How do you simplify $[\frac{{(3 + 2 i)}^{3}}{{(- 2 + 3 i)}^{4}}]$ $[(3+2i)^ 3 / (-2+3i)^4]$ ?

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How do you simplify $[\frac{{(3 + 2 i)}^{3}}{{(- 2 + 3 i)}^{4}}]$ $[(3+2i)^ 3 / (-2+3i)^4]$ ?

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Answer 1 · 2016-06-27T16:42:31

$\frac{3 - 2 i}{13}$ $(3-2i)/13$

Explanation:

$3 + 2 i = \sqrt{3^{2} + 2^{2}} e^{i ϕ}$ $3+2i=sqrt(3^2+2^2)e^{i phi}$
$- 2 + 3 i = \sqrt{3^{2} + 2^{2}} e^{i (ϕ + \frac{π}{2})}$ $-2+3i=sqrt(3^2+2^2)e^{i( phi+pi/2)}$

with $ϕ = arctan (\frac{2}{3})$ $phi = arctan(2/3)$

$[\frac{{(3 + 2 i)}^{3}}{{(- 2 + 3 i)}^{4}}] = \frac{{(\frac{3 + 2 i}{- 2 + 3 i})}^{3}}{(- 2 + 3 i)} = \frac{e^{- i \frac{3 π}{2}}}{\sqrt{3^{2} + 2^{2}} e^{i (ϕ + \frac{π}{2})}}$ $[(3+2i)^ 3 / (-2+3i)^4] =((3+2i)/(-2+3i))^3/( (-2+3i))=e^{-i (3pi)/2}/(sqrt(3^2+2^2)e^{i( phi+pi/2)})$
$= \frac{1}{\sqrt{3^{2} + 2^{2}}} e^{- i (ϕ + 2 π)} = \frac{1}{\sqrt{3^{2} + 2^{2}}} e^{- i ϕ} =$ $=1/sqrt(3^2+2^2)e^{-i(phi+2pi)} = 1/sqrt(3^2+2^2)e^{-i phi} =$
$\frac{\sqrt{3^{2} + 2^{2}}}{3^{2} + 2^{2}} e^{- i ϕ} = \frac{3 - 2 i}{13}$ $sqrt(3^2+2^2)/(3^2+2^2)e^{-i phi} = (3-2i)/13$

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How do you simplify $[\frac{{(3 + 2 i)}^{3}}{{(- 2 + 3 i)}^{4}}]$ $[(3+2i)^ 3 / (-2+3i)^4]$ ?

1 Answer

Explanation:

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How do you simplify [(3+2i)^ 3 / (-2+3i)^4] [(3+2i)3(−2+3i)4] [(3+2i)^ 3 / (-2+3i)^4] ?

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How do you simplify $[\frac{{(3 + 2 i)}^{3}}{{(- 2 + 3 i)}^{4}}]$ $[(3+2i)^ 3 / (-2+3i)^4]$ ?