Find the solutions of the following equations in complex?
Z^6 + (7+i)Z^3 -8+8i =0Z6+(7+i)Z3−8+8i=0
1 Answer
See explanation...
Explanation:
First note that:
sin(pi/12) = sin(pi/3 - pi/4)sin(π12)=sin(π3−π4)
color(white)(sin(pi/12)) = sin(pi/3)cos(pi/4)-sin(pi/4)cos(pi/3)sin(π12)=sin(π3)cos(π4)−sin(π4)cos(π3)
color(white)(sin(pi/12)) = sqrt(3)/2 sqrt(2)/2-sqrt(2)/2 1/2sin(π12)=√32√22−√2212
color(white)(sin(pi/12)) = 1/4(sqrt(6)-sqrt(2))sin(π12)=14(√6−√2)
cos(pi/12) = cos(pi/3 - pi/4)cos(π12)=cos(π3−π4)
color(white)(cos(pi/12)) = cos(pi/3)cos(pi/4)+sin(pi/3)sin(pi/4)cos(π12)=cos(π3)cos(π4)+sin(π3)sin(π4)
color(white)(cos(pi/12)) = 1/2 sqrt(2)/2+sqrt(3)/2 sqrt(2)/2cos(π12)=12√22+√32√22
color(white)(cos(pi/12)) = 1/4 (sqrt(6)+sqrt(2))cos(π12)=14(√6+√2)
de Moivre's formula tells us that:
(cos theta + i sin theta)^n = cos n theta + i sin n theta(cosθ+isinθ)n=cosnθ+isinnθ
By extension, it tells us that the cube roots of:
cos theta + i sin thetacosθ+isinθ
are:
cos (theta/3) + i sin (theta/3)cos(θ3)+isin(θ3)
cos ((theta+2pi)/3) + i sin ((theta+2pi)/3)cos(θ+2π3)+isin(θ+2π3)
cos ((theta+4pi)/3) + i sin ((theta+4pi)/3)cos(θ+4π3)+isin(θ+4π3)
Next note that:
{ (7+i = 8+i-1), (-8+8i = 8(i-1)) :}
Hence:
z^6+(7+i)z^3-8+8i = (z^3+8)(z^3-1+i)
If
z = -2
or:
z = -2omega = 1-sqrt(3)i
or:
z = -2omega^2 = 1+sqrt(3)i
where
If
z^3 = 1-i = sqrt(2) e^(-pi/4 i)
So:
z = 2^(1/6) e^(-pi/12 i) = root(6)(2)/4 ((sqrt(6)+sqrt(2)) - (sqrt(6)-sqrt(2))i)
or:
z = 2^(1/6) e^(-(3pi)/4 i) = -2^(2/3)/2 (1+i)
or:
z = 2^(1/6) e^((7pi)/12 i) = root(6)(2)/4 (-(sqrt(6)-sqrt(2))+(sqrt(6)+sqrt(2))i)