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Lösung 3.1:4f

Aus Online Mathematik Brückenkurs 2

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 1: Zeile 1:
Because the equation contains both <math>z</math> and <math>\bar{z}</math>, we cannot use <math>z</math> (or <math>\bar{z}</math>) alone as an unknown, so we are forced to set
Because the equation contains both <math>z</math> and <math>\bar{z}</math>, we cannot use <math>z</math> (or <math>\bar{z}</math>) alone as an unknown, so we are forced to set
-
{{Displayed math||<math>z=x+iy</math>}}
+
{{Abgesetzte Formel||<math>z=x+iy</math>}}
and use the real part <math>x</math> and the imaginary part <math>y</math> as unknowns.
and use the real part <math>x</math> and the imaginary part <math>y</math> as unknowns.
Zeile 7: Zeile 7:
With this approach, the left-hand side of the equation becomes
With this approach, the left-hand side of the equation becomes
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
(1+i)(x-iy)+i(x+iy)
(1+i)(x-iy)+i(x+iy)
&= 1\cdot x -1\cdot iy +i\cdot x -i^2y + i\cdot x + i^2y\\[5pt]
&= 1\cdot x -1\cdot iy +i\cdot x -i^2y + i\cdot x + i^2y\\[5pt]
Zeile 16: Zeile 16:
and the whole equation becomes
and the whole equation becomes
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{{Displayed math||<math>x+(2x-y)i=3+5i\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>x+(2x-y)i=3+5i\,\textrm{.}</math>}}
The two complex numbers on the left- and right-hand sides are equal when both real and imaginary parts are equal, i.e.
The two complex numbers on the left- and right-hand sides are equal when both real and imaginary parts are equal, i.e.
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{{Displayed math||<math>\left\{\begin{align}x\phantom{{}-y}{}&=3\,,\\[5pt] 2x-y&=5\,\textrm{.}\end{align}\right.</math>}}
+
{{Abgesetzte Formel||<math>\left\{\begin{align}x\phantom{{}-y}{}&=3\,,\\[5pt] 2x-y&=5\,\textrm{.}\end{align}\right.</math>}}
This gives <math>x=3</math> and <math>y=2x-5=2\cdot 3-5=1</math>. Thus, the equation has the solution <math>z=3+i</math>.
This gives <math>x=3</math> and <math>y=2x-5=2\cdot 3-5=1</math>. Thus, the equation has the solution <math>z=3+i</math>.
Zeile 26: Zeile 26:
A quick check shows that <math>z=3+i</math> satisfies the equation in the exercise,
A quick check shows that <math>z=3+i</math> satisfies the equation in the exercise,
-
{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\text{LHS}
\text{LHS}
&= (1+i)\bar{z}+iz\\[5pt]
&= (1+i)\bar{z}+iz\\[5pt]

Version vom 13:07, 10. Mär. 2009

Because the equation contains both z and z, we cannot use z (or z) alone as an unknown, so we are forced to set

z=x+iy

and use the real part x and the imaginary part y as unknowns.

With this approach, the left-hand side of the equation becomes

(1+i)(x−iy)+i(x+iy)=1x−1iy+ix−i2y+ix+i2y=x−iy+ix+y+ix−y=x+(2x−y)i

and the whole equation becomes

x+(2x−y)i=3+5i.

The two complex numbers on the left- and right-hand sides are equal when both real and imaginary parts are equal, i.e.

x2x−y=3=5. 

This gives x=3 and y=2x−5=23−5=1. Thus, the equation has the solution z=3+i.

A quick check shows that z=3+i satisfies the equation in the exercise,

LHS=(1+i)z+iz=(1+i)(3−i)+i(3+i)=3−i+3i+1+3i−1=3+5i=RHS.
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.1:4f“