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Aus Online Mathematik Brückenkurs 2

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 root.

We complete the square on the left-hand side:

z−1+i2−1+i2+2i−1=0z−1+i2−1+2i+i2+2i−1=0z−1+i2−1−2i+1+2i−1=0z−1+i2−1=0

Now, we see that the equation has the solutions

z−1+i=12+ii  

We test the solutions:

z=2+i:z2−21+iz+2i−1=2+i2−21+i2+i+2i−1=4+4i+i2−22+i+2i+i2+2i−1=4+4i−1−4−6i+2+2i−1=0

z=i:z2−21+iz+2i−1=i2−21+ii+2i−1=−1−2i+i2+2i−1=−1−2i+2+2i−1=0