Knock Knock! How many?

This is, by definition, the number of primitive $k$ th roots of unity for $k\leq 10$ , which is $\sum_{k=1}^{10}\phi(k)=32$ .

We know that the solutions to $x^n = 1$ are $x = e^{i \cdot 2\pi \cdot \frac{k}{n}}$ , for $k = 0,1,\ldots,n-1$ . However, when $k,n$ are not relatively prime, we can simplify the fraction to give $x = e^{i \cdot 2\pi \cdot \frac{k'}{n'}}$ , which is now also a solution for $x^{n'} = 1$ . Thus, in order to avoid double counting such solutions, we may only consider solutions where the numerator is relatively prime to the denominator, to make sure we count each solution only once. For a fixed $n$ , by definition there are $\phi(n)$ such numbers $k$ , and thus also $\phi(n)$ solutions. Thus, over all $n = 1,2,\ldots,10$ , the number of solutions is given by $\displaystyle\sum_{n=1}^{10} \phi(n) = \boxed{32}$ .

The complete set of solutions is given as $x = exp(j 2\pi \frac{p}{q}), q = 1..10, p=0..q-1$

However, many of the roots are repeated (Eg : p=1,q=2 is the same as p=2,q=4).

Hence, there would be only 32 distinct values of $\frac{p}{q}$ as $0$ , $\frac{1}{10}$ , $\frac{1}{2}$ , $\frac{1}{3}$ , $\frac{1}{4}$ , $\frac{1}{5}$ , $\frac{1}{6}$ , $\frac{1}{7}$ , $\frac{1}{8}$ , $\frac{1}{9}$ , $\frac{2}{3}$ , $\frac{2}{5}$ , $\frac{2}{7}$ , $\frac{2}{9}$ , $\frac{3}{10}$ , $\frac{3}{4}$ , $\frac{3}{5}$ , $\frac{3}{7}$ , $\frac{3}{8}$ , $\frac{4}{5}$ , $\frac{4}{7}$ , $\frac{4}{9}$ , $\frac{5}{6}$ , $\frac{5}{7}$ , $\frac{5}{8}$ , $\frac{5}{9}$ , $\frac{6}{7}$ , $\frac{7}{10}$ , $\frac{7}{8}$ , $\frac{7}{9}$ , $\frac{8}{9}$ , $\frac{9}{10}$

Added new from 1 to 10:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11

1   1                                   
1   -1                                  
2   -0.5+0.866025403784439i -0.5-0.866025403784439i                             
2   i   -i                              
4   0.309016994374947+0.951056516295154i    -0.809016994374948+0.587785252292473i   -0.809016994374947-0.587785252292474i   0.309016994374949-0.951056516295154i                        
2   0.5+0.866025403784439i  0.5-0.866025403784438i                              
6   0.623489801858734+0.78183148246803i -0.222520933956314+0.974927912181825i   -0.90096886790242+0.43388373911756i -0.900968867902422-0.433883739117558i   -0.222520933956317-0.974927912181826i   0.623489801858734-0.781831482468034i                
4   0.707106781186548+0.707106781186547i    -0.707106781186546+0.707106781186549i   -0.70710678118655-0.707106781186545i    0.707106781186544-0.707106781186551i                        
6   0.766044443118978+0.642787609686539i    0.173648177666931+0.984807753012208i    -0.939692620785908+0.34202014332567i    -0.939692620785909-0.342020143325667i   0.173648177666927-0.984807753012208i    0.766044443118975-0.642787609686541i                
4   0.809016994374947+0.587785252292473i    -0.309016994374947+0.951056516295152i   -0.309016994374946-0.951056516295151i   0.809016994374945-0.58778525229247i                     
32                                      

It is $\cos \frac{2 k \pi}{n} + i \sin \frac{2 k \pi}{n}$ that lists them all.

Answer: $\boxed{32}$

my first answer was 30, as I forgot that real numbers are complex too. then included 1,-1 in final counting.

When i=10 we get " 10 " points on the unit circle. Each one at the given fraction of the circumference. The fractions are 1/10, 1/5, 3/10, 2/5, 1/2, 3/5, 7/10, 4/5, 9/10, 1.......For i=9, we get 9 points, but 1 is same as for i=10, so it is " 8 " new points........Similarly, for 8, 1/2 and 1 are same from total 8 points. So new" 6 " ....For 7, only 1 is same so we get " 6 " more points. ...... For i=6, only 1/6 and 5/6 are not covered. So only " 2 " new points. All point for i=5, 4, 3, 2, are covered by i = 10, 8, 9, 10. So the total is 10+8+6+6+2=32.