As we know, we define $$\sqrt{-1}=i$$ But I always wondered, what about $\sqrt{i}$? As far as i can see, it is not an integer power of $i$. Every odd root has a solution in an integer power of $i$, but I can't find one for my question. Is there something I am missing here?
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Oh, I didn't (saw)see that question. I will also vote to close for duplicate. – chubakueno Aug 06 '13 at 04:17
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1@chubakueno "...didn't see..." – Pedro Aug 06 '13 at 04:19
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1As a spanish native, I hope you understand me, @PeterTamaroff :) – chubakueno Aug 06 '13 at 04:19
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Of course, but one should always strive for correctness. – Pedro Aug 06 '13 at 04:23
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Consider $\frac{1+i}{\sqrt{2}}$. The square of this is $i$. It is a fact that the complex numbers are algebraically closed, and so in particular, every number has a square root.
Alexander
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1"Recall" is a strong term - it implies a person might have forgotten something, when, given the question, it is quite possible that the OP hasn't encountered the notion of "algebraically closed." – Thomas Andrews Aug 06 '13 at 04:18
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That is a fair point. I accidentally find myself using "recall" all the time when I shouldn't! – Alexander Aug 06 '13 at 04:20
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Complex numbers have a nice geometric interpretation -- the complex number $a + bi$ is visualized as a point in a plane with coordinates $(a,b)$. When you multiply two complex numbers, geometrically you just "add the angles and multiply the lengths". With this geometric interpretation you can easily find a square root of $i$. $\frac{1 + i}{\sqrt 2}$ works.
littleO
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