Solution 3.1:4d

In the equation, \displaystyle z occurs only as \displaystyle \bar{z} and, to begin with, we can therefore treat \displaystyle \bar{z} as unknown.

Divide both sides by \displaystyle 2+i,

\displaystyle \bar{z}=\frac{1+i}{2+i}

and calculate the quotient on the right-hand side by multiplying top and bottom by the complex conjugate of the numerator:

\displaystyle \begin{align}\bar{z}&=\frac{(1+i)(2-i)}{(2+i)(2-i)}=\frac{1\cdot 2-1\cdot i +i \cdot 2 - i\cdot i}{2^2-i^2}\\ &=\frac{2-i+2i+1}{4+1}=\frac{3+i}{5}=\frac{3}{5}+\frac{1}{5}i.\end{align}

This means that \displaystyle z=\frac{3}{5}-\frac{1}{5}i.

We check that \displaystyle z=\frac{3}{5}-\frac{1}{5}i satisfies the original equation:

\displaystyle \begin{align}LHS &= (2+i)\bar{z} = (2+i)\overline{(\frac{3}{5}-\frac{1}{5}i)}=(2+i)(\frac{3}{5}+\frac{1}{5}i)\\ &=2\cdot\frac{3}{5}+2\cdot\frac{1}{5}i+i\cdot\frac{3}{5}+i\cdot\frac{1}{5}i=\frac{6}{5}+\frac{2}{5}i+\frac{3}{5}i-\frac{1}{5}\\ &=\frac{6-1}{5}+\frac{2+3}{5}i=1+i= RHS\end{align}