Lösung 3.2:1b

We can easily calculate z+u and z−u,

z+uz−u=2+i+(−1−2i)=2−1+(1−2)i=1−i=2+i−(−1−2i)=2+1+(1+2)i=3+3i

and then mark them on the complex plane.

An alternative is to view z and u as vectors and z+u as a vector addition of z and u.

We can either view the vector subtraction z−u as z+(−u),

or interpret z−u from the vector relation

z=(z−u)+u

i.e. z−u is the vector we add to u to arrive at z.