How do you multiply $(7 - 3 i) (5 - i)$ $(7-3i)(5-i)$ in trigonometric form?

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How do you multiply $(7 - 3 i) (5 - i)$ $(7-3i)(5-i)$ in trigonometric form?

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Answer 1 · 2017-03-08T10:25:45

$(7 - 3 i) (5 - i) = 2 \sqrt{377} (cos θ + i sin θ)$ $(7-3i)(5-i)=2sqrt377(costheta+isintheta)$ , where $θ = {tan}^{- 1} (- \frac{11}{16})$ $theta=tan^(-1)(-11/16)$

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 multipication leads us to

${r_{1} \cdot r_{2}} {(cos α + i sin α) \cdot (cos β + i sin β)}$ ${r_1*r_2}{(cosalpha+isinalpha)*(cosbeta+isinbeta)}$ or

${r_{1} \cdot r_{2}} {(cos α cos β + i^{2} sin α sin β) + i (cos α sin β + cos β sin α))$ ${r_1*r_2}{(cosalphacosbeta+i^2sinalphasinbeta)+i(cosalphasinbeta+cosbetasinalpha))$ or

${r_{1} \cdot r_{2}} {(cos α cos β - sin α sin β) + i (cos α sin β + cos β sin α))$ ${r_1*r_2}{(cosalphacosbeta-sinalphasinbeta)+i(cosalphasinbeta+cosbetasinalpha))$ or

$(r_{1} \cdot r_{2}) \cdot (cos (α + β) + i sin (α + β))$ $(r_1*r_2)*(cos(alpha+beta)+isin(alpha+beta))$ 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 os $r_{1} \cdot r_{2}$ $r_1*r_2$ of the modulus of two numbers.

Here $7 - 3 i$ $7-3i$ can be written as $r_{1} (cos α + i sin α)$ $r_1(cosalpha+isinalpha)$ where $r_{1} = \sqrt{7^{2} + {(- 3)}^{2}} = \sqrt{58}$ $r_1=sqrt(7^2+(-3)^2)=sqrt58$ and $α = {tan}^{- 1} (\frac{- 3}{7})$ $alpha=tan^(-1)((-3)/7)$

and $5 - i$ $5-i$ can be written as $r_{2} (cos β + i sin β)$ $r_2(cosbeta+isinbeta)$ where $r_{2} = \sqrt{5^{2} + {(- 1)}^{2}} = \sqrt{26}$ $r_2=sqrt(5^2+(-1)^2)=sqrt26$ and $β = {tan}^{- 1} (- \frac{1}{5})$ $beta=tan^(-1)(-1/5)$

and $z_{1} \cdot z_{2} = \sqrt{58} \cdot (\sqrt{26}) (cos θ + i sin θ)$ $z_1*z_2=sqrt58*(sqrt26)(costheta+isintheta)$ , where $θ = α + β$ $theta=alpha+beta$

Hence, $tan θ = tan (α + β) = \frac{tan α + tan β}{1 - tan α tan β} = \frac{\frac{- 3}{7} + (- \frac{1}{5})}{1 - (\frac{- 3}{7} \times (- \frac{1}{5}))} = \frac{- \frac{22}{35}}{\frac{32}{35}} = - \frac{22}{32} = - \frac{11}{16}$ $tantheta=tan(alpha+beta)=(tanalpha+tanbeta)/(1-tanalphatanbeta)=((-3)/7+(-1/5))/(1-((-3)/7xx(-1/5)))=(-22/35)/(32/35)=-22/32=-11/16$ .

Hence, $(7 - 3 i) (5 - i) = \sqrt{58 \times 26} (cos θ + i sin θ)$ $(7-3i)(5-i)=sqrt(58xx26)(costheta+isintheta)$

= $2 \sqrt{377} (cos θ + i sin θ)$ $2sqrt377(costheta+isintheta)$ , where $θ = {tan}^{- 1} (- \frac{11}{16})$ $theta=tan^(-1)(-11/16)$

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