How do you divide # (7-4i)/(3+2i) # in trigonometric form?

1 Answer
Shwetank Mauria
Mar 2, 2016

#sqrt5xxe^(i(alpha-beta)#, where #alpha=tan^-1(-4/7)# and #beta=tan^-1(2/3)#

Explanation:

#(a+bi)# an be written in as #sqrt(a^2+b^2)e^(i(tan^-1(b/a))#

Hence, #(7-4i)=sqrt(7^2+4^2)(e^(itan^-1(-4/7)))=sqrt65e^(itan^-1(-4/7))#

Similarly, #(3+2i)=sqrt(3^2+2^2)(e^(itan^-1(2/3)))=sqrt13e^(itan^-1(2/3))#

Hence #(7-4i)/(3+2i)=(sqrt65e^(itan^-1(-4/7)))/(sqrt13e^(itan^-1(2/3)))#

or #sqrt5xxe^(i(tan^-1(-4/7))-(tan^-1(2/3)))# or

#sqrt5xxe^(i(alpha-beta)#, where #alpha=tan^-1(-4/7)# and #beta=tan^-1(2/3)#

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