How do you divide # (8-2i) / (5-3i) # in trigonometric form?

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Answer 1 · 2016-01-27T09:26:39

#sqrt2(cos(tan^-1 (7/23))+isin(tan^-1 (7/23))# OR

#sqrt2(cos(16.9275^@)+isin(16.9275^@))#

Start from the given complex number

#(8-2i)/(5-3i)#

Convert the numerator

#r_1=sqrt(8^2+(-2)^2)=sqrt68#

#theta_1=tan^-1 ((-2)/8)=tan^-1 ((-1)/4)#

then

#8-2i=sqrt68[cos(tan^-1 ((-1)/4))+isin(tan^-1 ((-1)/4))]#

Convert the denominator

#r_2=sqrt(5^2+(-3)^2)=sqrt34#

#theta_2=tan^-1 ((-3)/5)=tan^-1 ((-3)/5)#

then

#5-3i=sqrt34[cos(tan^-1 ((-3)/5))+isin(tan^-1 ((-3)/5))]#

Let us divide now, from the given with the equivalent

#(8-2i)/(5-3i)=(sqrt68[cos(tan^-1 ((-1)/4))+isin(tan^-1 ((-1)/4))])/(sqrt34[cos(tan^-1 ((-3)/5))+isin(tan^-1 ((-3)/5))])#

Divide using the following formula

#(r_1/r_2)*[cos (theta_1-theta_2)+isin(theta1-theta_2)]#

#(8-2i)/(5-3i)= sqrt(68/34)[cos(tan^-1 ((-1)/4)-tan^-1 ((-3)/5))+isin(tan^-1 ((-1)/4)-tan^-1 ((-3)/5))]#

Take note: that
#tan (theta_1-theta_2)=(tan theta_1-tan theta_2) /(1+tan theta_1*tan theta_2)=(-1/4-(-3/5))/(1+(-1/4)*(-3/5))#

#tan (theta_1-theta_2)=7/23#

#theta_1-theta_2=tan^-1 (7/23)#

and

#(8-2i)/(5-3i)=sqrt2(cos(tan^-1 (7/23))+isin(tan^-1 (7/23))#

Have a nice day!!!.