SOLUTION: Put in standard form {{{(6+5i)/(3i-2)}}}

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Put in standard form {{{(6+5i)/(3i-2)}}}       Log On


   

Question 1068402: Put in standard form
%286%2B5i%29%2F%283i-2%29
Answer by Lightning_Fast(78) About Me  (Show Source):
You can put this solution on YOUR website!
= +%286%2B5i%29%2F%283i-2%29
= +%28%286%2B5i%29%2F%283i-2%29%29%2A%28%283i%2B2%29%2F%283i%2B2%29%29
= %28%286%2B5i%29%2A%283i%2B2%29%29%2F%28%283i-2%29%2A%283i%2B2%29%29 Rationalizing
= %2818i%2B12%2B15i%5E2%2B10i%29%2F%289i%5E2-4%29+
= %2815i%5E2%2B28i%2B12%29%2F%289i%5E2-4%29+
= %2815%28-1%29%2B28i%2B12%29%2F%289%28-1%29-4%29+
= %2828i-3%29%2F%28-13%29+
By standard form, you mean put in format of a+bi?
= %283%2F13%29-%2828i%2F13%29+