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Lösung 3.3:3d

Aus Online Mathematik Brückenkurs 2

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 2: Zeile 2:
<math>i</math> in front of <math>z^2</math>,
<math>i</math> in front of <math>z^2</math>,
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{{Displayed math||<math>i\Bigl(z^2+\frac{2+3i}{i}z-\frac{1}{i}\Bigr)\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>i\Bigl(z^2+\frac{2+3i}{i}z-\frac{1}{i}\Bigr)\,\textrm{.}</math>}}
Then, simplify the complex fractions by multiplying top and bottom by <math>-i</math> (the denominator's complex conjugate),
Then, simplify the complex fractions by multiplying top and bottom by <math>-i</math> (the denominator's complex conjugate),
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
i\Bigl(z^2+\frac{(2+3i)\cdot (-i)}{i\cdot (-i)}z-\frac{1\cdot (-i)}{i\cdot (-i)}\Bigr)
i\Bigl(z^2+\frac{(2+3i)\cdot (-i)}{i\cdot (-i)}z-\frac{1\cdot (-i)}{i\cdot (-i)}\Bigr)
&= i\Bigl(z^2+\frac{-2i+3}{1}z-\frac{-i}{1}\Bigr)\\[5pt]
&= i\Bigl(z^2+\frac{-2i+3}{1}z-\frac{-i}{1}\Bigr)\\[5pt]
Zeile 14: Zeile 14:
Now we are ready to complete the square of the second-degree expression inside the bracket,
Now we are ready to complete the square of the second-degree expression inside the bracket,
-
{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
i\bigl(z^2+(3-2i)z+i\bigr)
i\bigl(z^2+(3-2i)z+i\bigr)
&= i\Bigl(\Bigl(z+\frac{3-2i}{2}\Bigr)^2 - \Bigl(\frac{3-2i}{2}\Bigr)^2+i\Bigr)\\[5pt]
&= i\Bigl(\Bigl(z+\frac{3-2i}{2}\Bigr)^2 - \Bigl(\frac{3-2i}{2}\Bigr)^2+i\Bigr)\\[5pt]

Version vom 13:12, 10. Mär. 2009

Before we can complete the square of the expression, we need to take out the factor i in front of z2,

iz2+i2+3iz−i1. 

Then, simplify the complex fractions by multiplying top and bottom by −i (the denominator's complex conjugate),

iz2+i(−i)(2+3i)(−i)z−i(−i)1(−i)=iz2+1−2i+3z−1−i=iz2+(3−2i)z+i.

Now we are ready to complete the square of the second-degree expression inside the bracket,

iz2+(3−2i)z+i=iz+23−2i2−23−2i2+i=iz+23−i2−23−i2+i=iz+23−i2−49+3i−i2+i=iz+23−i2−45+4i=iz+23−i2−45i+4i2=iz+23−i2−4−45i.
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.3:3d“