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

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

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Answer 1 · 2016-03-02T11:56:20

$\sqrt{5} \times e^{i (α - β)}$ $sqrt5xxe^(i(alpha-beta)$ , where $α = {tan}^{- 1} (- \frac{4}{7})$ $alpha=tan^-1(-4/7)$ and $β = {tan}^{- 1} (\frac{2}{3})$ $beta=tan^-1(2/3)$

Explanation:

$(a + b i)$ $(a+bi)$ an be written in as $\sqrt{a^{2} + b^{2}} e^{i ({tan}^{- 1} (\frac{b}{a}))}$ $sqrt(a^2+b^2)e^(i(tan^-1(b/a))$

Hence, $(7 - 4 i) = \sqrt{7^{2} + 4^{2}} (e^{i {tan}^{- 1} (- \frac{4}{7})}) = \sqrt{65} e^{i {tan}^{- 1} (- \frac{4}{7})}$ $(7-4i)=sqrt(7^2+4^2)(e^(itan^-1(-4/7)))=sqrt65e^(itan^-1(-4/7))$

Similarly, $(3 + 2 i) = \sqrt{3^{2} + 2^{2}} (e^{i {tan}^{- 1} (\frac{2}{3})}) = \sqrt{13} e^{i {tan}^{- 1} (\frac{2}{3})}$ $(3+2i)=sqrt(3^2+2^2)(e^(itan^-1(2/3)))=sqrt13e^(itan^-1(2/3))$

Hence $\frac{7 - 4 i}{3 + 2 i} = \frac{\sqrt{65} e^{i {tan}^{- 1} (- \frac{4}{7})}}{\sqrt{13} e^{i {tan}^{- 1} (\frac{2}{3})}}$ $(7-4i)/(3+2i)=(sqrt65e^(itan^-1(-4/7)))/(sqrt13e^(itan^-1(2/3)))$

or $\sqrt{5} \times e^{i ({tan}^{- 1} (- \frac{4}{7})) - ({tan}^{- 1} (\frac{2}{3}))}$ $sqrt5xxe^(i(tan^-1(-4/7))-(tan^-1(2/3)))$ or

$\sqrt{5} \times e^{i (α - β)}$ $sqrt5xxe^(i(alpha-beta)$ , where $α = {tan}^{- 1} (- \frac{4}{7})$ $alpha=tan^-1(-4/7)$ and $β = {tan}^{- 1} (\frac{2}{3})$ $beta=tan^-1(2/3)$

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

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