Let z be a non-zero complex number with ${\mathrm{z}}^{23}=1$ 

If $\sum _{\mathrm{k}=0}^{22}\frac{1}{1+{\mathrm{z}}^{\mathrm{k}}+{\mathrm{z}}^{2\mathrm{k}}}=\frac{\mathrm{m}}{\mathrm{n}}$, where gcd of (m, n) = 1 then m+n=______

Using the geometric series formula, rewrite the given sum as

$\sum _{\mathrm{k}=0}^{22}\frac{1}{1+{\mathrm{z}}^{\mathrm{k}}+{\mathrm{z}}^{2\mathrm{k}}}=\frac{1}{3}+\sum _{\mathrm{k}=1}^{22}\frac{1-{\mathrm{z}}^{\mathrm{k}}}{1-{\mathrm{z}}^{3\mathrm{k}}}=\frac{1}{3}+\sum _{\mathrm{k}=1}^{22}\frac{1-{\left({\mathrm{z}}^{24}\right)}^{\mathrm{k}}}{1-{\mathrm{z}}^{3\mathrm{k}}}$

where the last equality employs the fact that ${\mathrm{z}}^{23}=1$. Using the geometric series formula in reverse, we can rewrite this as

$\frac{1}{3}+\sum _{\mathrm{k}=1}^{22}\frac{1-{\left({\mathrm{z}}^{24}\right)}^{\mathrm{k}}}{1-{\mathrm{z}}^{3\mathrm{k}}}=\frac{1}{3}+\sum _{\mathrm{k}=1}^{22}\sum _{\mathrm{j}=0}^7{\mathrm{z}}^{3\mathrm{kj}}$

Note that for $\mathrm{j}=1,2,\dots ,7$, we can write using the geometric series formula

$\sum _{\mathrm{k}=1}^{22}{\mathrm{z}}^{3\mathrm{kj}}={\mathrm{z}}^{3\mathrm{j}}+{\mathrm{z}}^{6\mathrm{j}}+\cdots +{\mathrm{z}}^{66\mathrm{j}}=\frac{{\mathrm{z}}^{69\mathrm{j}}-{\mathrm{z}}^3}{{\mathrm{z}}^{3\mathrm{j}}-1}$

We can now evaluate the desired sum:

$\frac{1}{3}+\sum _{\mathrm{k}=1}^{22}\sum _{\mathrm{j}=0}^7{\mathrm{z}}^{3\mathrm{kj}}=\frac{1}{3}+22+.7·(-1)=\frac{46}{3}$