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Lösung 3.3:4c

Aus Online Mathematik Brückenkurs 2

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&= (-1)^2 + 2\cdot i\sqrt{2} + i^2\bigl(\sqrt{2}\,\bigr)^2 - 2 - 2i\sqrt{2} + 3\\[5pt]
&= (-1)^2 + 2\cdot i\sqrt{2} + i^2\bigl(\sqrt{2}\,\bigr)^2 - 2 - 2i\sqrt{2} + 3\\[5pt]
&= 1+2\cdot i\sqrt{2} - 2 - 2 - 2\sqrt{2}i + 3\\[5pt]
&= 1+2\cdot i\sqrt{2} - 2 - 2 - 2\sqrt{2}i + 3\\[5pt]
-
&= 0.
+
&= 0\,\textrm{.}
\end{align}</math>
\end{align}</math>

Version vom 14:44, 30. Okt. 2008

We complete the square on the left-hand side,

(z+1)2−12+3(z+1)2+2=0=0.

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

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

z=−1+i2:z2+2z+3z=−1−i2:z2+2z+3=−1+i22+2−1+i2+3=(−1)2−2i2+i222−2+2i2+3=1−2i2−2−2+2i2+3=0=−1−i22+2−1−i2+3=(−1)2+2i2+i222−2−2i2+3=1+2i2−2−2−22i+3=0.

Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.3:4c“