If α,β∈R are such that 1−2i (here i2=−1) is a root of z2+αz+β=0, then (α−β) is equal to
A
7
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B
−3
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C
3
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D
−7
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Solution
The correct option is D−7 1−2i is a root of z2+αz+β=0.
So, (1−2i)2+α(1−2i)+β=0 ⇒1−4−4i+α−2iα+β=0 ⇒(α+β−3)−i(4+2α)=0 ⇒α+β−3=0 and 4+2α=0
So, α=−2,β=5 ∴α−β=−7