How do you multiply $(4 + 5 i) (- 3 + 7 i)$ $(4+5i)(-3+7i)$ in trigonometric form?

Question

How do you multiply $(4 + 5 i) (- 3 + 7 i)$ $(4+5i)(-3+7i)$ in trigonometric form?

1 Answer

Impact of this question

1680 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-28T14:05:48

$(4 + 5 i) (- 3 + 7 i) = 48.765 c i s 2.317$ $(4+5i)(-3+7i)=48.765cis2.317$
$(4 + 5 i) (- 3 + 7 i) = - 33.111 + 35.842 i$ $(4+5i)(-3+7i)=-33.111+35.842i$

Explanation:

Let
$z_{1} = 4 + 5 i$ $z_1=4+5i$
$z_{2} = - 3 + 7 i$ $z_2=-3+7i$
We need to find

$z = z_{1} z_{2}$ $z=z_1z_2$
$r_{1} = \sqrt{4^{2} + 5^{2}} = \sqrt{16 + 25} = \sqrt{41}$ $r_1=sqrt(4^2+5^2)=sqrt(16+25)=sqrt41$
$r 1 = \sqrt{41}$ $r1=sqrt41$
$θ_{1} = {tan}^{- 1} (\frac{5}{4})$ $theta_1=tan^-1(5/4)$
$z_{1} = r_{1} c i s θ_{1}$ $z_1=r_1cistheta_1$
$z_{1} = \sqrt{41} c i s ({tan}^{- 1} (\frac{5}{4}))$ $z_1=sqrt41cis(tan^-1(5/4))$

$r_{2} = \sqrt{{(- 3)}^{2} + 7^{2}} = \sqrt{9 + 49} = \sqrt{58}$ $r_2=sqrt((-3)^2+7^2)=sqrt(9+49)=sqrt58$
$r_{2} = \sqrt{58}$ $r_2=sqrt58$
$θ_{2} = {tan}^{- 1} (\frac{7}{- 3})$ $theta_2=tan^-1(7/-3)$
$z_{2} = r_{2} c i s θ_{2}$ $z_2=r_2cistheta_2$
$z_{2} = \sqrt{58} c i s ({tan}^{- 1} (\frac{7}{- 3}))$ $z_2=sqrt58cis(tan^-1(7/-3))$

$z = z_{1} z_{2}$ $z=z_1z_2$
$z = (\sqrt{41} c i s ({tan}^{- 1} (\frac{5}{4}))) (\sqrt{58} c i s ({tan}^{- 1} (\frac{7}{- 3})))$ $z=(sqrt41cis(tan^-1(5/4)))(sqrt58cis(tan^-1(7/-3)))$

$z = \sqrt{41} \sqrt{58} c i s ({tan}^{- 1} (\frac{5}{4})) c i s ({tan}^{- 1} (\frac{7}{- 3}))$ $z=sqrt41sqrt58cis(tan^-1(5/4))cis(tan^-1(7/-3))$
$\sqrt{41} \sqrt{58} = \sqrt{41 \times 58} = \sqrt{2378}$ $sqrt41sqrt58=sqrt(41xx58)=sqrt2378$
By De-Moivre's Theorem

$c i s ({tan}^{- 1} (\frac{5}{4})) c i s ({tan}^{- 1} (\frac{7}{- 3})) = c i s ({tan}^{- 1} (\frac{5}{4}) + {tan}^{- 1} (\frac{7}{- 3}))$ $cis(tan^-1(5/4))cis(tan^-1(7/-3))=cis(tan^-1(5/4)+tan^-1(7/-3))$
${tan}^{- 1} (\frac{5}{4}) + {tan}^{- 1} (\frac{7}{- 3}) = {tan}^{- 1} (\frac{\frac{5}{4} + \frac{7}{- 3}}{1 - \frac{5}{4} \times \frac{7}{- 3}})$ $tan^-1(5/4)+tan^-1(7/-3)=tan^-1((5/4+7/-3)/(1-5/4xx7/-3))$

$\frac{5}{4} + \frac{7}{- 3} = \frac{5 \times - 3 + 7 \times 4}{4 \times - 3} = \frac{- 15 + 28}{- 12} = \frac{13}{- 12}$ $5/4+7/-3=(5xx-3+7xx4)/(4xx-3)=(-15+28)/-12=13/-12$
$z = r c i s θ$ $z=rcistheta$

$r = \sqrt{2378}$ $r=sqrt2378$
$θ = {tan}^{- 1} (\frac{13}{- 12})$ $theta=tan^-1(13/-12)$
$z = \sqrt{2378} c i s ({tan}^{- 1} (\frac{13}{- 12}))$ $z=sqrt2378cis(tan^-1(13/-12))$
$r = 48.765$ $r=48.765$
${tan}^{- 1} (\frac{13}{- 12})$ $tan^-1(13/-12)$
adjacent side is negative,
the angle lies in the second quadrant
${tan}^{- 1} (\frac{13}{- 12}) = π - {tan}^{- 1} (\frac{13}{12})$ $tan^-1(13/-12)=pi-tan^-1(13/12)$
${tan}^{- 1} (\frac{13}{12}) = 0.825$ $tan^-1(13/12)=0.825$
expressed in radians
${tan}^{- 1} (\frac{13}{12}) = π - 0.825$ $tan^-1(13/12)=pi-0.825$
$θ = 2.317$ $theta=2.317$

$cos 0.825 = - 0.679$ $cos0.825=-0.679$
since the complex number is in 2nd quadrant
$sin 0.825 = 0.735$ $sin0.825=0.735$

$r = 48.765$ $r=48.765$
$x = r cos θ = 48.765 \times (- 0.679)$ $x=rcostheta=48.765xx(-0.679)$
$x = - 33.111$ $x=-33.111$

$y = r sin θ = 48.765 \times 0.735$ $y=rsintheta=48.765xx0.735$
$y = 35.842$ $y=35.842$
$z = x + i y$ $z=x+iy$

$z = (- 33.111) + i (35.842)$ $z=(-33.111)+i(35.842)$

Thus,
$(4 + 5 i) (- 3 + 7 i) = 48.765 c i s 2.317$ $(4+5i)(-3+7i)=48.765cis2.317$
$(4 + 5 i) (- 3 + 7 i) = - 33.111 + 35.842 i$ $(4+5i)(-3+7i)=-33.111+35.842i$

Science

Math

Humanities

... and beyond

How do you multiply $(4 + 5 i) (- 3 + 7 i)$ $(4+5i)(-3+7i)$ in trigonometric form?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you multiply (4+5i)(−3+7i) (4+5i)(-3+7i) in trigonometric form?

1 Answer

Explanation:

Related questions

Impact of this question

How do you multiply $(4 + 5 i) (- 3 + 7 i)$ $(4+5i)(-3+7i)$ in trigonometric form?