Processing Math: Done
Lösung 3.1:4f
Aus Online Mathematik Brückenkurs 2
(Unterschied zwischen Versionen)
K (Lösning 3.1:4f moved to Solution 3.1:4f: Robot: moved page) |
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- | < | + | 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 |
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- | < | + | <math>z=x+iy</math> |
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+ | and use the real part <math>x</math> and the imaginary part <math>y</math> as unknowns. | ||
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+ | With this approach, the left-hand side of the equation becomes | ||
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+ | <math>(1+i)(x-iy)+i(x+iy) | ||
+ | \begin{align} &=1\cdot x -1\cdot iy +i\cdot x -i^2y + i\cdot x + i^2y\\ | ||
+ | &=x-iy+ix+y+ix-y\\ | ||
+ | &=x+(2x-y)i\end{align}</math> | ||
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+ | and the whole equation becomes | ||
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+ | <math>x+(2x-y)i=3+5i</math>. | ||
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+ | 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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+ | <math>\begin{cases}x=3\\2x-y=5\end{cases}</math>. | ||
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+ | 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>. | ||
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+ | A quick check shows that <math>z=3+i</math> satisfies the equation in the exercise: | ||
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+ | <math>\begin{align}LHS &= (1+i)\bar{z}+iz=(1+i)(3-i)+i(3+i)\\ | ||
+ | &= 3-i+3i+1+3i-1 = 3+5i = RHS\end{align}</math> | ||
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Version vom 11:45, 23. Sep. 2008
Because the equation contains both
and use the real part
With this approach, the left-hand side of the equation becomes
x−1
iy+i
x−i2y+i
x+i2y=x−iy+ix+y+ix−y=x+(2x−y)i
and the whole equation becomes
The two complex numbers on the left- and right-hand sides are equal when both real and imaginary parts are equal, i.e.
x=32x−y=5
This gives 3−5=1
A quick check shows that
+iz=(1+i)(3−i)+i(3+i)=3−i+3i+1+3i−1=3+5i=RHS