anonymous
  • anonymous
Solve.
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[\frac{4+3i}{(7-2i)(5+4i)}\]
barrycarter
  • barrycarter
http://www.wolframalpha.com/input/?i=%284%2B3i%29%2F%28%287-2i%29*%285%2B4i%29%29
anonymous
  • anonymous
4+3i/(7-2i)(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: -226-57i/-1867

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ash2326
  • ash2326
we have \[\frac{4+3i}{(7-2i)(5+4i)}\] multiply numerator and denominator by the conjugate of (7-2i) and (5+4i), that is (7+2i) and (5-4i) so we have \[\frac{(4+3i)(7+2i)(5-4i)}{(7-2i)(7+2i)(5-4i)(5+4i)}\] now (a+bi)(a-bi)=a^2+b^2 so we have now \[\frac{(4+3i)(7+2i)(5-4i)}{(7^2+2^2)(5^2+4^2)}\] now let's simplify the numerator by multiplying and remembering that i^2=-1 \[\frac{(28+8i+21i+6i^2)(5-4i)}{(53)(41)}\] we get now \[\frac{(28+8i+21i-6)(5-4i)}{(53)(41)}\] now we have \[\frac{(22+29i)(5-4i)}{(53)(41)}\] now multiply the other two brackets \[\frac{(110-88i+145i-116i^2)}{(53)(41)}\] now simplifying the terms and substituting i^2=-1 \[\frac{(110-116+57i)}{(53)(41)}\] we get \[\frac{(-6+57i)}{(53)(41)}\] or \[\frac{-6+57i}{2173}\]
anonymous
  • anonymous
Thanks man!!!!
ash2326
  • ash2326
sorry 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 :(
anonymous
  • anonymous
It is fine man!

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