Lösung 3.2:2e

Aus Online Mathematik Brückenkurs 2

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

• Rez=x
• i+z=i+(x−iy)=x+(1−y)i

and the condition becomes

x=x+(1−y)i0=(1−y)i

which means that y=1.

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