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Lösung 3.3:4c

Aus Online Mathematik Brückenkurs 2

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We complete the square on the left-hand side,

(z+1)212+3(z+1)2+2=0=0.

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

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

z=1+i2:z2+2z+3z=1i2:z2+2z+3=1+i22+21+i2+3=(1)22i2+i2222+2i2+3=12i222+2i2+3=0=1i22+21i2+3=(1)2+2i2+i22222i2+3=1+2i22222i+3=0