Solve $$z^\sqrt5 =1$$ for $z$ and state how many unique solutions are possible.
I tried to convert $1$ to polar form and got $z=\exp^\left((2k\pi+2\pi)i/\sqrt5\right)$. Could someone please help me out?
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archangel89
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https://math.stackexchange.com/questions/2027183/de-moivre-theorem-for-irrational-exponents – lab bhattacharjee Nov 15 '17 at 05:35
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And nobody is mentioning that there is no such thing as a canonically defined function $$z\mapsto z^a$$ defined on the complex plane, when $a$ is not an integer, and even worse, when $a$ is not a rational? Well... Sorry but one must ask: what is the source for this exercise? – Did Nov 15 '17 at 07:59
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Let $z=re^{i\theta}$.
Since $z^\sqrt5=r^\sqrt5e^{i\sqrt5\theta}=1=1e^{\left(0+2k\pi\right)i}$ for $k \in \mathbb{Z}$, $r=1$.
Now we have $e^{i\sqrt5\theta}=e^{2k\pi i}$, which is exactly where you have reached.
Then we have $\sqrt5\theta=2k\pi$
So according to you, you are using $0 \leq \theta \leq 2\pi$. We know from earlier that $\theta=\frac{2k\pi}{\sqrt5}$.
So here is the list of the values of $\frac{2k}{\sqrt5}$ corresponding to some $k$'s.
\begin{align} &k=-1, &&\frac{2k}{\sqrt5}=-0.894427191 \\ &k=0, &&\frac{2k}{\sqrt5}=0 \\ &k=1, &&\frac{2k}{\sqrt5}=0.894427191 \\ &k=2, &&\frac{2k}{\sqrt5}=1.788854382 \\ &k=3, &&\frac{2k}{\sqrt5}=2.683281573 \\ \end{align}
Which values of them are within the range now?
Karn Watcharasupat
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Actually I thought K is an integer, so there is no chance of values repeating between 0 and 2pi.....so is it infinite unique solutions? – archangel89 Nov 15 '17 at 05:43
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Between 0 and 2pi.....actually comes under the chapter of finding the roots of unity – archangel89 Nov 15 '17 at 05:52
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Alright. You can take a look at the list above. Can you figure out the answer now? – Karn Watcharasupat Nov 15 '17 at 05:53
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Can you please look into the edit ....I made a mistake initially.. I think the question should be clear now – archangel89 Nov 15 '17 at 05:55
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What happens when the angle crosses 2pi...since there is no repetition..wil it be possible to get infinite solutions with increasing k? – archangel89 Nov 15 '17 at 05:57
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Think of a cirlcle, once it has go beyond $2\pi$ it just come back and overlap. – Karn Watcharasupat Nov 15 '17 at 05:58
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That will happen if k is an integer and instead of sqrt(5) we have some integer – archangel89 Nov 15 '17 at 06:01
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$e^xy=(e^x)^y$ doesn't really hold for complex numbers outside principle range. That's why the rest of the values don't work. – Karn Watcharasupat Nov 15 '17 at 06:12
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