Question

(4i+2)(2i-3)/3i+5
Express in a+in form

Answer 1

✍✍HERE IS YOUR ANSWER..⬇⬇
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$\underline{SOLUTION}$ ➡

$We \: know \: that,$

$\boxed{i = \sqrt{ - 1} \: \: = > i {}^{2} = ( \sqrt{ - 1} ) { }^{2} = - 1}$

$Now, \\ \\ \\ \frac{(4i + 2)(2i -3 )}{3i + 5} \\ \\ = \frac{4i(2i - 3) + 2(2i - 3)}{3i + 5} \\ \\ = \frac{8i {}^{2} -1 2i + 4i - 6}{3i + 5} \\ \\ = \frac{8( - 1) - 8i - 6}{3i + 5} \\ \\ = \frac{ - 8 - 8i - 6}{3i + 5} \\ \\ = \frac{ - 8i-1 4}{3i + 5} \\ \\ = \frac{( - 8i -1 4)(3i - 5)}{(3i + 5)(3i - 5)} \\ \\ = \frac{ - 8i(3i - 5) - 14(3i - 5)}{(3i ) {}^{2} - ( 5) {}^{2} } \\ \\ = \frac{ - 24i {}^{2} + 40 i - 42i + 70}{9i {}^{2} - 25} \\ \\ = \frac{ - 24( - 1) - 2i + 70}{9( - 1) - 25} \\ \\ = \frac{24 - 2i + 70}{ - 9 - 25} \\ \\ = \frac{94 - 2i}{ - 34} \\ \\ = \frac{94 }{ - 34} - \frac{2i}{ (- 34)} \\ \\ = (- \frac{47}{17}) + ( i.\frac{1}{17} )$

$This \: is \: in \: \: a + in \: \: form \: \: \\ \\ where, \\ \\ a \: = - \frac{ 47}{17} \\ \\n = \frac{1}{17}$

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