Solution 3.3:5a

From Förberedande kurs i matematik 2

(Difference between revisions)
Jump to: navigation, search
(Ny sida: {{NAVCONTENT_START}} <center> Bild:3_3_5a.gif </center> {{NAVCONTENT_STOP}})
Current revision (15:20, 30 October 2008) (edit) (undo)
m
 
(3 intermediate revisions not shown.)
Line 1: Line 1:
-
{{NAVCONTENT_START}}
+
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.
-
<center> [[Bild:3_3_5a.gif]] </center>
+
 
-
{{NAVCONTENT_STOP}}
+
We complete the square on the left-hand side,
 +
 
 +
{{Displayed math||<math>\begin{align}
 +
(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+1+2i-1 &= 0\,,\\[5pt]
 +
(z-(1+i))^2-1 &= 0\,\textrm{.}
 +
\end{align}</math>}}
 +
 
 +
Now, we see that the equation has the solutions
 +
 
 +
{{Displayed math||<math>z-(1+i) = \pm 1\quad \Leftrightarrow \quad z=\left\{ \begin{align}
 +
&2+i\,,\\
 +
&i\,\textrm{.}
 +
\end{align}\right.</math>}}
 +
 
 +
We test the solutions,
 +
 
 +
<math>\begin{align}
 +
z=2+i:\quad z^2-2(1+i)z+2i-1
 +
&= (2+i)^2 - 2(1+i)(2+i)+2i-1\\[5pt]
 +
&= 4+4i+i^2-2(2+i+2i+i^2)+2i-1\\[5pt]
 +
&= 4+4i-1-4-6i+2+2i-1\\[5pt]
 +
&= 0\,,\\[10pt]
 +
z={}\rlap{i:}\phantom{2+i:}{}\quad z^2-2(1+i)z+2i-1
 +
&= i^2-2(1+i)i+2i-1\\[5pt]
 +
&= -1-2(i+i^2)+2i-1\\[5pt]
 +
&= -1-2i+2+2i-1\\[5pt]
 +
&= 0\,\textrm{.}
 +
\end{align}</math>

Current revision

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,

\displaystyle \begin{align}

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

Now, we see that the equation has the solutions

\displaystyle z-(1+i) = \pm 1\quad \Leftrightarrow \quad z=\left\{ \begin{align}

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

We test the solutions,

\displaystyle \begin{align} z=2+i:\quad z^2-2(1+i)z+2i-1 &= (2+i)^2 - 2(1+i)(2+i)+2i-1\\[5pt] &= 4+4i+i^2-2(2+i+2i+i^2)+2i-1\\[5pt] &= 4+4i-1-4-6i+2+2i-1\\[5pt] &= 0\,,\\[10pt] z={}\rlap{i:}\phantom{2+i:}{}\quad z^2-2(1+i)z+2i-1 &= i^2-2(1+i)i+2i-1\\[5pt] &= -1-2(i+i^2)+2i-1\\[5pt] &= -1-2i+2+2i-1\\[5pt] &= 0\,\textrm{.} \end{align}

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