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 Find the real number x and y in the expression ( 1 + i )( x-1 ) + ( 1 - i )( y + 1 )...

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jjverrite | Student, Grade 10 | eNotes Newbie

Posted April 1, 2009 at 1:25 AM via web

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 Find the real number x and y in the expression

( 1 + i )( x-1 ) + ( 1 - i )( y + 1 ) = 1 - 3i

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deviselva75 | College Teacher | eNotes Newbie

Posted April 4, 2009 at 5:26 PM (Answer #1)

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x=0   ; y=1

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted December 12, 2011 at 10:57 PM (Answer #2)

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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.

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