Solution 3.1:1d

From Förberedande kurs i matematik 2

(Difference between revisions)
Jump to: navigation, search
m (Lösning 3.1:1d moved to Solution 3.1:1d: Robot: moved page)
Current revision (14:54, 29 October 2008) (edit) (undo)
m
 
(One intermediate revision not shown.)
Line 1: Line 1:
-
{{NAVCONTENT_START}}
+
We expand the expression by multiplying each term in the first bracket with every term in the second bracket,
-
<center> [[Image:3_1_1d.gif]] </center>
+
 
-
{{NAVCONTENT_STOP}}
+
{{Displayed math||<math>\begin{align}
 +
(3-2i)(7+5i)
 +
&= 3\cdot 7 + 3 \cdot 5i + \cdots\\[5pt]
 +
&= 3\cdot 7 + 3 \cdot 5i - 2i\cdot 7 - 2i \cdot 5i\,\textrm{.}
 +
\end{align}</math>}}
 +
 
 +
Then, we use that <math>i^2=-1</math> and write the real and imaginary parts together,
 +
 
 +
{{Displayed math||<math>\begin{align}
 +
(3-2i)(7+5i)
 +
&=21+15i-14i-10i^2\\[5pt]
 +
&=21+15i-14i-10\cdot(-1)\\[5pt]
 +
&=(21+10)+(15i-14i)\\[5pt]
 +
&=31+(15-14)i\\[5pt]
 +
&=31+i\,\textrm{.}
 +
\end{align}</math>}}

Current revision

We expand the expression by multiplying each term in the first bracket with every term in the second bracket,

\displaystyle \begin{align}

(3-2i)(7+5i) &= 3\cdot 7 + 3 \cdot 5i + \cdots\\[5pt] &= 3\cdot 7 + 3 \cdot 5i - 2i\cdot 7 - 2i \cdot 5i\,\textrm{.} \end{align}

Then, we use that \displaystyle i^2=-1 and write the real and imaginary parts together,

\displaystyle \begin{align}

(3-2i)(7+5i) &=21+15i-14i-10i^2\\[5pt] &=21+15i-14i-10\cdot(-1)\\[5pt] &=(21+10)+(15i-14i)\\[5pt] &=31+(15-14)i\\[5pt] &=31+i\,\textrm{.} \end{align}

Retrieved from "http://wiki.sommarmatte.se/wikis/2008/bridgecourse2-ImperialCollege/index.php/Solution_3.1:1d"