z=x+iy=\rho e^{i \theta} $ where $\rho =|z|$ and tan$\theta=\frac xy$

$w=-1=(1).e^{i(\pi+2k\pi)} $

According to me, there is a simple way to prove there is a solution to this equation. I may very well be wrong, but here I go.

$$ \frac{3\theta}{4}=\pi+2k\pi $$

$$ \theta =\frac{4\pi}{3}+\frac{8k\pi}{3}$$ Which gives me 3 unique solutions for $\theta$: $\frac{4\pi}{3},\frac{2\pi}{3},0$

I'd then like to check that they verify $|z|=1$, I can disqualify $\theta=0$ in a jiffy. I then check: $$tan\frac{4\pi}{3}=\frac{y}{x}$$ $$x tan\frac{4\pi}{3}=y$$ So from $|z|=1$: $$ \sqrt{(x tan\frac{4\pi}{3})^2 +x^2}=1$$ $$x=\frac{+}{-}\frac{1}{ \sqrt{ (tan^2\frac{4\pi}{3} +1)} } $$

So nothing seems out of place to me so far... Wolfram Alpha insists there are no solutions and my calculator doesn't make the cut for complex numbers.

FemtoComm
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    You can't disqualify $\theta = 0$ in a jiffy, one of the four possible values of $1^{3/4}$ is $-1$. As far as Wolfram Alpha is concerned, have you asked it about $(-1)^{4/3}$? – Daniel Fischer Dec 10 '16 at 21:45
  • What is going on is that there is no way to define a ower function $z^{3/4}$ on $\mathbb C$ without falling in this kind of trap. To be more specific, $z^3$ is all right but $z^{1/4}$ is not at all. – Did Dec 10 '16 at 21:45
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    For example... through computations quite similar to the ones in your question, one could be led to write absurd things such as $e^{i0}=e^{i2\pi}$ (completely true) hence $(e^{i0})^{3/4}=(e^{i2\pi})^{3/4}$ (too bad, from this point on, your soul is lost...) hence (I guess) $e^{i0\cdot 3/4}=e^{i2\pi\cdot3/4}$, that is, $1=e^{i0}=e^{i3\pi/2}=-i$. :-) – Did Dec 10 '16 at 21:48
  • Hmm I see. I tried to avoid this pitfall by writing out the $2k\pi$, Wouldn't this account for periodicity? If not, how do I find solutions? Would the wisest thing to do be to proceed as Daniel has said? Find $z=(-1)^{4/3}$... Sounds like it even may be simpler. – FemtoComm Dec 10 '16 at 21:51
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    You have found the solutions $z=e^{\frac{4}{3}\pi(1+2k)}$, for $k \in \mathbb{Z}$. I don't see the solution $\theta=0$ here. – Emilio Novati Dec 10 '16 at 22:08
  • @EmilioNovati Alright, so writing it out like that does seem easier. I can only assume what you mean when you say you don't see that solution is that you'd expect to see it (since it satisfies $z=e^{\frac{4}{3}\pi (1+2k)}$ ). Thank you for your help! – FemtoComm Dec 10 '16 at 22:19

2 Answers2

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Let $z=r(\cos(\theta)+i\sin(\theta))$, by De-Moivre's theorem, we have: $z^3=r^3(\cos(3\theta)+i\sin(3\theta))$ with $0\leq\theta\leq2\pi$. Now we have to compute the value of: $$\left(z^3\right)^{1/4}=\left(r^3(\cos(3\theta)+i\sin(3\theta)\right)^{1/4}$$ with $0\leq\phi_k\leq2\pi$. Using again De-Moivre' theorem, we have: $$w^4=z^3\leftrightarrow R^4(cos(4\phi)+i\sin(4\phi))=r^3(\cos(3\theta)+i\sin(3\theta))$$ From here, we deduce that: $$\phi_k=\frac{1}{4}(3\theta+2k\pi), k=0,1,2,3$$ and $R^4=r^3$. Now we can impose $w=-1$ and obtain: $$(r^3)^{1/4}\left(\cos\left(\frac{3\theta+2k\pi}{4}\right)+i\sin\left(\frac{3\theta+2k\pi}{4}\right)\right)=1(\cos(\pi)+i\sin(\pi))$$

This leads to: $$3\theta=4\pi+8k_1\pi-2k\pi,k_1=0,1,2$$

The solutions are: $$\theta=\pi/3 \: \vee \theta=5\pi/3 \: \vee \theta=\pi$$

In conclusion, I think that the solutions to this equations are the all $z$ such that $w=z^3=1$ because in that case the equation $w^{1/4}=-1$ has four solutions that in Gauss plane are the vertex of a square.

Matteo
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It's usually better to direct less attention to the unknown and use Euler's equation on them, or in this case Euler's identity.

$z=(-1)^{4\over3}=\displaystyle e^{(i\pi) \cdot \frac{4}{3}}$

Basically the exponent is $\dfrac{4i\pi}{3}$

And this is a point on the unit circle in the complex plane, so the rectangular coordinates are $\Re=\cos \frac{4\pi}{3}, \Im=i\sin \frac{4\pi}{3}$

The number in rectangular coordinates is thus $-0.5-\frac{\sqrt{3}}{2}i$

Edit: You could also consider the period of $8\pi\over3$ but the denominator in $z^{3\over4}$ makes this more complicated. If you do consider the period, only one of the fourth roots of $z$ will be $-1$ due to the roots of unity thing.

Nεo Pλατo
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