## anonymous 4 years ago Solve.

1. anonymous

$\frac{4+3i}{(7-2i)(5+4i)}$

2. barrycarter
3. 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

4. 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 $\frac{-6+57i}{2173}$

5. anonymous

Thanks man!!!!

6. 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 :(

7. anonymous

It is fine man!