How do you divide #( 12i -3) / ( 7 i -2 )# in trigonometric form?

1 Answer
Mar 11, 2018

#color(red)(90/53-(3i)/53 = (3/53)sqrt(901){cos(-a tan(1/30)+2pi}+i sin{-a tan(1/30)+2pi}#

Explanation:

Given:

Divide the complex numbers in trigonometric form:

#color(blue)((12i-3)/(7i-2)#

Rewrite in #color(green)(z=(a+bi)# form:

#color(blue)((-3+12i)/(-2+7i)#

Multiply and divide by the complex conjugate of the denominator

#rArr (-3+12i)/(-2+7i)*(-2-7i)/(-2-7i)#

#rArr {(-3+12i)(-2-7i)}/{(-2+7i)(-2-7i)#

#rArr {(-3+12i)(-2-7i)}/{(-2)^2-(7i)^2)#

Note that #i = sqrt(-1) and i^2 = (-1)# in complex number arithmetic.

#rArr (6+21i-24i-84i^2)/(4-(49i^2)#

#rArr (90-3i)/(4-49*(-1))#

#rArr (90-3i)/(4+49)#

#rArr (90-3i)/53#

#rArr 90/53 -3/53i#

Next, we will find the Polar form

Important:

For any complex number #color(blue)(a+bi#,

Polar form is given by

#color(blue)(r[cos(theta)+i sin(theta)],# where

#color(brown)(r = sqrt(a^2+b^2),# and

#color(brown)(theta = tan^(-1)(b/a)#

We have, for #a+bi#

#a=(90/53) and b = (-3/53)#

#r=sqrt((90/53)^2+(-3/53)^2)#

#r=sqrt(8100/2809+9/2809)#

#r=sqrt(8109/2809)#

#r=sqrt((8109)/(53*53))#

#r=sqrt((901*9)/(53*53))#

#r=sqrt(901/53*9/53)#

#color(brown)(r=(3sqrt(901))/53#

Also,

#theta = tan^-1[(-3/53)/(90/53)]#

#rArr theta = -tan^-1(1/30)#

Add #2pi#, since #theta# is negative.

#color(brown)[:. theta = -tan^-1(1/30)+2pi#

Hence, the final answer is given by

#color(blue)(90/53-(3i)/53 = (3/53)sqrt(901){cos(-a tan(1/30)+2pi}+i sin{-a tan(1/30)+2pi}#

Hope you find this solution useful.