Question

Simplify the expression. -6+1/-5+i.

Answer 1

Multiply both denominator and numerator with the complex conjugate of (-5+i)

Answer 2

So I multiply it all by (-5+i)?

Answer 3

Complex conjugate of (-5+i) is (-5-i)

Answer 4

Im still confused...could you explain in more detail?

Answer 5

The first step of simplification is probably find a way to move the imaginary number from denominator to numerator, so to do so, we use that fact that \[(a+ib)(a-ib)=a^2-(ib)^2=a^2+b^2\]

Answer 6

Btw, do you mean \[\frac{-6+i}{-5+i}\]instead of\[\frac{-6+1}{-5+i}\]?

Answer 7

Yea lol thats what i meant . Just typed it wrong.

Answer 8

So to simplify, we gotta remove imaginary numbers from denominator. Hence we do the following\[\frac{-6+i}{-5+i}=\frac{-6+i}{-5+i}\frac{-5-i}{-5-i}=\frac{(-6+i)(-5-i)}{(-5)^2-(i)^2}\]

Answer 9

So then once simplified we mutlipy all the numerator together?

Answer 10

Yeah

Answer 11

so -6*-5 is 30 and how would you multiply i would it just cancel out cause one is negative and one is positive?

Answer 12

Yep

Answer 13

er well 30/25-(i)^2?

Answer 14

but the answers are 31+i/26, 31+i/24, 29+i/26, and 31+11i/26...

Answer 15

the denominator should be \(26\) because \(5^2+1^2=25+1=26\)

Answer 16

the numerator is \((-6+i)(-5-i)\)

Answer 17

multiply this out and get \((30+1)+(6-5)i\) or \(31+i\)

Answer 18

so i is like a fake 1?

Answer 19

lol

Answer 20

\(i=\sqrt{-1}\) i.e. \(i^2=-1\)

Answer 21

Thank you! :)

Answer 22

\(a+bi)(a-bi)=a^2+b^2\) is what you use for the denominator
in \(-5+i\) you have \(a=-5,b=1\)

Answer 23

yw