Let z be the complex number such that | z |+ z =3+ i where i= -1 , then | z |=

Step-by-step solution

Let z = x + iy where x, y R with modulus |z| = x^2 + y^2 . Given the equation |z| + z = 3 + i , substitute to obtain: x^2 + y^2 + x + iy = 3 + i Equating real and imaginary parts yields y = 1 and x^2 + 1 + x = 3 . Solving the real equation: x^2 + 1 = 3 - x . Squaring both sides gives x^2 + 1 = 9 - 6x + x^2 , which simplifies to 6x = 8 , so x = 4 3 . Thus z = 4 3 + i and |z| = ( 4 3 )^2 + 1^2 = 16 9 + 9 9 = 25 9 = 5 3 .