Lösung 3.1:4a

A general strategy when solving equations is to try to get the unknown variable by itself on one side.

In this case, we start by subtracting z from both sides,

\displaystyle z+3i-z=2z-2-z.

Then we have a z left on the right-hand side,

\displaystyle 3i=z-2.

We add \displaystyle 2 to both sides to remove the \displaystyle -2 from the right hand side

\displaystyle 3i+2=z-2+2

and after that we can just read off the solution:

\displaystyle 2+3i=z

To check that we have calculated correctly, we substitute z=2+3i into the original equation and see that it is satisfied:

\displaystyle \begin{align}LHS &= z +3i = 2+3i+3i=2+6i,\\ RHS &= 2z-2 = 2(2+3i)-2 = 4 + 6i -2 = 2+6i.\end{align}