How do you divide $\frac{3 i - 1}{8 i - 4}$ $( 3i-1) / (8i -4 )$ in trigonometric form?

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How do you divide $\frac{3 i - 1}{8 i - 4}$ $( 3i-1) / (8i -4 )$ in trigonometric form?

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Answer 1 · 2016-03-02T12:08:08

$\frac{1}{2 \sqrt{2}} \times e^{i (α - β)}$ $1/(2sqrt2)xxe^(i(alpha-beta)$ , where $α = {tan}^{- 1} (- 3)$ $alpha=tan^-1(-3)$ and $β = {tan}^{- 1} (- 2)$ $beta=tan^-1(-2)$

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, $(3 i - 1) = (- 1 + 3 i) = \sqrt{1^{2} + 3^{2}} (e^{i {tan}^{- 1} (- \frac{3}{1})}) = \sqrt{10} e^{i {tan}^{- 1} (- 3)}$ $(3i-1)=(-1+3i)=sqrt(1^2+3^2)(e^(itan^-1(-3/1)))=sqrt10e^(itan^-1(-3))$

Similarly, $(8 i - 4) = (- 4 + 8 i) = \sqrt{4^{2} + 8^{2}} (e^{i {tan}^{- 1} (- \frac{8}{4})}) = \sqrt{80} e^{i {tan}^{- 1} (- 2)}$ $(8i-4)=(-4+8i)=sqrt(4^2+8^2)(e^(itan^-1(-8/4)))=sqrt80e^(itan^-1(-2))$

Hence $\frac{3 i - 1}{8 i - 4} = \frac{\sqrt{10} e^{i {tan}^{- 1} (- 3)}}{\sqrt{80} e^{i {tan}^{- 1} (- 2)}}$ $(3i-1)/(8i-4)=(sqrt10e^(itan^-1(-3)))/(sqrt80e^(itan^-1(-2)))$

or $\sqrt{\frac{1}{8}} \times e^{i ({tan}^{- 1} (- 3)) - ({tan}^{- 1} (- 2))}$ $sqrt(1/8)xxe^(i(tan^-1(-3))-(tan^-1(-2))$ or

$\frac{1}{2 \sqrt{2}} \times e^{i (α - β)}$ $1/(2sqrt2)xxe^(i(alpha-beta)$ , where $α = {tan}^{- 1} (- 3)$ $alpha=tan^-1(-3)$ and $β = {tan}^{- 1} (- 2)$ $beta=tan^-1(-2)$

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How do you divide $\frac{3 i - 1}{8 i - 4}$ $( 3i-1) / (8i -4 )$ in trigonometric form?

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How do you divide $\frac{3 i - 1}{8 i - 4}$ $( 3i-1) / (8i -4 )$ in trigonometric form?