Solution 3.3:5b

From Förberedande kurs i matematik 2

(Difference between revisions)
Jump to: navigation, search
m (Lösning 3.3:5b moved to Solution 3.3:5b: Robot: moved page)
Current revision (15:32, 30 October 2008) (edit) (undo)
m
 
(One intermediate revision not shown.)
Line 1: Line 1:
-
{{NAVCONTENT_START}}
+
Complete the square of the left-hand side,
-
<center> [[Image:3_3_5b.gif]] </center>
+
 
-
{{NAVCONTENT_STOP}}
+
{{Displayed math||<math>\begin{align}
 +
\Bigl(z-\frac{2-i}{2}\Bigr)^2-\Bigl(\frac{2-i}{2}\Bigr)^2+3-i &= 0\,,\\[5pt]
 +
\Bigl(z-\frac{2-i}{2}\Bigr)^2-\Bigl(1-i+\frac{1}{4}i^2\Bigr)+3-i&=0\,,\\[5pt]
 +
\Bigl(z-\frac{2-i}{2}\Bigr)^2-1+i+\frac{1}{4}+3-i&=0\,,\\[5pt]
 +
\Bigl(z-\frac{2-i}{2}\Bigr)^2+\frac{9}{4}&=0\,\textrm{.}
 +
\end{align}</math>}}
 +
 
 +
Taking the square root then gives that the solutions are
 +
 
 +
{{Displayed math||<math>z-\frac{2-i}{2} = \pm\frac{3}{2}\,i\quad \Leftrightarrow \quad z=\left\{ \begin{align}
 +
&1+i\,,\\
 +
&1-2i\,\textrm{.}
 +
\end{align}\right.</math>}}
 +
 
 +
Finally, we substitute the solutions into the equation and check that it is satisfied
 +
 
 +
<math>\begin{align}
 +
z={}\rlap{1+i:}\phantom{1-2i:}{}\quad z^2-(2-i)z+(3-i)
 +
&= (1+i)^2-(2-i)(1+i)+3-i\\[5pt]
 +
&= 1+2i+i^2-(2+2i-i-i^2)+3-i\\[5pt]
 +
&= 1+2i-1-2-i-1+3-i\\[5pt]
 +
&=0\,,\\[10pt]
 +
z=1-2i:\quad z^2-(2-i)z+(3-i)
 +
&= (1-2i)^2-(2-i)(1-2i)+3-i\\[5pt]
 +
&= 1-4i+4i^2-(2-4i-i+2i^2)+3-i\\[5pt]
 +
&= 1-4i-4-2+5i+2+3-i\\[5pt]
 +
&= 0\,\textrm{.}
 +
\end{align}</math>

Current revision

Complete the square of the left-hand side,

\displaystyle \begin{align}

\Bigl(z-\frac{2-i}{2}\Bigr)^2-\Bigl(\frac{2-i}{2}\Bigr)^2+3-i &= 0\,,\\[5pt] \Bigl(z-\frac{2-i}{2}\Bigr)^2-\Bigl(1-i+\frac{1}{4}i^2\Bigr)+3-i&=0\,,\\[5pt] \Bigl(z-\frac{2-i}{2}\Bigr)^2-1+i+\frac{1}{4}+3-i&=0\,,\\[5pt] \Bigl(z-\frac{2-i}{2}\Bigr)^2+\frac{9}{4}&=0\,\textrm{.} \end{align}

Taking the square root then gives that the solutions are

\displaystyle z-\frac{2-i}{2} = \pm\frac{3}{2}\,i\quad \Leftrightarrow \quad z=\left\{ \begin{align}

&1+i\,,\\ &1-2i\,\textrm{.} \end{align}\right.

Finally, we substitute the solutions into the equation and check that it is satisfied

\displaystyle \begin{align} z={}\rlap{1+i:}\phantom{1-2i:}{}\quad z^2-(2-i)z+(3-i) &= (1+i)^2-(2-i)(1+i)+3-i\\[5pt] &= 1+2i+i^2-(2+2i-i-i^2)+3-i\\[5pt] &= 1+2i-1-2-i-1+3-i\\[5pt] &=0\,,\\[10pt] z=1-2i:\quad z^2-(2-i)z+(3-i) &= (1-2i)^2-(2-i)(1-2i)+3-i\\[5pt] &= 1-4i+4i^2-(2-4i-i+2i^2)+3-i\\[5pt] &= 1-4i-4-2+5i+2+3-i\\[5pt] &= 0\,\textrm{.} \end{align}

Retrieved from "http://wiki.sommarmatte.se/wikis/2008/bridgecourse2-ImperialCollege/index.php/Solution_3.3:5b"