Question 29843: how do you find the nth term of a negative complex number?
for example the fourth roots of -4
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! how do you find the nth term of a negative complex number?
I HOPE YOU MEAN N th. ROOT NOT AND NOT N th. TERM..OF A COMPLEX NUMBER ..CAN BE POSITIVE OR NEGATIVE...
LET THE COMPLEX NUMBER BE Z = A+Bi IN GENERAL WHICH CAN ALWAYS BE WRITTEN AS
Z = R*(COS X + i SIN X ) WHERE R = SQRT(A^2+B^2) AND COS X = A/(SQRT(A^2+B^2)
AND SIN X = B/(SQRT(A^2+B^2)
SINCE N th.ROOT OF R IS EASY TO FIND (R^1/N),R BEING POSITIVE, WE NEED TO ONLY FIND Nth.ROOT OF SAY Z = COS X +i SIN X
Z^1/N = (COS X +i SIN X)^1/N...AS PER DEMOVIERS THEOREM THIS IS EQUAL TO
= COS (X/N) +i SIN (X/N)..SO NOW WE USE THE GENERAL FORMULA FOR TRIGNOMETRIC EQNS ..THAT IS ..IF COS X = P SAY THEN X = COS ^-1(P)..ITS GENERAL SOLUTION IS
X=2K*(PI)+COS^-1(P)...GIVE K=1,2,3 ETC.. TO GET ALL THE DIFFERENT ROOTS.
for example the fourth roots of -4
Z = -4 =4*(-1)...SINCE 4 TH. ROOT OF 4 CAN BE EASILY FOUND OUT .LET US FIND FOURTH ROOTS OF -1.WE CAN MULTIPLY THAT ANSWER WITH 4TH ROOT OF +4 TO GET THE TOTAL ANSWER.
Z = COS (0) +i SIN (0)..
= COS (2K*PI+0)+iSIN (2K*PI+0)=COS (2K*PI)+i SIN (2K*PI)
4TH.ROOTS OF THIS ARE AS PER ABOVE FORMULA...
Z^1/4 =COS (2K*PI)/4+i SIN (2K*PI)/4...PUTTING K=1,2,3 AND 4 TO GET THE 4 ROOTS WE GET...
COS (2*1*PI)/4+i SIN (2*1*PI)/4...,
COS (2*2*PI)/4+i SIN (2*2*PI)/4...,
COS (2*3*PI)/4+i SIN (2*3*PI)/4...,
COS (2*4*PI)/4+i SIN (2*4*PI)/4...AS THE 4 ROOTS.
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