How do you divide #( 2i -4) / ( 7 i -2 )# in trigonometric form?

1 Answer
Leland Adriano Alejandro
Jul 25, 2016

#(2i-4)/(7i-2)=(2sqrt(265))/53[cos 47.48^@+i*sin 47.48^@]#

Explanation:

The solution:

#2i-4=#
#sqrt(4+16)[cos (tan^-1 (-1/2))+i*sin (tan^-1 (-1/2))]#
#sqrt(20)[cos (tan^-1 (-1/2))+i*sin (tan^-1 (-1/2))]#

#7i-2=#
#sqrt(4+49)[cos (tan^-1 (-7/2))+i*sin (tan^-1 (-7/2))]#

#(2i-4)/(7i-2)=#
#sqrt(20)/sqrt(53)[cos (tan^-1 (-1/2)-tan^-1 (-1/2))+i*sin (tan^-1 (-1/2)-tan^-1 (-1/2))]#

#(2i-4)/(7i-2)=(2sqrt(265))/53[cos 47.48^@+i*sin 47.48^@]#

God bless.....I hope the explanation is useful.

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