How do you multiply $(1 - 2 i) (2 - 3 i)$ $(1-2i)(2-3i)$ in trigonometric form?

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How do you multiply $(1 - 2 i) (2 - 3 i)$ $(1-2i)(2-3i)$ in trigonometric form?

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Answer 1 · 2016-07-23T06:51:11

$(1 - 2 i) \times (2 - 3 i) = \sqrt{65} (cos θ + i sin θ)$ $(1-2i)xx(2-3i)=sqrt65(costheta+isintheta)$ , where $θ = {tan}^{- 1} (\frac{7}{4})$ $theta=tan^(-1)(7/4)$

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 $1 - 2 i$ $1-2i$ can be written as $r_{1} (cos α + i sin α)$ $r_1(cosalpha+isinalpha)$ where $r_{1} = \sqrt{1^{2} + {(- 2)}^{2}} = \sqrt{5}$ $r_1=sqrt(1^2+(-2)^2)=sqrt5$ and $α = {tan}^{- 1} (- \frac{2}{1}) = {tan}^{- 1} (- 2)$ $alpha=tan^(-1)(-2/1)=tan^(-1)(-2)$

and $2 - 3 i$ $2-3i$ can be written as $r_{2} (cos β + i sin β)$ $r_2(cosbeta+isinbeta)$ where $r_{2} = \sqrt{2^{2} + {(- 3)}^{2}} = \sqrt{4 + 9} = \sqrt{13}$ $r_2=sqrt(2^2+(-3)^2)=sqrt(4+9)=sqrt13$ and $β = {tan}^{- 1} (- \frac{3}{2})$ $beta=tan^(-1)(-3/2)$

and $z_{1} \cdot z_{2} = \sqrt{5} \times \sqrt{13} (cos θ + i sin θ)$ $z_1*z_2=sqrt5xxsqrt13(costheta+isintheta)$ , where $θ = α + β$ $theta=alpha+beta$

Hence, as $tan θ = tan (α + β) = \frac{tan α + tan β}{1 - tan α tan β} = \frac{(- 2) + (- \frac{3}{2})}{1 - (- 2) \times (- \frac{3}{2})} = \frac{- \frac{7}{2}}{1 - 3} = \frac{- \frac{7}{2}}{- 2} = \frac{7}{4}$ $tantheta=tan(alpha+beta)=(tanalpha+tanbeta)/(1-tanalphatanbeta)=((-2)+(-3/2))/(1-(-2)xx(-3/2))=(-7/2)/(1-3)=(-7/2)/(-2)=7/4$ .

and $z = \sqrt{65}$ $z=sqrt65$

Hence, $(1 - 2 i) \times (2 - 3 i) = \sqrt{65} (cos θ + i sin θ)$ $(1-2i)xx(2-3i)=sqrt65(costheta+isintheta)$ , where $θ = {tan}^{- 1} (\frac{7}{4})$ $theta=tan^(-1)(7/4)$

Note that both $α$ $alpha$ and $β$ $beta$ are in fourth quadrant based on signs of sine and cosine functions, hence $θ = α + β$ $theta=alpha+beta$ ought to be between $3 π$ $3pi$ and $4 π$ $4pi$ . Hence as $tan θ$ $tantheta$ is positive, $θ$ $theta$ is in third quadrant.

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How do you multiply $(1 - 2 i) (2 - 3 i)$ $(1-2i)(2-3i)$ in trigonometric form?

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