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Question

Find the value of x and y for which (2+3i)x2(32i)y=2x3y+5i where x,yϵR.

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Solution

(2+3i)x2(32i)y=2x3y+5i
2x23y=2x3y
x2x=0
x=0,1 and 3x2+2y=5
if x=0,y=52 and if x=1,y=1
x=0,y=52 and x=1,y=1
are two solutions of the given equation which can also be represented as (0,52) & (1,1),(0,52)(1,1)

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