Lösung 3.1:4e

If we multiply both sides by z+i, we avoid having z in the denominator:

\displaystyle iz+1=(3+i)(z+i).

At the same time, this means that we are now working with a new equation which is not necessarily entirely equivalent with the original equation. Should the new equation show itself to have z=-i as a solution, then we must ignore that solution, because our initial equation cannot possibly have \displaystyle z=-i as a solution (the denominator of the left-hand side becomes zero).

We expand the right-hand side in the new equation,

\displaystyle iz+1=3z+3i+iz-1

and move all the terms in z over to the left-hand side and the constants to the right-hand side:

\displaystyle \begin{align}iz-3z-iz &= 3i-1-1,\\ -3z &= -2+3i.\end{align}

Then, we obtain

\displaystyle z=\frac{-2+3i}{-3}=\frac{2}{3}-i.

It is a little troublesome to divide two complex numbers, so we will therefore not check whether \displaystyle z=\frac{2}{3}-i is a solution to the original equation, but satisfy ourselves with substituting into the equation \displaystyle iz+1=(3+i)(z+i):

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