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Question
Answer the following:
Find the real numbers x and y such that `x/(1 + 2"i") + y/(3 + 2"i") = (5 + 6"i")/(-1 + 8"i")`
Solution
`x/(1 + 2"i") + y/(3 + 2"i") = (5 + 6"i")/(-1 + 8"i")`
∴ `((3 + 2"i")x + (1 + 2"i")y)/((1 + 2"i")(3 + 2"i")) = (5 + 6"i")/(-1 + 8"i")`
∴ `((3 + 2"i")x + (1 + 2"i")y)/(3 + 2"i" + 6"i" + 4"i"^2) = (5 + 6"i")/(-1 + 8"i")`
∴ `((3 + 2"i")x + (1 + 2"i")y)/(3 + 8"i" - 4) = (5 + 6"i")/(-1 + 8"i")` ...[∵ i2 = – 1]
∴ `(3x + 2"i"x + y + 2"i"y)/(-1 + 8"i") = (5 + 6"i")/(-1 + 8"i")`
∴ (3x + y) + (2x + 2y)i = 5 + 6i
Equating the real and imaginary parts separately, we get,
3x + y = 5 ...(1)
and 2x + 2y = 6
i.e., x + y = 3 ...(2)
On subtracting, we get,
2x = 2
∴ x = 1
∴ from (2), 1 + y = 3
∴ y = 2
Hence, x = 1, y = 2.
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