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Lösung 3.3:5a

Aus Online Mathematik Brückenkurs 2

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Version vom 10:48, 11. Mär. 2009

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+2i−1(z−(1+i))2−(1+2i+i2)+2i−1(z−(1+i))2−1−2i+1+2i−1(z−(1+i))2−1=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:z2−2(1+i)z+2i−1z=i:z2−2(1+i)z+2i−1=(2+i)2−2(1+i)(2+i)+2i−1=4+4i+i2−2(2+i+2i+i2)+2i−1=4+4i−1−4−6i+2+2i−1=0=i2−2(1+i)i+2i−1=−1−2(i+i2)+2i−1=−1−2i+2+2i−1=0.

Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.3:5a“