Question

second question , it is a question from class 11 ch- complex No. and please don't spam​

Answer 1

Answer:

your solution is as follows

pls mark it as brainliest

Step-by-step explanation:

$to \: find : \\ value \: of \: A {}^{2} +B {}^{2} \\ \\ given \: complex \: number \: is \: \\ A+iB = \frac{2 + i}{2 - 3i} \\ \\ = \frac{2 + i}{2 - 3i} \times \frac{2 + 3i}{2 + 3i} \\ \\ = \frac{(2 + i)(2 + 3i)}{(2) {}^{2} - (3i) {}^{2} } \\ \\ = \frac{4 + 6i + 2i + 3i {}^{2} }{4 - 9i {}^{2} } \\ \\ = \frac{4 + 3( - 1) + 8i}{4 - 9( - 1)} \\ \\ = \frac{4 - 3 + 8i}{4 + 9} \\ \\ = \frac{1 + 8i}{13} \\ \\ = \frac{1}{13} + \frac{8}{13} \: i$

$equating \: real \: and \: imaginary \: parts \\ we \: get \\ A = \frac{1}{13} \\ \\ B = \frac{8}{13} \\ \\ A {}^{2} +B {}^{2} = ( \frac{1}{13} ) {}^{2} + ( \frac{8}{13} ) {}^{2} \\ \\ = \frac{1}{169} + \frac{64}{169} \\ \\ = \frac{1 + 64}{169} \\ \\ = \frac{65}{169} \\ \\ A {}^{2} +B {}^{2} = \frac{5}{13}$