Lösung 3.1:4c

If we subtract \displaystyle 2z from both sides,

\displaystyle iz+2-2z=-3

and then subtract \displaystyle 2 from both sides, we have \displaystyle z terms left only on the left-hand side,

\displaystyle iz-2z=-3-2

After taking out a factor \displaystyle z from the left-hand side,

\displaystyle (i-2)z=-5

we obtain, after dividing by \displaystyle -2+i,

\displaystyle \begin{align}z&=\frac{-5}{-2+i}=\frac{-5(-2-i)}{(-2+i)(-2-i)}=\frac{(-5)\cdot(-2)-5\cdot(-i)}{(-2)^2-i^2}\\ &=\frac{10+5i}{4+1}=\frac{10+5i}{5}=2+i.\end{align}

A quick check shows that \displaystyle z=2+i satisfies the original equation

\displaystyle \begin{align}LHS &= iz+2=i(2+i)+2=2i-1+2=1+2i\\ RHS &= 2z-3 = 2(2+i)-3=4+2i-3=1+2i.\end{align}