How do you divide # (-3-4i)/(5+2i) # in trigonometric form?

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Answer 1 · 2018-04-26T07:37:03

#5/sqrt(29)(cos(0.540)+isin(0.540))~~0.79+0.48i#

#(-3-4i)/(5+2i)=-(3+4i)/(5+2i)#

#z=a+bi# can be written as #z=r(costheta+isintheta)#, where

For #z_1=3+4i#:
#r=sqrt(3^2+4^2)=5#
#theta=tan^-1(4/3)=~~0,927#

For #z_2=5+2i#:
#r=sqrt(5^2+2^2)=sqrt29#
#theta=tan^-1(2/5)=~~0.381#

For #z_1/z_2#:
#z_1/z_2=r_1/r_2(cos(theta_1-theta_2)+isin(theta_1-theta_2))#

#z_1/z_2=5/sqrt(29)(cos(0.921-0.381)+isin(0.921-0.381))#

#z_1/z_2=5/sqrt(29)(cos(0.540)+isin(0.540))=0.79+0.48i#

Proof:
#-(3+4i)/(5+2i)*(5-2i)/(5-2i)=-(15+20i-6i+8)/(25+4)=(23+14i)/29=0.79+0.48i#