Aus Online Mathematik Brückenkurs 2

Because the expression contains both \displaystyle z and \displaystyle \bar{z}, we write out \displaystyle z=x+iy, where \displaystyle x is the real part of \displaystyle z and \displaystyle y is the imaginary part. Thus, we have

• \displaystyle \mathop{\rm Re}z = x
• \displaystyle i+\bar{z} = i+(x-iy) = x+(1-y)i

and the condition becomes

\displaystyle x=x+(1-y)i\quad\Leftrightarrow\quad 0=(1-y)i

which means that \displaystyle y=1.

The set therefore consists of all complex numbers with imaginary part 1.