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

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