Processing Math: Done

Aus Online Mathematik Brückenkurs 2

Version vom 15:20, 30. Okt. 2008 (bearbeiten) Tek (Diskussion | Beiträge) K ← Zum vorherigen Versionsunterschied		Version vom 13:13, 10. Mär. 2009 (bearbeiten) (rückgängig) Tekbot (Diskussion | Beiträge) K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)) Zum nächsten Versionsunterschied →
Zeile 3:		Zeile 3:
	We complete the square on the left-hand side,		We complete the square on the left-hand side,
-	{{Displayed math||<math>\begin{align}	+	{{Abgesetzte Formel||<math>\begin{align}
	(z-(1+i))^2-(1+i)^2+2i-1 &= 0\,,\\[5pt]		(z-(1+i))^2-(1+i)^2+2i-1 &= 0\,,\\[5pt]
	(z-(1+i))^2-(1+2i+i^2)+2i-1&=0\,,\\[5pt]		(z-(1+i))^2-(1+2i+i^2)+2i-1&=0\,,\\[5pt]
Zeile 12:		Zeile 12:
	Now, we see that the equation has the solutions		Now, we see that the equation has the solutions
-	{{Displayed math||<math>z-(1+i) = \pm 1\quad \Leftrightarrow \quad z=\left\{ \begin{align}	+	{{Abgesetzte Formel||<math>z-(1+i) = \pm 1\quad \Leftrightarrow \quad z=\left\{ \begin{align}
	&2+i\,,\\		&2+i\,,\\
	&i\,\textrm{.}		&i\,\textrm{.}

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Version vom 13:13, 10. 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.