Find the solutions of the following equations in complex?

#Z^6 + (7+i)Z^3 -8+8i =0#

1 Answer
George C.
Nov 28, 2017

See explanation...

Explanation:

First note that:

#sin(pi/12) = sin(pi/3 - pi/4)#

#color(white)(sin(pi/12)) = sin(pi/3)cos(pi/4)-sin(pi/4)cos(pi/3)#

#color(white)(sin(pi/12)) = sqrt(3)/2 sqrt(2)/2-sqrt(2)/2 1/2#

#color(white)(sin(pi/12)) = 1/4(sqrt(6)-sqrt(2))#

#cos(pi/12) = cos(pi/3 - pi/4)#

#color(white)(cos(pi/12)) = cos(pi/3)cos(pi/4)+sin(pi/3)sin(pi/4)#

#color(white)(cos(pi/12)) = 1/2 sqrt(2)/2+sqrt(3)/2 sqrt(2)/2#

#color(white)(cos(pi/12)) = 1/4 (sqrt(6)+sqrt(2))#

de Moivre's formula tells us that:

#(cos theta + i sin theta)^n = cos n theta + i sin n theta#

By extension, it tells us that the cube roots of:

#cos theta + i sin theta#

are:

#cos (theta/3) + i sin (theta/3)#

#cos ((theta+2pi)/3) + i sin ((theta+2pi)/3)#

#cos ((theta+4pi)/3) + i sin ((theta+4pi)/3)#

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^3+8 = 0# then:

#z = -2#

or:

#z = -2omega = 1-sqrt(3)i#

or:

#z = -2omega^2 = 1+sqrt(3)i#

where #omega = -1/2+sqrt(3)/2i# is the primitive complex cube root of #1#.

If #z^3-1+i# then:

#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)#

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