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Aus Online Mathematik Brückenkurs 2

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

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 z=−i as a solution (the denominator of the left-hand side becomes zero).

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

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,

iz−3z−iz−3z=3i−1−1=−2+3i.

Then, we obtain

z=−3−2+3i=32−i.

It is a little troublesome to divide two complex numbers, so we will therefore not check whether z=32−i is a solution to the original equation, but satisfy ourselves with substituting into the equation iz+1=(3+i)(z+i),

LHSRHS=iz+1=i(32−i)+1=32i+1+1=2+32i=(3+i)(z+i)=(3+i)(32−i+i)=(3+i)32=2+32i.