How do you divide #( 2i -7) / ( 3 i -2 )# in trigonometric form?

Question

How do you divide #( 2i -7)    /   (  3 i -2  )# in trigonometric form?

Asher Friedman · Accepted Answer

#(2i-7)/(3i-2)=sqrt(53/13)(costheta+isintheta)#, where #theta=tan^(-1)(17/20)#

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

#z_1=r_1(cosalpha+isinalpha)# and #z_2=r_2(cosbeta+isinbeta)#

Here, if two complex numbers are #a_1+ib_1# and #a_2+ib_2# #r_1=sqrt(a_1^2+b_1^2)#, #r_2=sqrt(a_2^2+b_2^2)# and #alpha=tan^(-1)(b_1/a_1)#, #beta=tan^(-1)(b_2/a_2)#

Their division leads us to

#{r_1/r_2}{(cosalpha+isinalpha)/(cosbeta+isinbeta)}# or

#{r_1/r_2}{(cosalpha+isinalpha)/(cosbeta+isinbeta)xx(cosbeta-isinbeta)/(cosbeta-isinbeta)}#

#(r_1/r_2){(cosalphacosbeta+sinalphasinbeta)+i(sinalphacosbeta-cosalphasinbeta))/((cos^2beta+sin^2beta))# or

#(r_1/r_2)*(cos(alpha-beta)+isin(alpha-beta))# or

#z_1/z_2# is given by #(r_1/r_2, (alpha-beta))#

So for division complex number #z_1# by #z_2# , take new angle as #(alpha-beta)# and modulus the ratio #r_1/r_2# of the modulus of two numbers.

Here #2i-7=-7+2i# can be written as #r_1(cosalpha+isinalpha)# where #r_1=sqrt((-7)^2+2^2)=sqrt53# and #alpha=tan^(-1)(-2/7)#

and #3i-2=-2+3i# can be written as #r_2(cosbeta+isinbeta)# where #r_2=sqrt((-2)^2+3^2)=sqrt13# and #beta=tan^(-1)(-3/2)#

and #z_1/z_2=sqrt53/sqrt13(costheta+isintheta)#, where #theta=alpha-beta#

Hence, #tantheta=tan(alpha-beta)=(tanalpha-tanbeta)/(1+tanalphatanbeta)=((-2/7)-(-3/2))/(1+(-2/7)xx(-3/2))=(-2/7+3/2)/(1+3/7)=(17/14)/(10/7)=17/14xx7/10=17/20#.

Hence, #(2i-7)/(3i-2)=sqrt(53/13)(costheta+isintheta)#, where #theta=tan^(-1)(17/20)#

Carter Anderson · Answer

undefined To divide (2i - 7) by (3i - 2) in trigonometric form, follow these steps:

1. Multiply both the numerator and denominator by the conjugate of the denominator. The conjugate of (3i - 2) is (3i + 2). So, the expression becomes:

(2i - 7) * (3i + 2) / (3i - 2) * (3i + 2)

2. Perform the multiplication in the numerator and denominator:

Numerator: (2i * 3i) + (2 * 2i) - (7 * 3i) - (7 * 2)
             = 6i^2 + 4i - 21i - 14
             = -6 - 17i

Denominator: (3i * 3i) + (3i * 2) - (2 * 3i) - (2 * 2)
               = 9i^2 + 6i - 6i - 4
               = -9 - 4
               = -13

3. Rewrite the division as a fraction:

(-6 - 17i) / -13

4. Divide both the real and imaginary parts by the denominator (-13):

Real part: -6 / -13 = 6/13
   Imaginary part: -17i / -13 = 17i/13

5. Express the result in trigonometric form. In trigonometric form, a complex number $a + bi$ is represented as $r(cosθ + isinθ)$, where $r$ is the modulus of the complex number and $θ$ is the argument.

To find $r$, the modulus:
   $r = \sqrt{a^2 + b^2}$

Substitute $a = 6/13$ and $b = 17/13$:
   $r = \sqrt{(6/13)^2 + (17/13)^2} = \sqrt{(36/169) + (289/169)} = \sqrt{325/169}$

To find $θ$, the argument:
   $θ = \tan^{-1}(\frac{b}{a})$

Substitute $a = 6/13$ and $b = 17/13$:
   $θ = \tan^{-1}(\frac{17/13}{6/13}) = \tan^{-1}(\frac{17}{6})$

So, the division of (2i - 7) by (3i - 2) in trigonometric form is $ \frac{\sqrt{325}}{13}(cos(\tan^{-1}(\frac{17}{6})) + i sin(\tan^{-1}(\frac{17}{6}))) $.