How do you multiply $(4 + 3 i) (- 3 + 2 i)$ $(4+3i)(-3+2i)$ in trigonometric form?

Question

How do you multiply $(4 + 3 i) (- 3 + 2 i)$ $(4+3i)(-3+2i)$ in trigonometric form?

1 Answer

Impact of this question

1643 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-08T09:28:19

$(4 + 3 i) \cdot (- 3 + 2 i) = 5 \sqrt{13} (cos θ + i sin θ)$ $(4+3i)*(-3+2i)=5sqrt13(costheta+isintheta)$ , where $θ = {tan}^{- 1} (\frac{1}{18})$ $theta=tan^(-1)(1/18)$

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 multiplication leads us to

${r_{1} r_{2}} {(cos α + i sin α) (cos β + i sin β)}$ ${r_1r_2}{(cosalpha+isinalpha)(cosbeta+isinbeta)}$ or

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

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

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

$z_{1} \cdot z_{2}$ $z_1*z_2$ is given by $(r_{1} r_{2}, (α + β))$ $(r_1r_2, (alpha+beta))$

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

Here $4 + 3 i$ $4+3i$ can be written as $r_{1} (cos α + i sin α)$ $r_1(cosalpha+isinalpha)$ where $r_{1} = \sqrt{4^{2} + 3^{2}} = \sqrt{25} = 5$ $r_1=sqrt(4^2+3^2)=sqrt25=5$ and $α = \frac{{tan}^{- 1} 3}{4}$ $alpha=tan^(-1)3/4$

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

and $z_{1} \cdot z_{2} = 5 \sqrt{13} (cos θ + i sin θ)$ $z_1*z_2=5sqrt13(costheta+isintheta)$ , where $θ = α + β$ $theta=alpha+beta$

Hence, $tan θ = tan (α + β) = \frac{tan α + tan β}{1 - tan α tan β} = \frac{\frac{3}{4} + (- \frac{2}{3})}{1 - \frac{3}{4} \times (- \frac{2}{3})} = \frac{\frac{1}{12}}{\frac{3}{2}} = \frac{1}{18}$ $tantheta=tan(alpha+beta)=(tanalpha+tanbeta)/(1-tanalphatanbeta)=(3/4+(-2/3))/(1-3/4xx(-2/3))=(1/12)/(3/2)=1/18$ .

Hence, $(4 + 3 i) \cdot (- 3 + 2 i) = 5 \sqrt{13} (cos θ + i sin θ)$ $(4+3i)*(-3+2i)=5sqrt13(costheta+isintheta)$ , where $θ = {tan}^{- 1} (\frac{1}{18})$ $theta=tan^(-1)(1/18)$

Science

Math

Humanities

... and beyond

How do you multiply $(4 + 3 i) (- 3 + 2 i)$ $(4+3i)(-3+2i)$ in trigonometric form?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you multiply (4+3i)(-3+2i) (4+3i)(−3+2i) (4+3i)(-3+2i) in trigonometric form?

1 Answer

Explanation:

Related questions

Impact of this question

How do you multiply $(4 + 3 i) (- 3 + 2 i)$ $(4+3i)(-3+2i)$ in trigonometric form?