Better Students Ask More Questions.
Find the real number x and y in the expression ( 1 + i )( x-1 ) + ( 1 - i )( y + 1 )...
2 Answers | add yours
x=0 ; y=1
Posted by deviselva75 on April 4, 2009 at 5:26 PM (Answer #1)
You need to open the brackets:
x - 1 + ix - i + y + 1 - iy - i = 1 - 3i
You need to rearrange the terms from the left side such that:
(x - 1 + y + 1) + i(x - 1 - y - 1) = 1 - 3i
Reducing the opposite terms inside the brackets yields:
x + y + i(x - y - 2) = 1 - 3i
Equating real parts yields:
x + y = 1
Equating imaginary parts yields:
x - y - 2 = -3 => x - y = -3+2 => x - y = -1
Adding the first equation to the second yields:
x + y + x - y = 1 - 1 => 2x = 0 => x = 0 => y = 1
The solution to the given identity is x = 0 ; y = 1.
Posted by sciencesolve on December 12, 2011 at 10:57 PM (Answer #2)
Join to answer this question
Join a community of thousands of dedicated teachers and students.