meganlee1
meganlee1 meganlee1
  • 17-01-2017
  • Mathematics
contestada

What is the standard form of (5+i)(6-5i)/ 2i (-3+7i) ?

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redxxangel
redxxangel redxxangel
  • 17-01-2017
Since all parts of 2i(-3+7i) is in the  denominator, you have to use an  extra set of parentheses around the whole denominator. this is what you do

(5+i)(6−5i) / (2i(−3+7i)) 
= (30−25i+6i−5i^2) / (−6i+14i^2) 
= (35−19i) / (−14−6i) 
= (35−19i)(−7+3i) / (2(−7−3i)(−7+3i)) 
= (−245+105i+133i−57i^2) / (2(49−9i^2)) 
= (−188+238i) / (2(58)) 
= (−188+238i) / 116 
= −47/29 + 119/58 i
so that is how you do it

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