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Lösung 3.3:5a

Aus Online Mathematik Brückenkurs 2

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Even if the equation contains complex numbers as coefficients, we treat is as an ordinary second-degree equation and solve it by completing the square taking the square root.

We complete the square on the left-hand side,

(z(1+i))2(1+i)2+2i1(z(1+i))2(1+2i+i2)+2i1(z(1+i))212i+1+2i1(z(1+i))21=0=0=0=0.

Now, we see that the equation has the solutions

z(1+i)=1z=2+ii. 

We test the solutions,

z=2+i:z22(1+i)z+2i1z=i:z22(1+i)z+2i1=(2+i)22(1+i)(2+i)+2i1=4+4i+i22(2+i+2i+i2)+2i1=4+4i146i+2+2i1=0=i22(1+i)i+2i1=12(i+i2)+2i1=12i+2+2i1=0.