<s>[INST] Find the real and imaginary parts of an equation 
Find the real and imaginary parts of $\frac{1}{3z+2}$

So I have expanded it out to get $\frac{1}{3x+3iy+2}$ 
Thus giving $Re(\frac{1}{3z+2})=\frac{1}{3x+2}$ and $Im(\frac{1}{3z+2})=\frac{1}{3y}$
However in my answer book it says: $Re(\frac{1}{3z+2})=\frac{3x+2}{(3x+2)^2+9y^2}$ and $Im(\frac{1}{3z+2})=\frac{-3y}{(3x+2)^2+9y^2}$
Is the book incorrect/outdated or if not could someone explain how to gain these answers, thanks
 [/INST] you find it as the form $z=a+ib$ so
$$z=\frac { 1 }{ 3x+3iy+2 } =\frac { 3x+2-3iy }{ \left( 3x+2+3iy \right) \left( 3x+2-3iy \right)  } =\frac { 3x+2 }{ { \left( 3x+2 \right)  }^{ 2 }+9{ y }^{ 2 } } +i\frac { -3y }{ { \left( 3x+2 \right)  }^{ 2 }+9{ y }^{ 2 } } \\  $$
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