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The cyclic group of $\mathbb{C}- \{ 0\}$ of complex numbers under multiplication generated by $(1+i)/\sqrt{2}$

I just wrote that this is $\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i$ making a polar angle of $\pi/4$. I am not sure what to do next. My book say there are 8 elements.

Working backwards, maybe the angle divides the the four quadrant into 8 areas? I have no idea

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it seems like all you have to do to verify that this is true is to compute all the powers of that up to 8. You should get the identity once you reach 8. The order of a cyclic group is the number of elements it has. – HowardRoark Oct 25 '12 at 5:59
up vote 4 down vote accepted

HINT Find the minimum $k$ such that $\left(\dfrac1{\sqrt{2}} + \dfrac{i}{\sqrt{2}} \right)^k = 1$. It will be of help to write $\dfrac1{\sqrt{2}} + \dfrac{i}{\sqrt{2}}$ as $\exp(i \pi/4)$.

The element in the group generated by $r = \dfrac1{\sqrt{2}} + \dfrac{i}{\sqrt{2}}$ will then be $$\{r^m : m = 0,1,2,\ldots,k-1\}$$

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Draw a sketch. Do you know how to multiply complex numbers in polar coordinates? Add the angles, multiply the lengths. Now compute the powers of your number until you reach 1.

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Firstly, $\mathbb{C}-\{0\}$ is not a cyclic group generated by that complex number (was that a typo?). The element $1/\sqrt{2}(1+i)=e^{\pi i/4}$ generates a cyclic subgroup of $\mathbb{C}-\{0\}$, so you need to find the smallest positive $k$ for which $(e^{\pi i/4})^k=e^{k\pi i/4}=1$ (since $1$ is the identity in your group). Why must $k$ be 8?

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