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Like this?I'll give you some hints:
7) You can do the sub u=x^2 then solve the quadratic, or:
Multiply both sides by (z^2+1)
8)Similarly do a substitution or multiply both sides by (z^2-1)
9) We re-arrange:
Then roots of unity.
I multiplied both sides of the equation by (z^2+1), this is because:@Sy123, I'm not sure what you did for (z^2 +- 1). Care to explain?
The substitution method is very clunky, its better to just multiply both sides by (z^2+1)ThanksI scanned my solution for Q7, but not sure if I did the right thing
There are four answers?
View attachment 28762
I'm still working on Q9
Yea something like thatU did sillies a1079atw. Please check z^2= 1+....... and then u continued on with z = 1+....
Like this?
((z-1)/(z+1))^6=-1
(z-1)/(z+1)= [cis((2kpi+pi)/6)]
Now z-1= (z+1)(cis ((2pi+pi)/6))
z-1 = zcis ((2kpi+pi)/6))+ cis(cis ((2pi+pi)/6)
Since z is factorable
z(1- cis((2kpi+pi)/6)) -1 = cis ((2kpi+pi)/6)
Therefore z= (1+(cis ((2pi+pi)/6))) / (1- (cis ((2pi+pi)/6)))
Now we change to polar form:
z= (1+ cos ((2pi+pi)/6) + i sin ((2pi+pi)/6))/ (1- cos ((2pi+pi)/6) - i sin ((2pi+pi)/6))
Now using double angle
z= (2cos^2((2kpi+pi)/12) + 2icos ((2kpi+pi)/12) x sin ((2kpi+pi)/12)) / (2sin^2((2kpi+pi)/12) - 2i cos((2kpi+pi)/12) x sin((2kpi+pi)/12))
Factorising we get :
(2cos((2kpi+pi)/12) x (cos((2kpi+pi)/12)+ sin((2kpi+pi)/12))/ ( 2sin ((2kpi+pi)/12) x (sin((2kpi+pi)/12) - i cos((2kpi+pi)/12)))
Now we can take out the -i from the denominator and set in it the polar form x + iy
This leads to :
(cos((2kpi+pi)/12) x (cos((2kpi+pi)/12)+ sin((2kpi+pi)/12))/ ( -isin ((2kpi+pi)/12) x ( cos ((2kpi+pi)/12) + sin((2kpi+pi)/12))))
This leads to cot ((2kpi+pi)/12)/(-i )
= i cot((2kpi+pi)/12)
let k=0,+_ 1, +_2, -3
Now can find solutions.
Np.Thanks for the working of Q9, but I don't understand what's wrong with the z=1+...... part.
Don't you use this equation where z^n = a+bi is z^2= [cos(pi/3)+isin(pi/3)]?
View attachment 28767