Processing Math: Done
To print higher-resolution math symbols, click the
Hi-Res Fonts for Printing button on the jsMath control panel.
jsMath

Lösung 3.1:4c

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 1: Zeile 1:
If we subtract <math>2z</math> from both sides,
If we subtract <math>2z</math> from both sides,
-
{{Displayed math||<math>iz+2-2z=-3</math>}}
+
{{Abgesetzte Formel||<math>iz+2-2z=-3</math>}}
and then subtract <math>2</math> from both sides, we have <math>z</math> terms left only on the left-hand side,
and then subtract <math>2</math> from both sides, we have <math>z</math> terms left only on the left-hand side,
-
{{Displayed math||<math>iz-2z=-3-2\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>iz-2z=-3-2\,\textrm{.}</math>}}
After taking out a factor <math>z</math> from the left-hand side,
After taking out a factor <math>z</math> from the left-hand side,
-
{{Displayed math||<math>(i-2)z=-5\,,</math>}}
+
{{Abgesetzte Formel||<math>(i-2)z=-5\,,</math>}}
we obtain, after dividing by <math>-2+i</math>,
we obtain, after dividing by <math>-2+i</math>,
-
{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
z &= \frac{-5}{-2+i}
z &= \frac{-5}{-2+i}
= \frac{-5(-2-i)}{(-2+i)(-2-i)}
= \frac{-5(-2-i)}{(-2+i)(-2-i)}
Zeile 23: Zeile 23:
A quick check shows that <math>z=2+i</math> satisfies the original equation,
A quick check shows that <math>z=2+i</math> satisfies the original equation,
-
{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\text{LHS} &= iz+2 = i(2+i)+2 = 2i-1+2 = 1+2i\,,\\[5pt]
\text{LHS} &= iz+2 = i(2+i)+2 = 2i-1+2 = 1+2i\,,\\[5pt]
\text{RHS} &= 2z-3 = 2(2+i)-3 = 4+2i-3 = 1+2i\,\textrm{.}
\text{RHS} &= 2z-3 = 2(2+i)-3 = 4+2i-3 = 1+2i\,\textrm{.}
\end{align}</math>}}
\end{align}</math>}}

Version vom 13:06, 10. Mär. 2009

If we subtract 2z from both sides,

iz+2−2z=−3

and then subtract 2 from both sides, we have z terms left only on the left-hand side,

iz−2z=−3−2.

After taking out a factor z from the left-hand side,

(i−2)z=−5

we obtain, after dividing by −2+i,

z=−5−2+i=−5(−2−i)(−2+i)(−2−i)=(−2)2−i2(−5)(−2)−5(−i)=4+110+5i=510+5i=2+i.

A quick check shows that z=2+i satisfies the original equation,

LHSRHS=iz+2=i(2+i)+2=2i−1+2=1+2i=2z−3=2(2+i)−3=4+2i−3=1+2i.
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.1:4c“