If z=1+2icosθ4+3isinθ, where θ∈(0,π2), then which of the following is/are INCORRECT?
A
there exist one value of θ for which Re(z)=0
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B
there exist 2 values of θ for which Re(z)=0
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C
there exist one value of θ for which Im(z)=0
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D
there exist 2 values of θ for which Im(z)=0
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Solution
The correct options are A there exist one value of θ for which Re(z)=0 B there exist 2 values of θ for which Re(z)=0 D there exist 2 values of θ for which Im(z)=0 Given : z=1+2icosθ4+3isinθ Now, z=1+2icosθ4+3isinθ×4−3isinθ4−3isinθ⇒z=(4+6sinθcosθ)+i(8cosθ−3sinθ)16+9sin2θ
When Re(z)=0, then 4+6sinθcosθ16+9sin2θ=0⇒4+3sin2θ=0⇒sin2θ=−43 Therefore Re(z)=0 not possible for any θ∈(0,π2)
When Im(z)=0, then 8cosθ−3sinθ16+9sin2θ=0⇒8cosθ−3sinθ=0⇒tanθ=83 Therefore there exist 1 value of θ∈(0,π2)