Question

$4i {}^{8} - 3i {}^{9} + 3 \\ \div 3i {}^{11} - 4i {}^{10} - 2$
complex number.
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Answer 1

Answer:

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Step-by-step explanation:

$to \: find : \\ \\ \frac{4i {}^{8} - 3i {}^{9} + 3 }{3i {}^{11} - 4i - 2 } \\ \\ we \: know \: that \\ \\ i {}^{2} = - 1 \\ \\ i {}^{3} = - i \\ \\ i {}^{4} \: and \: \: i {}^{4n} = 1$

$thus \: then \\ \\ = \frac{4i {}^{8} - 3i {}^{9} + 3}{3i {}^{11} - 4i {}^{10} - 2 } \\ \\ = \frac{(4 \times 1) - (3i {}^{8} \times i) + 3}{(3i {}^{8 + 3 \ {}^{} } {}^{} ) - (4i {}^{8 + 2} {} ) - 2 } \\ \\ = \frac{4 - 3i + 3}{3( - i) - 4( - 1)(1) - 2} \\ \\ = \frac{7 - 3i}{ - 3i + 4 - 2} \\ \\ = \frac{7 - 3i}{2 - 3i}$

$so \: conjugate \: of \: (2 - 3i) \: \: is \\ (2 + 3i) \\ \\ thus \: then \\ rationalsing \: denominator \\ we \: have \\ \\ = \frac{7 - 3i}{2 - 3i} \times \frac{2 + 3i}{2 + 3i} \\ \\ = \frac{(7 - 3i)(2 + 3i)}{(2) {}^{2} - (3i) {}^{2} } \\ \\ = \frac{14 + 21i - 6i - 9i {}^{2} }{4 - 9i {}^{2} } \\ \\ = \frac{14 - 9( - 1) + 15i}{4 - 9( - 1)} \\ \\ = \frac{14 + 9 + 15i}{4 + 9} \\ \\ = \frac{23 + 15i}{13} \\ \\ a + ib = \frac{23}{13} + \frac{15i}{13} \\ \\ a + ib = \frac{23}{13} + \frac{15}{13} i$

$equating \: real \: and \: imaginary \: parts \\ we \: have \: then \\ \\ a = \frac{23}{13} \\ \\ b = \frac{15}{13}$