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

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Answer 1 · 2016-03-02T11:56:20

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

$(a+bi)$ 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 $beta=tan^-1(2/3)$

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