If z satisfies the equation |z| -z=1+2i, then z is equal to
A
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B
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C
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D
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Solution
The correct option is B Let z = x + iy ∴|z|−z=1+2i ⇒√x2+y2−(x+iy)=1+2i ⇒(√x2+y2−x)+i(−y)=1+2i ⇒√x2+y2−x=1andy=−2 Ify=−2,√x2+4−x=1 ⇒x2+4=(1+x)2 ⇒2x=3;x=32 ∴z=x+iy=32−2i