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anonymous
 4 years ago
Solve.
anonymous
 4 years ago
Solve.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{4+3i}{(72i)(5+4i)}\]

barrycarter
 4 years ago
Best ResponseYou've already chosen the best response.1http://www.wolframalpha.com/input/?i=%284%2B3i%29%2F%28%2872i%29*%285%2B4i%29%29

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.04+3i/(72i)(5+4i) We multiply the 2 complex numbers: And we get= 4+3i/18i + 43 From here, we can just rationalize the denominator so: We get the final answer as: 22657i/1867

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.0we have \[\frac{4+3i}{(72i)(5+4i)}\] multiply numerator and denominator by the conjugate of (72i) and (5+4i), that is (7+2i) and (54i) so we have \[\frac{(4+3i)(7+2i)(54i)}{(72i)(7+2i)(54i)(5+4i)}\] now (a+bi)(abi)=a^2+b^2 so we have now \[\frac{(4+3i)(7+2i)(54i)}{(7^2+2^2)(5^2+4^2)}\] now let's simplify the numerator by multiplying and remembering that i^2=1 \[\frac{(28+8i+21i+6i^2)(54i)}{(53)(41)}\] we get now \[\frac{(28+8i+21i6)(54i)}{(53)(41)}\] now we have \[\frac{(22+29i)(54i)}{(53)(41)}\] now multiply the other two brackets \[\frac{(11088i+145i116i^2)}{(53)(41)}\] now simplifying the terms and substituting i^2=1 \[\frac{(110116+57i)}{(53)(41)}\] we get \[\frac{(6+57i)}{(53)(41)}\] or \[\frac{6+57i}{2173}\]

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.0sorry i made a mistake in the third last step it'd be \[\frac{110+116+57i}{53*41}\] so it's \[\frac{226+57i}{53*41}\] finally we get \[\frac{226+57i}{2173}\] sorry I made a mistake :(
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