Multiplying and Simplifying Complex BinomialsDate: 8/19/96 at 1:40:24 From: Anonymous Subject: Multiplying and Simplifying Complex Binomials Multiply and simplify: (2+3i)(5-i) I could not see how they simplified this to what they say is the answer. Could you help? I used FOIL to get: 10-2i+15i-3i^2 I then simplified to get: 10+13i-3i^2 The program said it could be simplified to: 13+13i The value of (i) was not given but the problem said that (i^2) = -1. I plugged numbers in for (i) to work out this equation with their answer and the orginial binomial and it didn't work. What did I miss? Thanks. Date: 8/19/96 at 2:42:18 From: Doctor Paul Subject: Re: Multiplying and Simplifying Complex Binomials You've almost got it right. Let's first define 'i'. i = sqrt(-1) so i^2 = -1 You have: 10 + 13i - 3i^2 Let's substitute for i^2... 10 + 13i - 3(-1) 10 + 13i + 3 and you can see how they got their answer. 13 + 13i I hope this clears things up for you. If you have any more questions, feel free to submit them to Dr. Math. Regards, -Doctor Paul, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/