If complex number z satisfies |z|+z=2+i, then z is
A
z=34+2i
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B
z=34+i
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C
z=1+i
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D
z=35+i
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Solution
The correct option is Bz=34+i Let z=x+iy ⇒|z|+z=(2+i)⇒√x2+y2+x+iy=(2+i)
By comparing real and imaginary part, we get √x2+y2+x=2 and y=1 ⇒(2−x)2=x2+y2⇒(2−x)2=x2+1⇒x=34∴y=1,x=34