Question 133839

7/(3-5i)  First, we'll multiply the numerator and denominator by the conjugate of 3-5i which is 3+5i.  This will get rid of the radical in the denominator.

(7(3+5i))/((3-5i)(3+5i)) =
(7(3+5i))/(9-25(i^2)) but i^2 is -1, so

(7(3+5i))/(9+25)=
7(3+5i)/34


Hope this helps---ptaylor