How do you multiply # (-1-7i)(-3-4i) # in trigonometric form?

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Answer 1 · 2018-02-23T12:30:41

Thus,
#(-1-7i)(-3-4i)=10sqrt2cis(7pi)/4#

#"Let,"z_1=-1-7i,#

#Re(z_1)=-1," "Im(z_1)=-7#

#r_1=sqrt((-1)^2+(-7)^2) = sqrt50=5sqrt2#

#theta_1=tan^-1((-7)/(-1))=pi+tan^-1(7)#

#"Let," z_2=-3-4i#

#Re(z_2)=-3," "Im(z_2)=-4#

#r_2=sqrt((-3)^2+(-4)^2) = sqrt25=5#

#theta_2=tan^-1((-4)/(-3))=pi+tan^-1(4/3)#

#z_1=r_1cistheta_1#
#z_2=r_2cistheta_2#

#z_1z_2=(r_1cistheta_1)(r_2cistheta_2)#

#z_1z_2=r_1r_2cistheta_1cistheta_2#
By De-Moivre's theorem

#cistheta_1cistheta_2=cis(theta_1+theta_2)#

Thus,

#z_1z_2=r_1r_2cis(theta_1+theta_2)#

Substituting,

#r_1r_2=5sqrt2xx5=10sqrt2#
#theta_1+theta_2=pi+tan^-1(7)+pi+tan^-1(4/3)#
#=2pi+tan^-1(7)+tan^-1(4/3)#

#tan^-1(7)+tan^-1(4/3)=tan^-1((7+4/3)/(1-7xx4/3))#

#(7+4/3)/(1-7xx4/3)=(7xx3+4)/(3-7xx4)=(21+4)/(3-28)=25/-25=1/-1#
#tan^-1(7)+tan^-1(4/3)=tan^-1(1/-1)=2pi-tan^-1(1)#

#tan^-1(1)=pi/4#

#2pi-tan^-1(1)=2pi-pi/4=(7pi)/4#

#r_1cistheta_1r_2cistheta_2=10sqrt2cis(7pi)/4#

Thus,
#(-1-7i)(-3-4i)=10sqrt2cis(7pi)/4#