How do you multiply $(- 1 + 8 i) (- 9 + 7 i)$ $(-1+8i)(-9+7i)$ in trigonometric form?

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How do you multiply $(- 1 + 8 i) (- 9 + 7 i)$ $(-1+8i)(-9+7i)$ in trigonometric form?

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Answer 1 · 2016-07-23T05:29:29

$(- 1 + 8 i) \times (- 9 + 7 i) = 65 \sqrt{2} (cos θ + i sin θ)$ $(-1+8i)xx(-9+7i)=65sqrt2(costheta+isintheta)$ , where $θ = {tan}^{- 1} (\frac{79}{47})$ $theta=tan^(-1)(79/47)$

Explanation:

Let us write the two complex numbers in polar coordinates and let them be

$z_{1} = r_{1} (cos α + i sin α)$ $z_1=r_1(cosalpha+isinalpha)$ and $z_{2} = r_{2} (cos β + i sin β)$ $z_2=r_2(cosbeta+isinbeta)$

Here, if two complex numbers are $a_{1} + i b_{1}$ $a_1+ib_1$ and $a_{2} + i b_{2}$ $a_2+ib_2$ $r_{1} = \sqrt{a_{1}^{2} + b_{1}^{2}}$ $r_1=sqrt(a_1^2+b_1^2)$ , $r_{2} = \sqrt{a_{2}^{2} + b_{2}^{2}}$ $r_2=sqrt(a_2^2+b_2^2)$ and $α = {tan}^{- 1} (\frac{b_{1}}{a_{1}})$ $alpha=tan^(-1)(b_1/a_1)$ , $β = {tan}^{- 1} (\frac{b_{2}}{a_{2}})$ $beta=tan^(-1)(b_2/a_2)$

Their multiplicaton leads us to

${r_{1} \times r_{2}} {(cos α + i sin α) \times (cos β + i sin β)}$ ${r_1xxr_2}{(cosalpha+isinalpha)xx(cosbeta+isinbeta)}$ or

${r_{1} r_{2}} (cos α cos β + i sin α cos β + i sin α cos β + i^{2} sin α sin β)$ ${r_1r_2}(cosalphacosbeta+isinalphacosbeta+isinalphacosbeta+i^2sinalphasinbeta)$

${r_{1} r_{2}} (cos α cos β + i sin α cos β + i sin α cos β - sin α sin β)$ ${r_1r_2}(cosalphacosbeta+isinalphacosbeta+isinalphacosbeta-sinalphasinbeta)$

${r_{1} r_{2}} [(cos α cos β - sin α sin β + i (sin α cos β + sin α cos β)]$ ${r_1r_2}[(cosalphacosbeta-sinalphasinbeta+i(sinalphacosbeta+sinalphacosbeta)]$ or

$(r_{1} r_{2}) (cos (α + β) + i sin (α + β))$ $(r_1r_2)(cos(alpha+beta)+isin(alpha+beta))$ or

$z_{1} \cdot z_{2}$ $z_1*z_2$ is given by $(r_{1} \cdot r_{2}, (α + β))$ $(r_1*r_2, (alpha+beta))$

So for multiplication of complex number $z_{1}$ $z_1$ and $z_{2}$ $z_2$ , take new angle as $(α + β)$ $(alpha+beta)$ and modulus $r_{1} \cdot r_{2}$ $r_1*r_2$ of the modulus of two numbers.

Here $- 1 + 8 i$ $-1+8i$ can be written as $r_{1} (cos α + i sin α)$ $r_1(cosalpha+isinalpha)$ where $r_{1} = \sqrt{{(- 1)}^{2} + 8^{2}} = \sqrt{65}$ $r_1=sqrt((-1)^2+8^2)=sqrt65$ and $α = {tan}^{- 1} (- \frac{8}{1}) = {tan}^{- 1} (- 8)$ $alpha=tan^(-1)(-8/1)=tan^(-1)(-8)$

and $- 9 + 7 i$ $-9+7i$ can be written as $r_{2} (cos β + i sin β)$ $r_2(cosbeta+isinbeta)$ where $r_{2} = \sqrt{{(- 9)}^{2} + 7^{2}} = \sqrt{81 + 49} = \sqrt{130}$ $r_2=sqrt((-9)^2+7^2)=sqrt(81+49)=sqrt130$ and $β = {tan}^{- 1} (\frac{7}{- 9}) = {tan}^{- 1} (- \frac{7}{9})$ $beta=tan^(-1)(7/(-9))=tan^(-1)(-7/9)$

and $z_{1} \cdot z_{2} = \sqrt{65} \times \sqrt{130} (cos θ + i sin θ)$ $z_1*z_2=sqrt65xxsqrt130(costheta+isintheta)$ , where $θ = α + β$ $theta=alpha+beta$

Hence, as $tan θ = tan (α + β) = \frac{tan α + tan β}{1 - tan α tan β} = \frac{(- 8) + (- \frac{7}{9})}{1 - (- 8) \times (- \frac{7}{9})} = \frac{- \frac{79}{9}}{1 - (\frac{56}{9})} = \frac{- \frac{79}{9}}{- \frac{47}{9}} = \frac{79}{47}$ $tantheta=tan(alpha+beta)=(tanalpha+tanbeta)/(1-tanalphatanbeta)=((-8)+(-7/9))/(1-(-8)xx(-7/9))=(-79/9)/(1-(56/9))=(-79/9)/(-47/9)=79/47$ .

and $z = 65 \sqrt{2}$ $z=65sqrt2$

Hence, $(- 1 + 8 i) \times (- 9 + 7 i) = 65 \sqrt{2} (cos θ + i sin θ)$ $(-1+8i)xx(-9+7i)=65sqrt2(costheta+isintheta)$ , where $θ = {tan}^{- 1} (\frac{79}{47})$ $theta=tan^(-1)(79/47)$

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How do you multiply $(- 1 + 8 i) (- 9 + 7 i)$ $(-1+8i)(-9+7i)$ in trigonometric form?

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