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9231/01/M/J/05
10(ii) (z + 2)^12 = z^12
[(z + 2)^12]/z^12 = 1
[(z + 2)/z]^12 = 1
[1 + 2/z]^12 = 1
Now, using the answer to part (i), we get
1 + 2/z = exp[(k*pi*i)/6]
2/z = exp[(k*pi*i)/6] - 1
z = 2/[exp{(k*pi*i)/6} - 1]
Multiplying the fraction by the conjugate of the denominator [i.e. multiplying by exp{-(k*pi*i)/6} - 1] in order to rationalize the denominator gives ----->
z = [2*exp{-(k*pi*i)/6} - 2]/[[1 - exp[(k*pi*i)/6] - exp[-(k*pi*i)/6] + 1]
Now using "exp(i*x) = cos(x) + i*sin(x)" and "exp(i*x) + exp(-ix) = 2cos(x)" we get
z = [2[cos{-(k*pi)/6} + i*sin{-(k*pi)/6}] - 2]/[2 - 2cos{(k*pi)/6}]
z = 2[cos{-(k*pi)/6} + i*sin{-(k*pi)/6} - 1]/2[1 - cos{(k*pi)/6}]
z = [cos{(k*pi)/6} - i*sin{(k*pi)/6} - 1]/[1 - cos{(k*pi)/6}]
OMG m HALLUCINATING ri8 now...... :shock: Anyways, moving on
Now we may use the identity sin(x)/[1 - cos(x)] = cot(x/2) to simplify the last expression which is of the form
z = [cos(x) - i*sin(x) - 1]/[1 - cos(x)] with x = (k*pi)/6
z = [-(1 - cos(x) - i*sin(x)]/[1 - cos(x)]
z = -1 - i*cot(x/2)
z = -1 - i*cot[(k*pi)/12] for k = 0, +/- 1, +/- 2, +/- 3, +/- 4, +/- 5, +6
But at k = +6, the value of the root is equal is to -1(real); while the root is undefined at k = 0
Therefore, k = +/- 1, +/- 2, +/- 3, +/- 4, +/- 5 i.e. 10 complex(non-real) roots which is consistent with the fact that each individual complex root has its conjugate also as another root of the given equation(since the given equation has only real-coefficients in it) i.e. there are 5 distinct pairs of complex roots. Also, if u look closely @ it, the given equation is actually an 11th degree equation although it seems to be a 12th degree one. So, there can't be 11 complex roots since then one complex root won't have its conjugate also as another root. If the equation was given instead as a 12th degree equation then there mi8 possibly be all 12 roots being complex....!!!
Lemme know if u find any difficulty in the following parts [(ii) and (iii)]
Hop this helps
9231/01/M/J/05
10(ii) (z + 2)^12 = z^12
[(z + 2)^12]/z^12 = 1
[(z + 2)/z]^12 = 1
[1 + 2/z]^12 = 1
Now, using the answer to part (i), we get
1 + 2/z = exp[(k*pi*i)/6]
2/z = exp[(k*pi*i)/6] - 1
z = 2/[exp{(k*pi*i)/6} - 1]
Multiplying the fraction by the conjugate of the denominator [i.e. multiplying by exp{-(k*pi*i)/6} - 1] in order to rationalize the denominator gives ----->
z = [2*exp{-(k*pi*i)/6} - 2]/[[1 - exp[(k*pi*i)/6] - exp[-(k*pi*i)/6] + 1]
Now using "exp(i*x) = cos(x) + i*sin(x)" and "exp(i*x) + exp(-ix) = 2cos(x)" we get
z = [2[cos{-(k*pi)/6} + i*sin{-(k*pi)/6}] - 2]/[2 - 2cos{(k*pi)/6}]
z = 2[cos{-(k*pi)/6} + i*sin{-(k*pi)/6} - 1]/2[1 - cos{(k*pi)/6}]
z = [cos{(k*pi)/6} - i*sin{(k*pi)/6} - 1]/[1 - cos{(k*pi)/6}]
OMG m HALLUCINATING ri8 now...... :shock: Anyways, moving on
Now we may use the identity sin(x)/[1 - cos(x)] = cot(x/2) to simplify the last expression which is of the form
z = [cos(x) - i*sin(x) - 1]/[1 - cos(x)] with x = (k*pi)/6
z = [-(1 - cos(x) - i*sin(x)]/[1 - cos(x)]
z = -1 - i*cot(x/2)
z = -1 - i*cot[(k*pi)/12] for k = 0, +/- 1, +/- 2, +/- 3, +/- 4, +/- 5, +6
But at k = +6, the value of the root is equal is to -1(real); while the root is undefined at k = 0
Therefore, k = +/- 1, +/- 2, +/- 3, +/- 4, +/- 5 i.e. 10 complex(non-real) roots which is consistent with the fact that each individual complex root has its conjugate also as another root of the given equation(since the given equation has only real-coefficients in it) i.e. there are 5 distinct pairs of complex roots. Also, if u look closely @ it, the given equation is actually an 11th degree equation although it seems to be a 12th degree one. So, there can't be 11 complex roots since then one complex root won't have its conjugate also as another root. If the equation was given instead as a 12th degree equation then there mi8 possibly be all 12 roots being complex....!!!
Lemme know if u find any difficulty in the following parts [(ii) and (iii)]
Hop this helps