Answer: [tex]\frac{2}{9}\left(\cos\left(\frac{\pi}{132}\right)+i*\sin\left(\frac{\pi}{132}\right)\right)[/tex]
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The answer written out in keyboard form is (2/9)*(cos(pi/132)+i*sin(pi/132))
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Work Shown:
z1 = a*(cos(b) + i*sin(b))
z1 = 2*(cos(pi/11) + i*sin(pi/11))
here we see that a = 2 and b = pi/11
z2 = c*(cos(d) + i*sin(d))
z2 = 9*(cos(pi/12) + i*sin(pi/12))
here we see that c = 9 and d = pi/12
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Now use the rule
If
z = a*(cos(b) + i*sin(b)) and w = c*(cos(d)+i*sin(d))
then
z/w = (a/c)*(cos(b-d)+i*sin(b-d))
We have
a = 2
b = pi/11
c = 9
d = pi/12
So...
z/w = (a/c)*(cos(b-d)+i*sin(b-d))
(z1)/(z2) = (a/c)*(cos(b-d)+i*sin(b-d))
(z1)/(z2) = (2/9)*(cos(pi/11-pi/12)+i*sin(pi/11-pi/12))
(z1)/(z2) = (2/9)*(cos(12pi/132-11pi/132)+i*sin(12pi/132-11pi/132))
(z1)/(z2) = (2/9)*(cos(pi/132)+i*sin(pi/132))
which is the answer in polar form
