C cormglakes New Member Joined Jan 4, 2021 Messages 18 Gender Undisclosed HSC 2021 Mar 19, 2021 #1 let z=cosx + isinx solve z^4 - 4z^3 + 2z^2 - 4z+1 = 0
Q Qeru Well-Known Member Joined Dec 30, 2020 Messages 364 Gender Male HSC 2021 Mar 19, 2021 #2 cormglakes said: let z=cosx + isinx solve z^4 - 4z^3 + 2z^2 - 4z+1 = 0 Click to expand... But So the problem becomes: which is a 3U problem (remember to take only the first 4 unique solutions). OR notice So the problem becomes: Which is simply a quadratic in
cormglakes said: let z=cosx + isinx solve z^4 - 4z^3 + 2z^2 - 4z+1 = 0 Click to expand... But So the problem becomes: which is a 3U problem (remember to take only the first 4 unique solutions). OR notice So the problem becomes: Which is simply a quadratic in
C cormglakes New Member Joined Jan 4, 2021 Messages 18 Gender Undisclosed HSC 2021 Mar 19, 2021 #3 Qeru said: But So the problem becomes: which is a 3U problem (remember to take only the first 4 unique solutions). Click to expand... and then after I get cosx=0 --> z= i & -i and cosx=2 --> how do I find the other 2 roots?
Qeru said: But So the problem becomes: which is a 3U problem (remember to take only the first 4 unique solutions). Click to expand... and then after I get cosx=0 --> z= i & -i and cosx=2 --> how do I find the other 2 roots?
C CM_Tutor Moderator Moderator Joined Mar 11, 2004 Messages 2,730 Gender Male HSC N/A Mar 19, 2021 #4 Qeru said: Click to expand... At this point, notice we have a quadratic in ? So... and the equation has two real roots, , and two purely imaginary roots
Qeru said: Click to expand... At this point, notice we have a quadratic in ? So... and the equation has two real roots, , and two purely imaginary roots
C CM_Tutor Moderator Moderator Joined Mar 11, 2004 Messages 2,730 Gender Male HSC N/A Mar 19, 2021 #5 cormglakes said: and then after I get cosx=0 --> z= i & -i and cosx=2 --> how do I find the other 2 roots? Click to expand... In taking there is an assumption made that all solutions in will have modulus of 1 and satisfy . This is not true for the two real solutions so you run into a problem.
cormglakes said: and then after I get cosx=0 --> z= i & -i and cosx=2 --> how do I find the other 2 roots? Click to expand... In taking there is an assumption made that all solutions in will have modulus of 1 and satisfy . This is not true for the two real solutions so you run into a problem.