How can I use De Moivre's theorem to show that #z=-2-2i# is a solution to #z^4-3z^3-38z^2-128z-144=0#?

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Answer 1 · 2018-04-30T17:11:35

Givn

#z=-2-2i#

#=>z=-2sqrt2(1/sqrt2+i/sqrt2)#
#=>z=-2sqrt2(cos(pi/4)+isin(pi/4))#

Now

#z^4=[-2sqrt2(cos(pi/4)+isin(pi/4))]^4#

#=>z^4=64(cos(pi/4*4)+isin(pi/4*4)=-64#

#z^3=[-2sqrt2(cos(pi/4)+isin(pi/4))]^3#

#=>z^3=-16sqrt2(cos(pi/4*3)+isin(pi/4*3))#

#=>z^3=-16sqrt2(cos(pi-pi/4)+isin(pi-pi/4))#

#=>z^3=-16sqrt2(-cos(pi/4)+isin(pi/4))#

#=>z^3=-16sqrt2(-1/sqrt2+i/sqrt2)#

#=>z^3=16(1-i)#

#z^2=[-2sqrt2(cos(pi/4)+isin(pi/4))]^2#

#=>z^2=8(cos(pi/4*2)+isin(pi/4*2))#

#=>z^2=8*0+i8=8i#

Now putting the values in the LHS we get

#z^4-3z^3-38z^2-128z-144#

#=-64-3*16(1-i)-38*8i-128(-2-2i)-144#

#=-64-48+48i-304i+256+256i-144#

#=-256-256i+256+256i=0#

So the given value of #z# satisfies the given equation of #z#.

Hence #z=-2-2i# is a solution of given equation.