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Aus Online Mathematik Brückenkurs 2

If we subtract 2z from both sides,

iz+2−2z=−3

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

iz−2z=−3−2.

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

(i−2)z=−5

we obtain, after dividing by −2+i,

z=−5−2+i=−5(−2−i)(−2+i)(−2−i)=(−2)2−i2(−5)(−2)−5(−i)=4+110+5i=510+5i=2+i.

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

LHSRHS=iz+2=i(2+i)+2=2i−1+2=1+2i=2z−3=2(2+i)−3=4+2i−3=1+2i.