How do you divide #( 9i- 2) / (- 5 i+ 2 )# in trigonometric form?

1 Answer
Dec 20, 2016

#=1/29 (41+8i)#

Explanation:

Rewrite 9i-2 as -2+9i. Now square and the numbers -2 and 9, which would be 85. Its square root is#sqrt85#. Multiply and divide the expression -2 +9i with #sqrt85# as follows:

#sqrt85 (-2/sqrt85 +i 9/sqrt85)#

If #theta# is some angle then let #cos theta= -2/sqrt85 and sin theta=9/sqrt85#

Thus #-2 +9i= sqrt85(cos theta +isin theta)#

Like wise -5i +2 would be 2 -5i =#sqrt29( 2/sqrt 29-i5/sqrt29)#

If #phi# is some angle, then let #cos phi= 2/sqrt29 and sin phi= -5/sqrt29#

Thus #2-5i=sqrt29(cos phi -sin phi)#

#(9i-2)/(-5i+2)= sqrt(85/29) (cos theta +isin theta)/(cosphi-isin phi)=sqrt(85/29) e^(itheta)/e^(-iphi) =sqrt(85/29) e^(i(theta +phi)#

#=sqrt(85/29)( cos(theta+phi)+i sin (theta+phi) )#

#=sqrt(85/29)[ (cos theta cos phi- sin theta sin phi) +i(sin theta cos phi + cos theta sin phi)]#

#=sqrt(85/29)[(-2)/sqrt85 * 2/sqrt29 -9/sqrt85 * (-5)/sqrt29 +i(9/sqrt85 * 2/sqrt29 +(-2)/sqrt85 * (-5)/sqrt29)]#

#sqrt(85/29) (41/(sqrt85 sqrt29)+i 28/(sqrt 85 sqrt29))#
#=1/29 (41+28i)#