Question

4+3i/(2+3i)(4-3i) in the form of a+ib

Answer 1

Answer:

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Step-by-step explanation:

$so \: here \: let \\ \\ \frac{4 + 3i}{(2 + 3i)(4 - 3i)} = a + ib \\ \\ = \frac{4 + 3i}{8 - 6i + 12i - 9i {}^{2} } \\ \\ = \frac{4 + 3i}{8 - 9( - 1) + 6i} \\ \\ = \frac{4 + 3i}{(8 + 9) + 6i} \\ \\ = \frac{4 + 3i}{17 + 6i} \\ \\ \\ = \frac{4 + 3i}{17 + 6i} \times \frac{17 - 6i}{17 - 6i} \\ \\ = \frac{(4 + 3i)(17 - 6i)}{(17) {}^{2} - (6i) {}^{2} } \\ \\ = \frac{68 - 24i + 51i - 18i {}^{2} }{289 - 36i {}^{2} } \\ \\ = \frac{68 - 18( - 1) + 27i}{289 - 36( - 1)} \\ \\ \\ = \frac{68 + 18 + 27i}{289 + 36} \\ \\ \\ = \frac{86 + 27i}{325} \\ \\ a + ib = \frac{86}{325} + \frac{27i}{325} \\ \\ \\ equating \: real \: and \: imaginary \: parts \\ we \: get \\ \\ \\ a = \frac{86}{325} \\ \\ b = \frac{27}{325}$