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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Is the book incorrect/outdated or if not could someone explain how to gain these answers, thanks

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  • $\begingroup$ Book is correct you have to make it of form $x + iy$ $\endgroup$ – Aakash Kumar Jul 27 '16 at 14:02
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Note quite, remember that $\frac{1}{a} + \frac{1}{b} \neq \frac{1}{a+b}$. In general, the real and imaginary part of $z$ is $x$ and $y$ where $z = x+iy$. That is, you must seek to turn your expression for $z$ into the form $x+iy$. Once you have $\frac{1}{3x + 3iy + 2}$ (excellent work!) multiply the fraction by the conjugate of the denominator. In general $$\frac{1}{a+ib} = \frac{a-ib}{(a+ib)(a-ib)} = \frac{a-ib}{a^2 + b^2}.$$

In your case, try $$\frac{1}{(3x+2) +3iy} \times \frac{(3x+2) - 3iy}{(3x+2) - 3iy} = \frac{3x+2 - 3iy}{(3x+2) + 9y^2}$$

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Your book is correct because in general $\frac1{a+b}\ne \frac 1a+\frac 1b$.

To solve the problem, you might have tried $$\frac1{3z+2} =\frac{3\bar z+2}{(3z+2)(3\bar z+2)}=\frac{3\bar z+2}{|3z+2|^2}$$

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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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