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Question

If k+k+z2=|z|2(kR), then possible argument of z is

A
0
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B
π
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C
π2
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D
none of these
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Solution

The correct option is C π2
k+k+z2=|z|2(kR) ...(1)
Let z=r(cosθ+isinθ)
Substitute z in equation (1)
k+k+r2(cosθ+isinθ)2=r2(k+r2cos2θ)2+(r2sin2θ)2=(r2k)2k2+r4(sin22θ+cos22θ)+2kr2cos2θ=k2+r42kr2cos2θ=1
One of many solution of above equation is 2θ=π.
θ=π2
Hence, option 'C' is correct.

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