Lösung 3.3:4c

Aus Online Mathematik Brückenkurs 2

We complete the square on the left-hand side,

(z+1)2−12+3(z+1)2+2=0=0.

Taking the root now gives z+1=i2 , i.e. z=−1+i2  and z=−1−i2 .

We test the solutions in the equation to ascertain that we have calculated correctly.

z=−1+i2:z2+2z+3z=−1−i2:z2+2z+3=−1+i22+2−1+i2+3=(−1)2−2i2+i222−2+2i2+3=1−2i2−2−2+2i2+3=0=−1−i22+2−1−i2+3=(−1)2+2i2+i222−2−2i2+3=1+2i2−2−2−22i+3=0.