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Aus Online Mathematik Brückenkurs 2

Complete the square of the left-hand side,

z−22−i2−22−i2+3−iz−22−i2−1−i+41i2+3−iz−22−i2−1+i+41+3−iz−22−i2+49=0=0=0=0.

Taking the square root then gives that the solutions are

z−22−i=23iz=1+i1−2i.

Finally, we substitute the solutions into the equation and check that it is satisfied

z=1+i:z2−(2−i)z+(3−i)z=1−2i:z2−(2−i)z+(3−i)=(1+i)2−(2−i)(1+i)+3−i=1+2i+i2−(2+2i−i−i2)+3−i=1+2i−1−2−i−1+3−i=0=(1−2i)2−(2−i)(1−2i)+3−i=1−4i+4i2−(2−4i−i+2i2)+3−i=1−4i−4−2+5i+2+3−i=0.